³1username³2³2name³2text_default³2version³21³2ts³2980537086³2ip³2209.69.30.10³2data³2newauthor³31³3minor³30³3text³3The Cantor Set is a remarkable construction involving only the real numbers between zero and one.  The concept is simple, you take the interval (0,1) and remove the "middle third" of this interval.  Then you take what is left (0, 1/3) U  (2/3, 1) and remove the "middle thirds" from these two intervals.  Continue this process ad infinitum.  Pictorially, the process would look something like this:

   0                 1/3                  2/3                 1
   ============================================================
   ====================                    ====================
   =======      =======                    =======      =======
   ==   ==      ==   ==                    ==   ==      ==   ==

The question becomes, what is left when you are done?  If you add up the length of segments removed, it would calculate out to be:

    1/3 + 2/9 + 4/27 + 8/81 + ... + 2^n/3^(n+1) = (1/3) SUM [n=1,infinity] (2^n/3^n)
                                                = (1/3) (1 / (1- 2/3))
                                                = (1/3) (1 / (1/3))
                                                = (1/3) (3)
                                                = 1

Using this calculation, you may be surprised if there were anything left - after all, the sum of the lengths of the removed intervals is equal to the length of the original interval.  However a closer look at the process reveals that we must have something left, since removing the "middle-thirds" of an interval involved removing open sets (sets that do not include their endpoints).  So removing the line segment (1/3, 2/3) from the original (0,1) interval leaves behind the points 1/3 and 2/3.  A little reflection will convince you quickly that they will never be removed, in fact none of the enpoints of any of the intervals at any stage in the process will ever be removed.  So we know for certainty that the Cantor set is not empty.

In fact, it can be shown that there as many points left behind in this process as there were those that were removed.  To see this, consider the points in the (0,1) interval in terms of base-three notation.  In this notation, 1/3 can be written as 0.1 and 2/3 can be written as 0.2.  If we remove everything from 1/3 and 2/3 we are really removing everything between 0.1 and 0.2, or in other words, '''everything with a 1 in the first position after the point''' (except for .1 itself).  The next step examines the intervals (0, 0.1] and [0.2, 1) and removes their middle thirds.  In this case we are removing everything between 0.01 and 0.02 in the first interval and between 0.21 and 0.22 in the second interval.
³3summary³3*³2tscreate³2980537086³2keepts³2980541036³2id³21019³2host³2209.69.30.xxx³2revision³21³1username³2³2name³2text_default³2version³21³2ts³2980541036³2ip³2209.69.30.10³2data³2text³3The Cantor Set is a remarkable construction involving only the real numbers between zero and one.  The concept is simple, you take the interval (0,1) and remove the "middle third" of this interval.  Then you take what is left (0, 1/3) U  (2/3, 1) and remove the "middle thirds" from these two intervals.  Continue this process ad infinitum.  Pictorially, the process would look something like this:

   0                 1/3                  2/3                 1
   ============================================================
   ====================                    ====================
   =======      =======                    =======      =======
   ==   ==      ==   ==                    ==   ==      ==   ==

The question becomes, what is left when you are done?  If you add up the length of segments removed, it would calculate out to be:

    1/3 + 2/9 + 4/27 + 8/81 + ... + 2^n/3^(n+1) = (1/3) SUM [n=1,infinity] (2^n/3^n)
                                                = (1/3) (1 / (1- 2/3))
                                                = (1/3) (1 / (1/3))
                                                = (1/3) (3)
                                                = 1

Using this calculation, you may be surprised if there were anything left - after all, the sum of the lengths of the removed intervals is equal to the length of the original interval.  However a closer look at the process reveals that we must have something left, since removing the "middle-thirds" of an interval involved removing open sets (sets that do not include their endpoints).  So removing the line segment (1/3, 2/3) from the original (0,1) interval leaves behind the points 1/3 and 2/3.  A little reflection will convince you quickly that they will never be removed, in fact none of the enpoints of any of the intervals at any stage in the process will ever be removed.  So we know for certainty that the Cantor set is not empty.

In fact, it can be shown that there as many points left behind in this process as there were those that were removed.  To see this, consider the points in the (0,1) interval in terms of base 3 (or ternial) notation.  In this notation, 1/3 can be written as 0.1 and 2/3 can be written as 0.2.  If we remove everything from 1/3 and 2/3 we are really removing everything between 0.1 and 0.2, or in other words, '''everything with a 1 in the first position after the point''' (except for .1 itself, but since .1 = 0.02222222..., we can represent it without using a one in any position).  The next step examines the intervals (0, 0.1] and [0.2, 1) and removes their middle thirds.  In this case we are removing everything between 0.01 and 0.02 in the first interval and between 0.21 and 0.22 in the second interval, or in other words, '''everything with a 1 in the second position after the point'''.

By the time you are done, the net result is that the only numbers removed from the interval are the numbers with ternary representations containing a '1' somewhere after the point.  So the numbers that remain are the ones that can be represented in ternary (base 3) notation with no '1' in any decimal (or ternial, if you prefer) place.  Stated another way the cantor set consists of all the numbers between 0 and 1 that can be represented using only 0's and 2's in ternary notation.  However, and here is the kicker, there exists a one-to-one correspondence between the numbers left in the Cantor set, and the all the real numbers between 0 and 1.

To see this, represent the real numbers in binary (instead of ternary) notation.  Then each number can be represented by a string of 0's and 1's (e.g. 0.101 = 5/8).  Now consider each number in the cantor set.  You can map each one to a number in the interval (0,1) by representing it as a ternary number, changing all 2's to 1's and considering it a base 2 number.  Likewise, each number in the interval (0,1) maps to a number in the Cantor set by representing the number in binary notation, changing all 1's to 2's and considering it a base 3 number.  Thus there is a 1-1 correspondance between the CantorSet and the real numbers in the interval (0,1).

³3minor³30³3newauthor³30³3summary³3*³2tscreate³2980537086³2keepts³2980541284³2id³21019³2host³2209.69.30.xxx³2revision³22³1username³2³2name³2text_default³2version³21³2ts³2980541284³2ip³2209.69.30.10³2data³2newauthor³30³3minor³31³3text³3The Cantor Set is a remarkable construction involving only the real numbers between zero and one.  The concept is simple, you take the interval (0,1) and remove the "middle third" of this interval.  Then you take what is left (0, 1/3) U  (2/3, 1) and remove the "middle thirds" from these two intervals.  Continue this process ad infinitum.  Pictorially, the process would look something like this:

   0                 1/3                  2/3                 1
   ============================================================
   ====================                    ====================
   =======      =======                    =======      =======
   ==   ==      ==   ==                    ==   ==      ==   ==

The question becomes, what is left when you are done?  If you add up the length of segments removed, it would calculate out to be:

    1/3 + 2/9 + 4/27 + 8/81 + ... + 2^n/3^(n+1) = (1/3) SUM [n=1,infinity] (2^n/3^n)
                                                = (1/3) (1 / (1- 2/3))
                                                = (1/3) (1 / (1/3))
                                                = (1/3) (3)
                                                = 1

Using this calculation, you may be surprised if there were anything left - after all, the sum of the lengths of the removed intervals is equal to the length of the original interval.  However a closer look at the process reveals that we must have something left, since removing the "middle-thirds" of an interval involved removing open sets (sets that do not include their endpoints).  So removing the line segment (1/3, 2/3) from the original (0,1) interval leaves behind the points 1/3 and 2/3.  A little reflection will convince you quickly that they will never be removed, in fact none of the endpoints of any of the intervals at any stage in the process will ever be removed.  So we know for certainty that the Cantor set is not empty.

In fact, it can be shown that there as many points left behind in this process as there were those that were removed.  To see this, consider the points in the (0,1) interval in terms of base 3 (or ternial) notation.  In this notation, 1/3 can be written as 0.1 and 2/3 can be written as 0.2.  If we remove everything from 1/3 and 2/3 we are really removing everything between 0.1 and 0.2, or in other words, '''everything with a 1 in the first position after the point''' (except for .1 itself, but since .1 = 0.02222222..., we can represent it without using a one in any position).  The next step examines the intervals (0, 0.1] and [0.2, 1) and removes their middle thirds.  In this case we are removing everything between 0.01 and 0.02 in the first interval and between 0.21 and 0.22 in the second interval, or in other words, '''everything with a 1 in the second position after the point'''.

By the time you are done, the net result is that the only numbers removed from the interval are the numbers with ternary representations containing a '1' somewhere after the point.  So the numbers that remain are the ones that can be represented in ternary (base 3) notation with no '1' in any decimal (or ternial, if you prefer) place.  Stated another way the cantor set consists of all the numbers between 0 and 1 that can be represented using only 0's and 2's in ternary notation.  However, and here is the kicker, there exists a one-to-one correspondence between the numbers left in the Cantor set, and the all the real numbers between 0 and 1.

To see this, represent the real numbers in binary (instead of ternary) notation.  Then each number can be represented by a string of 0's and 1's (e.g. 0.101 = 5/8).  Now consider each number in the cantor set.  You can map each one to a number in the interval (0,1) by representing it as a ternary number, changing all 2's to 1's and considering it a base 2 number.  Likewise, each number in the interval (0,1) maps to a number in the Cantor set by representing the number in binary notation, changing all 1's to 2's and considering it a base 3 number.  Thus there is a 1-1 correspondance between the CantorSet and the real numbers in the interval (0,1).

³3summary³3*³2tscreate³2980537086³2keepts³2980541751³2id³21019³2host³2209.69.30.xxx³2revision³23³1username³2³2name³2text_default³2version³21³2ts³2980541751³2ip³2209.69.30.10³2data³2text³3The Cantor Set is a remarkable construction involving only the real numbers between zero and one.  The concept is simple, you take the interval (0,1) and remove the "middle third" of this interval.  Then you take what is left (0, 1/3) U  (2/3, 1) and remove the "middle thirds" from these two intervals.  Continue this process ad infinitum.  Pictorially, the process would look something like this:

   0                 1/3                  2/3                 1
   ============================================================
   ====================                    ====================
   =======      =======                    =======      =======
   ==   ==      ==   ==                    ==   ==      ==   ==

The question becomes, what is left when you are done?  If you add up the length of segments removed, it would calculate out to be:

    1/3 + 2/9 + 4/27 + 8/81 + ... + 2^n/3^(n+1) = (1/3) SUM [n=0,infinity] (2^n/3^n)
                                                = (1/3) (1 / (1- 2/3))
                                                = (1/3) (1 / (1/3))
                                                = (1/3) (3)
                                                = 1

Using this calculation, you may be surprised if there were anything left - after all, the sum of the lengths of the removed intervals is equal to the length of the original interval.  However a closer look at the process reveals that we must have something left, since removing the "middle-thirds" of an interval involved removing open sets (sets that do not include their endpoints).  So removing the line segment (1/3, 2/3) from the original (0,1) interval leaves behind the points 1/3 and 2/3.  A little reflection will convince you quickly that they will never be removed, in fact none of the endpoints of any of the intervals at any stage in the process will ever be removed.  So we know for certainty that the Cantor set is not empty.

In fact, it can be shown that there as many points left behind in this process as there were those that were removed.  To see this, consider the points in the (0,1) interval in terms of base 3 (or ternial) notation.  In this notation, 1/3 can be written as 0.1 and 2/3 can be written as 0.2.  If we remove everything from 1/3 and 2/3 we are really removing everything between 0.1 and 0.2, or in other words, '''everything with a 1 in the first position after the point''' (except for .1 itself, but since .1 = 0.02222222..., we can represent it without using a one in any position).  The next step examines the intervals (0, 0.1] and [0.2, 1) and removes their middle thirds.  In this case we are removing everything between 0.01 and 0.02 in the first interval and between 0.21 and 0.22 in the second interval, or in other words, '''everything with a 1 in the second position after the point'''.

By the time you are done, the net result is that the only numbers removed from the interval are the numbers with ternary representations containing a '1' somewhere after the point.  So the numbers that remain are the ones that can be represented in ternary (base 3) notation with no '1' in any decimal (or ternial, if you prefer) place.  Stated another way the cantor set consists of all the numbers between 0 and 1 that can be represented using only 0's and 2's in ternary notation.  However, and here is the kicker, there exists a one-to-one correspondence between the numbers left in the Cantor set, and the all the real numbers between 0 and 1.

To see this, represent the real numbers in binary (instead of ternary) notation.  Then each number can be represented by a string of 0's and 1's (e.g. 0.101 = 5/8).  Now consider each number in the cantor set.  You can map each one to a number in the interval (0,1) by representing it as a ternary number, changing all 2's to 1's and considering it a base 2 number.  Likewise, each number in the interval (0,1) maps to a number in the Cantor set by representing the number in binary notation, changing all 1's to 2's and considering it a base 3 number.  Thus there is a 1-1 correspondance between the CantorSet and the real numbers in the interval (0,1).

³3minor³31³3newauthor³30³3summary³3*³2tscreate³2980537086³2keepts³2980542215³2id³21019³2host³2209.69.30.xxx³2revision³24³1username³2³2name³2text_default³2version³21³2ts³2980542215³2ip³2209.69.30.10³2data³2newauthor³30³3minor³30³3text³3The Cantor Set is a remarkable construction involving only the real numbers between zero and one.  The concept is simple, you take the interval (0,1) and remove the "middle third" of this interval.  Then you take what is left (0, 1/3) U  (2/3, 1) and remove the "middle thirds" from these two intervals.  Continue this process ad infinitum.  Pictorially, the process would look something like this:

   0                 1/3                  2/3                 1
   ============================================================
   ====================                    ====================
   =======      =======                    =======      =======
   ==   ==      ==   ==                    ==   ==      ==   ==

The question becomes, what is left when you are done?  If you add up the length of segments removed, it would calculate out to be:

    1/3 + 2/9 + 4/27 + 8/81 + ... + 2^n/3^(n+1) = (1/3) SUM [n=0,infinity] (2^n/3^n)
                                                = (1/3) (1 / (1- 2/3))
                                                = (1/3) (1 / (1/3))
                                                = (1/3) (3)
                                                = 1

Using this calculation, you may be surprised if there were anything left - after all, the sum of the lengths of the removed intervals is equal to the length of the original interval.  However a closer look at the process reveals that we must have something left, since removing the "middle-thirds" of an interval involved removing open sets (sets that do not include their endpoints).  So removing the line segment (1/3, 2/3) from the original (0,1) interval leaves behind the points 1/3 and 2/3.  A little reflection will convince you quickly that they will never be removed, in fact none of the endpoints of any of the intervals at any stage in the process will ever be removed.  So we know for certainty that the Cantor set is not empty.

In fact, it can be shown that there as many points left behind in this process as there were those that were removed.  To see this, consider the points in the (0,1) interval in terms of base 3 (or ternial) notation.  In this notation, 1/3 can be written as 0.1 and 2/3 can be written as 0.2.  If we remove everything from 1/3 and 2/3 we are really removing everything between 0.1 and 0.2, or in other words, '''everything with a 1 in the first position after the point''' (except for .1 itself, but since .1 = 0.02222222..., we can represent it without using a one in any position).  The next step examines the intervals (0, 0.1] and [0.2, 1) and removes their middle thirds.  In this case we are removing everything between 0.01 and 0.02 in the first interval and between 0.21 and 0.22 in the second interval, or in other words, '''everything with a 1 in the second position after the point'''.

By the time you are done, the net result is that the only numbers removed from the interval are the numbers with ternary representations containing a '1' somewhere after the point.  So the numbers that remain are the ones that can be represented in ternary (base 3) notation with no '1' in any decimal (or ternial, if you prefer) place.  Stated another way the cantor set consists of all the numbers between 0 and 1 that can be represented using only 0's and 2's in ternary notation.  However, and here is the kicker, there exists a one-to-one correspondence between the numbers left in the CantorSet, and the all the real numbers between 0 and 1.

To see this, represent the real numbers in binary (instead of ternary) notation.  Then each number can be represented by a string of 0's and 1's (e.g. 0.101 = 5/8).  Now consider each number in the cantor set.  You can map each one to a number in the interval (0,1) by representing it as a ternary number, changing all 2's to 1's and considering it a base 2 number.  Likewise, each number in the interval (0,1) maps to a number in the Cantor set by representing the number in binary notation, changing all 1's to 2's and considering it a base 3 number.  Thus there is a 1-1 correspondance between the CantorSet and the real numbers in the interval (0,1).

----

Some topographically interesting facts about the Cantor set.
    * Since it is the complement of a union of disjoint open sets, it itself is closed.
    * Pick any point in the CantorSet.  In any arbitrarily small neighborhood, there is some other number that can be represented as a ternary number with only 0's and 2's.  Hence every point in the CantorSet is an accumulation point, and the set is a PerfectSet in the topological sense.
    * Again, pick any point in the CantorSet (which is itself a subset of the unit interval).  Any arbitrarily small neighborhood around that point contains an open set in the unit interval that is disjoint from the CantorSet.  Thus the CantorSet is NowhereDense.


³3summary³3*³2tscreate³2980537086³2keepts³2980542245³2id³21019³2host³2209.69.30.xxx³2revision³25³1username³2³2name³2text_default³2version³21³2ts³2980542245³2ip³2209.69.30.10³2data³2text³3The Cantor Set is a remarkable construction involving only the real numbers between zero and one.  The concept is simple, you take the interval (0,1) and remove the "middle third" of this interval.  Then you take what is left (0, 1/3) U  (2/3, 1) and remove the "middle thirds" from these two intervals.  Continue this process ad infinitum.  Pictorially, the process would look something like this:

   0                 1/3                  2/3                 1
   ============================================================
   ====================                    ====================
   =======      =======                    =======      =======
   ==   ==      ==   ==                    ==   ==      ==   ==

The question becomes, what is left when you are done?  If you add up the length of segments removed, it would calculate out to be:

    1/3 + 2/9 + 4/27 + 8/81 + ... + 2^n/3^(n+1) = (1/3) SUM [n=0,infinity] (2^n/3^n)
                                                = (1/3) (1 / (1- 2/3))
                                                = (1/3) (1 / (1/3))
                                                = (1/3) (3)
                                                = 1

Using this calculation, you may be surprised if there were anything left - after all, the sum of the lengths of the removed intervals is equal to the length of the original interval.  However a closer look at the process reveals that we must have something left, since removing the "middle-thirds" of an interval involved removing open sets (sets that do not include their endpoints).  So removing the line segment (1/3, 2/3) from the original (0,1) interval leaves behind the points 1/3 and 2/3.  A little reflection will convince you quickly that they will never be removed, in fact none of the endpoints of any of the intervals at any stage in the process will ever be removed.  So we know for certainty that the Cantor set is not empty.

In fact, it can be shown that there as many points left behind in this process as there were those that were removed.  To see this, consider the points in the (0,1) interval in terms of base 3 (or ternial) notation.  In this notation, 1/3 can be written as 0.1 and 2/3 can be written as 0.2.  If we remove everything from 1/3 and 2/3 we are really removing everything between 0.1 and 0.2, or in other words, '''everything with a 1 in the first position after the point''' (except for .1 itself, but since .1 = 0.02222222..., we can represent it without using a one in any position).  The next step examines the intervals (0, 0.1] and [0.2, 1) and removes their middle thirds.  In this case we are removing everything between 0.01 and 0.02 in the first interval and between 0.21 and 0.22 in the second interval, or in other words, '''everything with a 1 in the second position after the point'''.

By the time you are done, the net result is that the only numbers removed from the interval are the numbers with ternary representations containing a '1' somewhere after the point.  So the numbers that remain are the ones that can be represented in ternary (base 3) notation with no '1' in any decimal (or ternial, if you prefer) place.  Stated another way the cantor set consists of all the numbers between 0 and 1 that can be represented using only 0's and 2's in ternary notation.  However, and here is the kicker, there exists a one-to-one correspondence between the numbers left in the CantorSet, and the all the real numbers between 0 and 1.

To see this, represent the real numbers in binary (instead of ternary) notation.  Then each number can be represented by a string of 0's and 1's (e.g. 0.101 = 5/8).  Now consider each number in the cantor set.  You can map each one to a number in the interval (0,1) by representing it as a ternary number, changing all 2's to 1's and considering it a base 2 number.  Likewise, each number in the interval (0,1) maps to a number in the Cantor set by representing the number in binary notation, changing all 1's to 2's and considering it a base 3 number.  Thus there is a 1-1 correspondance between the CantorSet and the real numbers in the interval (0,1).

----

Some topographically interesting facts about the Cantor set.
    o Since it is the complement of a union of disjoint open sets, it itself is closed.
    o Pick any point in the CantorSet.  In any arbitrarily small neighborhood, there is some other number that can be represented as a ternary number with only 0's and 2's.  Hence every point in the CantorSet is an accumulation point, and the set is a PerfectSet in the topological sense.
    o Again, pick any point in the CantorSet (which is itself a subset of the unit interval).  Any arbitrarily small neighborhood around that point contains an open set in the unit interval that is disjoint from the CantorSet.  Thus the CantorSet is NowhereDense.


³3minor³30³3newauthor³30³3summary³3*³2tscreate³2980537086³2keepts³2980542266³2id³21019³2host³2209.69.30.xxx³2revision³26³1username³2³2name³2text_default³2version³21³2ts³2980542266³2ip³2209.69.30.10³2data³2newauthor³30³3minor³30³3text³3The Cantor Set is a remarkable construction involving only the real numbers between zero and one.  The concept is simple, you take the interval (0,1) and remove the "middle third" of this interval.  Then you take what is left (0, 1/3) U  (2/3, 1) and remove the "middle thirds" from these two intervals.  Continue this process ad infinitum.  Pictorially, the process would look something like this:

   0                 1/3                  2/3                 1
   ============================================================
   ====================                    ====================
   =======      =======                    =======      =======
   ==   ==      ==   ==                    ==   ==      ==   ==

The question becomes, what is left when you are done?  If you add up the length of segments removed, it would calculate out to be:

    1/3 + 2/9 + 4/27 + 8/81 + ... + 2^n/3^(n+1) = (1/3) SUM [n=0,infinity] (2^n/3^n)
                                                = (1/3) (1 / (1- 2/3))
                                                = (1/3) (1 / (1/3))
                                                = (1/3) (3)
                                                = 1

Using this calculation, you may be surprised if there were anything left - after all, the sum of the lengths of the removed intervals is equal to the length of the original interval.  However a closer look at the process reveals that we must have something left, since removing the "middle-thirds" of an interval involved removing open sets (sets that do not include their endpoints).  So removing the line segment (1/3, 2/3) from the original (0,1) interval leaves behind the points 1/3 and 2/3.  A little reflection will convince you quickly that they will never be removed, in fact none of the endpoints of any of the intervals at any stage in the process will ever be removed.  So we know for certainty that the Cantor set is not empty.

In fact, it can be shown that there as many points left behind in this process as there were those that were removed.  To see this, consider the points in the (0,1) interval in terms of base 3 (or ternial) notation.  In this notation, 1/3 can be written as 0.1 and 2/3 can be written as 0.2.  If we remove everything from 1/3 and 2/3 we are really removing everything between 0.1 and 0.2, or in other words, '''everything with a 1 in the first position after the point''' (except for .1 itself, but since .1 = 0.02222222..., we can represent it without using a one in any position).  The next step examines the intervals (0, 0.1] and [0.2, 1) and removes their middle thirds.  In this case we are removing everything between 0.01 and 0.02 in the first interval and between 0.21 and 0.22 in the second interval, or in other words, '''everything with a 1 in the second position after the point'''.

By the time you are done, the net result is that the only numbers removed from the interval are the numbers with ternary representations containing a '1' somewhere after the point.  So the numbers that remain are the ones that can be represented in ternary (base 3) notation with no '1' in any decimal (or ternial, if you prefer) place.  Stated another way the cantor set consists of all the numbers between 0 and 1 that can be represented using only 0's and 2's in ternary notation.  However, and here is the kicker, there exists a one-to-one correspondence between the numbers left in the CantorSet, and the all the real numbers between 0 and 1.

To see this, represent the real numbers in binary (instead of ternary) notation.  Then each number can be represented by a string of 0's and 1's (e.g. 0.101 = 5/8).  Now consider each number in the cantor set.  You can map each one to a number in the interval (0,1) by representing it as a ternary number, changing all 2's to 1's and considering it a base 2 number.  Likewise, each number in the interval (0,1) maps to a number in the Cantor set by representing the number in binary notation, changing all 1's to 2's and considering it a base 3 number.  Thus there is a 1-1 correspondance between the CantorSet and the real numbers in the interval (0,1).

----

Some topographically interesting facts about the Cantor set.
o Since it is the complement of a union of disjoint open sets, it itself is closed.
o Pick any point in the CantorSet.  In any arbitrarily small neighborhood, there is some other number that can be represented as a ternary number with only 0's and 2's.  Hence every point in the CantorSet is an accumulation point, and the set is a PerfectSet in the topological sense.
o Again, pick any point in the CantorSet (which is itself a subset of the unit interval).  Any arbitrarily small neighborhood around that point contains an open set in the unit interval that is disjoint from the CantorSet.  Thus the CantorSet is NowhereDense.


³3summary³3*³2tscreate³2980537086³2keepts³2980542288³2id³21019³2host³2209.69.30.xxx³2revision³27³1username³2³2name³2text_default³2version³21³2ts³2980542288³2ip³2209.69.30.10³2data³2text³3The Cantor Set is a remarkable construction involving only the real numbers between zero and one.  The concept is simple, you take the interval (0,1) and remove the "middle third" of this interval.  Then you take what is left (0, 1/3) U  (2/3, 1) and remove the "middle thirds" from these two intervals.  Continue this process ad infinitum.  Pictorially, the process would look something like this:

   0                 1/3                  2/3                 1
   ============================================================
   ====================                    ====================
   =======      =======                    =======      =======
   ==   ==      ==   ==                    ==   ==      ==   ==

The question becomes, what is left when you are done?  If you add up the length of segments removed, it would calculate out to be:

    1/3 + 2/9 + 4/27 + 8/81 + ... + 2^n/3^(n+1) = (1/3) SUM [n=0,infinity] (2^n/3^n)
                                                = (1/3) (1 / (1- 2/3))
                                                = (1/3) (1 / (1/3))
                                                = (1/3) (3)
                                                = 1

Using this calculation, you may be surprised if there were anything left - after all, the sum of the lengths of the removed intervals is equal to the length of the original interval.  However a closer look at the process reveals that we must have something left, since removing the "middle-thirds" of an interval involved removing open sets (sets that do not include their endpoints).  So removing the line segment (1/3, 2/3) from the original (0,1) interval leaves behind the points 1/3 and 2/3.  A little reflection will convince you quickly that they will never be removed, in fact none of the endpoints of any of the intervals at any stage in the process will ever be removed.  So we know for certainty that the Cantor set is not empty.

In fact, it can be shown that there as many points left behind in this process as there were those that were removed.  To see this, consider the points in the (0,1) interval in terms of base 3 (or ternial) notation.  In this notation, 1/3 can be written as 0.1 and 2/3 can be written as 0.2.  If we remove everything from 1/3 and 2/3 we are really removing everything between 0.1 and 0.2, or in other words, '''everything with a 1 in the first position after the point''' (except for .1 itself, but since .1 = 0.02222222..., we can represent it without using a one in any position).  The next step examines the intervals (0, 0.1] and [0.2, 1) and removes their middle thirds.  In this case we are removing everything between 0.01 and 0.02 in the first interval and between 0.21 and 0.22 in the second interval, or in other words, '''everything with a 1 in the second position after the point'''.

By the time you are done, the net result is that the only numbers removed from the interval are the numbers with ternary representations containing a '1' somewhere after the point.  So the numbers that remain are the ones that can be represented in ternary (base 3) notation with no '1' in any decimal (or ternial, if you prefer) place.  Stated another way the cantor set consists of all the numbers between 0 and 1 that can be represented using only 0's and 2's in ternary notation.  However, and here is the kicker, there exists a one-to-one correspondence between the numbers left in the CantorSet, and the all the real numbers between 0 and 1.

To see this, represent the real numbers in binary (instead of ternary) notation.  Then each number can be represented by a string of 0's and 1's (e.g. 0.101 = 5/8).  Now consider each number in the cantor set.  You can map each one to a number in the interval (0,1) by representing it as a ternary number, changing all 2's to 1's and considering it a base 2 number.  Likewise, each number in the interval (0,1) maps to a number in the Cantor set by representing the number in binary notation, changing all 1's to 2's and considering it a base 3 number.  Thus there is a 1-1 correspondance between the CantorSet and the real numbers in the interval (0,1).

----

Some topographically interesting facts about the Cantor set.

o Since it is the complement of a union of disjoint open sets, it itself is closed.

o Pick any point in the CantorSet.  In any arbitrarily small neighborhood, there is some other number that can be represented as a ternary number with only 0's and 2's.  Hence every point in the CantorSet is an accumulation point, and the set is a PerfectSet in the topological sense.

o Again, pick any point in the CantorSet (which is itself a subset of the unit interval).  Any arbitrarily small neighborhood around that point contains an open set in the unit interval that is disjoint from the CantorSet.  Thus the CantorSet is NowhereDense.
³3minor³30³3newauthor³30³3summary³3*³2tscreate³2980537086³2keepts³2980542307³2id³21019³2host³2209.69.30.xxx³2revision³28³1username³2³2name³2text_default³2version³21³2ts³2980542307³2ip³2209.69.30.10³2data³2newauthor³30³3minor³30³3text³3The Cantor Set is a remarkable construction involving only the real numbers between zero and one.  The concept is simple, you take the interval (0,1) and remove the "middle third" of this interval.  Then you take what is left (0, 1/3) U  (2/3, 1) and remove the "middle thirds" from these two intervals.  Continue this process ad infinitum.  Pictorially, the process would look something like this:

   0                 1/3                  2/3                 1
   ============================================================
   ====================                    ====================
   =======      =======                    =======      =======
   ==   ==      ==   ==                    ==   ==      ==   ==

The question becomes, what is left when you are done?  If you add up the length of segments removed, it would calculate out to be:

    1/3 + 2/9 + 4/27 + 8/81 + ... + 2^n/3^(n+1) = (1/3) SUM [n=0,infinity] (2^n/3^n)
                                                = (1/3) (1 / (1- 2/3))
                                                = (1/3) (1 / (1/3))
                                                = (1/3) (3)
                                                = 1

Using this calculation, you may be surprised if there were anything left - after all, the sum of the lengths of the removed intervals is equal to the length of the original interval.  However a closer look at the process reveals that we must have something left, since removing the "middle-thirds" of an interval involved removing open sets (sets that do not include their endpoints).  So removing the line segment (1/3, 2/3) from the original (0,1) interval leaves behind the points 1/3 and 2/3.  A little reflection will convince you quickly that they will never be removed, in fact none of the endpoints of any of the intervals at any stage in the process will ever be removed.  So we know for certainty that the Cantor set is not empty.

In fact, it can be shown that there as many points left behind in this process as there were those that were removed.  To see this, consider the points in the (0,1) interval in terms of base 3 (or ternial) notation.  In this notation, 1/3 can be written as 0.1 and 2/3 can be written as 0.2.  If we remove everything from 1/3 and 2/3 we are really removing everything between 0.1 and 0.2, or in other words, '''everything with a 1 in the first position after the point''' (except for .1 itself, but since .1 = 0.02222222..., we can represent it without using a one in any position).  The next step examines the intervals (0, 0.1] and [0.2, 1) and removes their middle thirds.  In this case we are removing everything between 0.01 and 0.02 in the first interval and between 0.21 and 0.22 in the second interval, or in other words, '''everything with a 1 in the second position after the point'''.

By the time you are done, the net result is that the only numbers removed from the interval are the numbers with ternary representations containing a '1' somewhere after the point.  So the numbers that remain are the ones that can be represented in ternary (base 3) notation with no '1' in any decimal (or ternial, if you prefer) place.  Stated another way the cantor set consists of all the numbers between 0 and 1 that can be represented using only 0's and 2's in ternary notation.  However, and here is the kicker, there exists a one-to-one correspondence between the numbers left in the CantorSet, and the all the real numbers between 0 and 1.

To see this, represent the real numbers in binary (instead of ternary) notation.  Then each number can be represented by a string of 0's and 1's (e.g. 0.101 = 5/8).  Now consider each number in the cantor set.  You can map each one to a number in the interval (0,1) by representing it as a ternary number, changing all 2's to 1's and considering it a base 2 number.  Likewise, each number in the interval (0,1) maps to a number in the Cantor set by representing the number in binary notation, changing all 1's to 2's and considering it a base 3 number.  Thus there is a 1-1 correspondance between the CantorSet and the real numbers in the interval (0,1).

----

Some topographically interesting facts about the Cantor set.

Since it is the complement of a union of disjoint open sets, it itself is closed.

Pick any point in the CantorSet.  In any arbitrarily small neighborhood, there is some other number that can be represented as a ternary number with only 0's and 2's.  Hence every point in the CantorSet is an accumulation point, and the set is a PerfectSet in the topological sense.

Again, pick any point in the CantorSet (which is itself a subset of the unit interval).  Any arbitrarily small neighborhood around that point contains an open set in the unit interval that is disjoint from the CantorSet.  Thus the CantorSet is NowhereDense.
³3summary³3*³2tscreate³2980537086³2keepts³2980542333³2id³21019³2host³2209.69.30.xxx³2revision³29³1username³2³2name³2text_default³2version³21³2ts³2980542333³2ip³2209.69.30.10³2data³2text³3The Cantor Set is a remarkable construction involving only the real numbers between zero and one.  The concept is simple, you take the interval (0,1) and remove the "middle third" of this interval.  Then you take what is left (0, 1/3) U  (2/3, 1) and remove the "middle thirds" from these two intervals.  Continue this process ad infinitum.  Pictorially, the process would look something like this:

   0                 1/3                  2/3                 1
   ============================================================
   ====================                    ====================
   =======      =======                    =======      =======
   ==   ==      ==   ==                    ==   ==      ==   ==

The question becomes, what is left when you are done?  If you add up the length of segments removed, it would calculate out to be:

    1/3 + 2/9 + 4/27 + 8/81 + ... + 2^n/3^(n+1) = (1/3) SUM [n=0,infinity] (2^n/3^n)
                                                = (1/3) (1 / (1- 2/3))
                                                = (1/3) (1 / (1/3))
                                                = (1/3) (3)
                                                = 1

Using this calculation, you may be surprised if there were anything left - after all, the sum of the lengths of the removed intervals is equal to the length of the original interval.  However a closer look at the process reveals that we must have something left, since removing the "middle-thirds" of an interval involved removing open sets (sets that do not include their endpoints).  So removing the line segment (1/3, 2/3) from the original (0,1) interval leaves behind the points 1/3 and 2/3.  A little reflection will convince you quickly that they will never be removed, in fact none of the endpoints of any of the intervals at any stage in the process will ever be removed.  So we know for certainty that the Cantor set is not empty.

In fact, it can be shown that there as many points left behind in this process as there were those that were removed.  To see this, consider the points in the (0,1) interval in terms of base 3 (or ternial) notation.  In this notation, 1/3 can be written as 0.1 and 2/3 can be written as 0.2.  If we remove everything from 1/3 and 2/3 we are really removing everything between 0.1 and 0.2, or in other words, '''everything with a 1 in the first position after the point''' (except for .1 itself, but since .1 = 0.02222222..., we can represent it without using a one in any position).  The next step examines the intervals (0, 0.1] and [0.2, 1) and removes their middle thirds.  In this case we are removing everything between 0.01 and 0.02 in the first interval and between 0.21 and 0.22 in the second interval, or in other words, '''everything with a 1 in the second position after the point'''.

By the time you are done, the net result is that the only numbers removed from the interval are the numbers with ternary representations containing a '1' somewhere after the point.  So the numbers that remain are the ones that can be represented in ternary (base 3) notation with no '1' in any decimal (or ternial, if you prefer) place.  Stated another way the cantor set consists of all the numbers between 0 and 1 that can be represented using only 0's and 2's in ternary notation.  However, and here is the kicker, there exists a one-to-one correspondence between the numbers left in the CantorSet, and the all the real numbers between 0 and 1.

To see this, represent the real numbers in binary (instead of ternary) notation.  Then each number can be represented by a string of 0's and 1's (e.g. 0.101 = 5/8).  Now consider each number in the cantor set.  You can map each one to a number in the interval (0,1) by representing it as a ternary number, changing all 2's to 1's and considering it a base 2 number.  Likewise, each number in the interval (0,1) maps to a number in the Cantor set by representing the number in binary notation, changing all 1's to 2's and considering it a base 3 number.  Thus there is a 1-1 correspondance between the CantorSet and the real numbers in the interval (0,1).

----

Some topographically interesting facts about the Cantor set.

Since it is the complement of a union of disjoint open sets, it itself is a ClosedSet.

Pick any point in the CantorSet.  In any arbitrarily small neighborhood, there is some other number that can be represented as a ternary number with only 0's and 2's.  Hence every point in the CantorSet is an accumulation point, and the set is a PerfectSet in the topological sense.

Again, pick any point in the CantorSet (which is itself a subset of the unit interval).  Any arbitrarily small neighborhood around that point contains an open set in the unit interval that is disjoint from the CantorSet.  Thus the CantorSet is NowhereDense.
³3minor³30³3newauthor³30³3summary³3*³2tscreate³2980537086³2keepts³2980542423³2id³21019³2host³2209.69.30.xxx³2revision³210³1username³2³2name³2text_default³2version³21³2ts³2980542423³2ip³2209.69.30.10³2data³2newauthor³30³3minor³30³3text³3The Cantor Set is a remarkable construction involving only the real numbers between zero and one.  The concept is simple, you take the interval (0,1) and remove the "middle third" of this interval.  Then you take what is left (0, 1/3) U  (2/3, 1) and remove the "middle thirds" from these two intervals.  Continue this process ad infinitum.  Pictorially, the process would look something like this:

   0                 1/3                  2/3                 1
   ============================================================
   ====================                    ====================
   =======      =======                    =======      =======
   ==   ==      ==   ==                    ==   ==      ==   ==

The question becomes, what is left when you are done?  If you add up the length of segments removed, it would calculate out to be:

    1/3 + 2/9 + 4/27 + 8/81 + ... + 2^n/3^(n+1) = (1/3) SUM [n=0,infinity] (2^n/3^n)
                                                = (1/3) (1 / (1- 2/3))
                                                = (1/3) (1 / (1/3))
                                                = (1/3) (3)
                                                = 1

Using this calculation, you may be surprised if there were anything left - after all, the sum of the lengths of the removed intervals is equal to the length of the original interval.  However a closer look at the process reveals that we must have something left, since removing the "middle-thirds" of an interval involved removing open sets (sets that do not include their endpoints).  So removing the line segment (1/3, 2/3) from the original (0,1) interval leaves behind the points 1/3 and 2/3.  A little reflection will convince you quickly that they will never be removed, in fact none of the endpoints of any of the intervals at any stage in the process will ever be removed.  So we know for certainty that the Cantor set is not empty.

In fact, it can be shown that there as many points left behind in this process as there were those that were removed.  To see this, consider the points in the (0,1) interval in terms of base 3 (or ternial) notation.  In this notation, 1/3 can be written as 0.1 and 2/3 can be written as 0.2.  If we remove everything from 1/3 and 2/3 we are really removing everything between 0.1 and 0.2, or in other words, '''everything with a 1 in the first position after the point''' (except for .1 itself, but since .1 = 0.02222222..., we can represent it without using a one in any position).  The next step examines the intervals (0, 0.1] and [0.2, 1) and removes their middle thirds.  In this case we are removing everything between 0.01 and 0.02 in the first interval and between 0.21 and 0.22 in the second interval, or in other words, '''everything with a 1 in the second position after the point'''.

By the time you are done, the net result is that the only numbers removed from the interval are the numbers with ternary representations containing a '1' somewhere after the point.  So the numbers that remain are the ones that can be represented in ternary (base 3) notation with no '1' in any decimal (or ternial, if you prefer) place.  Stated another way the cantor set consists of all the numbers between 0 and 1 that can be represented using only 0's and 2's in ternary notation.  However, and here is the kicker, there exists a one-to-one correspondence between the numbers left in the CantorSet, and the all the real numbers between 0 and 1.

To see this, represent the real numbers in binary (instead of ternary) notation.  Then each number can be represented by a string of 0's and 1's (e.g. 0.101 = 5/8).  Now consider each number in the cantor set.  You can map each one to a number in the interval (0,1) by representing it as a ternary number, changing all 2's to 1's and considering it a base 2 number.  Likewise, each number in the interval (0,1) maps to a number in the Cantor set by representing the number in binary notation, changing all 1's to 2's and considering it a base 3 number.  Thus there is a 1-1 correspondance between the CantorSet and the real numbers in the interval (0,1).

----

Some topographically interesting facts about the Cantor set.

Since it is the complement of a union of disjoint open sets, it itself is a ClosedSet.

Pick any point in the CantorSet.  In any arbitrarily small neighborhood, there is some other number that can be represented as a ternary number with only 0's and 2's.  Hence every point in the CantorSet is an accumulation point.

The last two statements combine to make the CantorSet a PerfectSet in the topological sense.

Again, pick any point in the CantorSet (which is itself a subset of the unit interval).  Any arbitrarily small neighborhood around that point contains an open set in the unit interval that is disjoint from the CantorSet.  Thus the CantorSet is NowhereDense.
³3summary³3*³2tscreate³2980537086³2keepts³2980710316³2id³21019³2host³2209.69.30.xxx³2revision³211