## The lemma that robert ignores

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**1**of**1**### The lemma that robert ignores

There are, generally speaking, two ways to outline a mathematical proof. You can put all of the steps in some sequential order, but that can be confusing if the path to the conclusion has branches in it. Or, you can divide the flow into several smaller proofs, each of which follows linearly along its own logical path. These smaller proofs are called lemmas. You then insert just the conclusion of the lemmas where they fit into the main proof, without needing branches.

Cantor's diagonalization theorem does not need lemmas, because its flow is more-or-less linear. That doesn't mean you can't use one, tho. Here it is:

Notice that this doesn't say XB is not in R; in fact, it says it is. In months of so-called "discussion," robert 46 has tried to claim Cantor's proof is invalid. Robert's argument is that XB is in the unexamined portion of something, but he refuses to comprehend that that something is not what Cantor talks about. IT IS NOT IN S. Robert also refuses to comprehend that even if S is infinite, B' is still formed using every element of B,and is still not in S

To repeat: the lemma says XB is not in S. Even if S is infinite.

Using this lemma, Cantor then shows that any countable set of reals in the range cannot be a complete set of reals in the range. It's simple, really: If the statement "If A then B" is true, so is the inverse statement "If not B, then not A." Inverting the lemma this way makes it:

This lemma can't be used in a similar proof about rational numbers, because step 6 would have to say that there was a rational number XB in [0,1) whose binary expansion was B'. As robert himself states (well, he states it incorrectly, but this is what he means), this distinction between irrational and rational reals cannot be made for infinite binary strings.

+++++

[1] If S has finite size N, use the first N integers in C

[2] Rational reals whose denominator is a power of 2 can be represented by two: one that terminates with infinite zeroes and one that terminates with infinite ones. For this proof, use just the one that terminates with infinite zeroes.

Cantor's diagonalization theorem does not need lemmas, because its flow is more-or-less linear. That doesn't mean you can't use one, tho. Here it is:

- Definition: a set is countable if you can make a 1:1 mapping of it to the integers in C={1,2,3,...}. [1]
- Every real number X in [0,1) can be represented by a unique[2], infinite binary string, where the Kth bit represents the factor of (1/2)^K in an infinite sequence that sums to X. Call this the binary expansion of X.
- Assume a set S of distinct real numbers in [0,1) exists, and is countable.
- Each element S(K) has a unique binary expansion. Call the set of them B. It is countable the same way as S.
- Form a new infinite string B', where the Kth bit of B' is the inverse of the Kth bit of B(K) USING EVERY ELEMENT OF B. If S has a finite size N, the bits beyond K=N are zeros.
- B' is not in B because the Kth bit of its binary expansion is different from the Kth bit of B(K),for every K in C.
- There is a real number XB in [0,1) whose binary expansion is B'.
- XB is not in S because its binary expansion is different from every binary expansion of a number in S.

The lemma is:

If a set S of distinct real numbers in the range is countable, then there exists a real number in the range [0,1) that is not in S.

Rough Proof:

Notice that this doesn't say XB is not in R; in fact, it says it is. In months of so-called "discussion," robert 46 has tried to claim Cantor's proof is invalid. Robert's argument is that XB is in the unexamined portion of something, but he refuses to comprehend that that something is not what Cantor talks about. IT IS NOT IN S. Robert also refuses to comprehend that even if S is infinite, B' is still formed using every element of B,and is still not in S

To repeat: the lemma says XB is not in S. Even if S is infinite.

Using this lemma, Cantor then shows that any countable set of reals in the range cannot be a complete set of reals in the range. It's simple, really: If the statement "If A then B" is true, so is the inverse statement "If not B, then not A." Inverting the lemma this way makes it:

- If the set of distinct real numbers S is not missing a real in the range [0,1), then it is not countable.

This lemma can't be used in a similar proof about rational numbers, because step 6 would have to say that there was a rational number XB in [0,1) whose binary expansion was B'. As robert himself states (well, he states it incorrectly, but this is what he means), this distinction between irrational and rational reals cannot be made for infinite binary strings.

+++++

[1] If S has finite size N, use the first N integers in C

[2] Rational reals whose denominator is a power of 2 can be represented by two: one that terminates with infinite zeroes and one that terminates with infinite ones. For this proof, use just the one that terminates with infinite zeroes.

- JeffJo
- Intellectual
**Posts:**2503**Joined:**Tue Mar 10, 2009 11:01 am

### Re: The lemma that robert ignores

Robert also refuses to comprehend that even if S is infinite [and countable], B' is still formed using every element of B,and is still not in S.

I'm not sure if B' can be constructed if S is not countable.

- Gofer
- Intellectual
**Posts:**165**Joined:**Mon May 09, 2016 8:24 am

### Re: The lemma that robert ignores

What part of "If a set S of distinct real numbers in the range is countable" are you having trouble understanding?Gofer wrote:I'm not sure if B' can be constructed if S is not countable.

One of the big mistakes that Cantor doubters always make, is to think that the set which gets diagonalized is the set that is the subject of the proof. It is not. It is ANY subset of the subject that is countable.

- JeffJo
- Intellectual
**Posts:**2503**Joined:**Tue Mar 10, 2009 11:01 am

### Re: The lemma that robert ignores

Jeff, you didn't explicitly state that, even though mentioned in the lemma. It's perfectly reasonable to assume that the only condition you placed on S was that it was infinite.

- Gofer
- Intellectual
**Posts:**165**Joined:**Mon May 09, 2016 8:24 am

### Re: The lemma that robert ignores

Gofer wrote:Jeff, you didn't explicitly state that, even though mentioned in the lemma. It's perfectly reasonable to assume that the only condition you placed on S was that it was infinite.

Gofer, I did. What do you think a lemma is?

Now, look at step #2 in the proof.

- JeffJo
- Intellectual
**Posts:**2503**Joined:**Tue Mar 10, 2009 11:01 am

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