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by **Giacomo** on March 16th, 2008, 5:02 pm

Part Four

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Lemma 4: The ordering of (0,1) using decimals does not prove that (0,1) is countable.

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We can order them, but this does not prove that they are countable.

1) Let us define the following ordering

0.1

...

0.9

0.01

0.02

...

0.09

0.11

...

0.19

0.21

...

0.001

...

So for example:

O(1) = 0.1

O(9) = 0.9

etc.

(2) It is clear that for all i, O(i) is a terminating decimal

(3) This means that for nonterminating decimals such as 1/3 or for irrational numbers such (pi - 1),

there is no i such that O(i) = 1/3 or O(i) = (pi - 1).

(4) Since not all reals in (0,1) have an i associated with them, using Lemma 1 above, this shows that O(i) is not a one-to-one correspondence between the reals in (0,1) and the whole numbers.

Q.E.D.

Take the real numbers in (0,1)

0.1

.....

0.9

0.01

.....

0.09

0.11

.....

0.19

....

0.91

.....

0.99

0.001

.....

0.009

0.010

.....

be called A.

It is clear that every element of A is a terminating decimal (no matter how far you go you will never get anything but a terminating decimal).

Not every real number is a terminating decimal. Therefore A is not the set of all real numbers.

It is true that A is countable, but since A is not the set of all real numbers it does not follow that the real numbers are countable.

Note: It is fairly easy to show that the set of all terminating decimals is countable. From time to time, someone notes this and declares that the reals are countable. Well, if someone says, "I have a proof that the reals are countable" you can say "you forgot about infinite decimals" without even looking at the proof. You will almost always be right. Note further that Cantor's argument deals with infinite decimals.