Continuum Hypothesis

Sep 2009
502
39
The author of my book stated

The Continuum Hypothesis: There exists no set \(\displaystyle S\) such that

\(\displaystyle \aleph_0<|S|<c.\) where \(\displaystyle c = |\mathbb{R}|\)

He then went on raising a question: Is there as set \(\displaystyle S\) such that \(\displaystyle |S|>c\)?

Following the question, he proved that \(\displaystyle |A|<|P(a)|\), and at the end of the proof, he said the proof shows that there is no largest set. In particular, there is a set S with \(\displaystyle |S|>c\).

Remark: I have not a clue what he meant in blue texts.

Question: Is there or is there not that \(\displaystyle |S|>c\)?
 
Sep 2009
502
39
Open interval (0,1)

Theorem: The set (0,1) and \(\displaystyle \mathbb{R}\) are numerically equivalent.


I know that if I write the open interval as

\(\displaystyle (a,b)\)

where \(\displaystyle a<b\), and that \(\displaystyle a,b\in \mathbb{R}\), I also can conclude that \(\displaystyle (a,b)\) and \(\displaystyle \mathbb{R}\) are numerically equivalent.

Question: What is the purpose of the theorem? Does it mean to say that any uncountable subset of \(\displaystyle \mathbb{R}\) is numerically equivalent with \(\displaystyle \mathbb{R}\)?
 

Plato

MHF Helper
Aug 2006
22,461
8,632
(it is) proved that \(\displaystyle |A|<|P(a)|\), and at the end of the proof, he said the proof shows that there is no largest set. In particular, there is a set S with \(\displaystyle |S|>c\).
Remark: I have not a clue what he meant in blue texts.
Question: Is there or is there not that \(\displaystyle |S|>c\)?
\(\displaystyle c = \left| \mathbb{R} \right| < \left| {\mathcal{P}\left( \mathbb{R} \right)} \right|\).
The statement in blue simply means: given any set its power set has greater cardinality so there is always a 'larger' set.
 
  • Like
Reactions: novice
Sep 2009
64
23
Edinburgh, UK
There are two ways of comparing the number of elements in a set. You either count them, and see which one is bigger, which is fine for finite sets, or you can try and set up a one to one correspondence between the elements.

Kolmogrov gives the example of comparing the number of students is a classroom to the number of chairs: if there are still chairs left over when eveyone is sat down, then there are more chairs than students, and if there are still students standing up when everyone is sat down, then there are more students than chairs. And, in fact, this works well for infinite sets.

The fact that \(\displaystyle |A| < |\mathcal{P}(A)|\) means that we can match every element of \(\displaystyle A\) to an element of \(\displaystyle \mathcal{P}(A)\) like students to chairs, with some elements in \(\displaystyle \mathcal{P}(A)\) to spare, which intuitively means that the set \(\displaystyle \mathcal{P}(A)\) is 'bigger' than \(\displaystyle A\). The fact that there is no biggest set means that there is no set which cannot be 'fit inside' another set in this way: no matter how many students we have, we can always find a classroom to sit them all with some chairs left over.

In particular, suppose that \(\displaystyle A\) is the biggest possible set, then \(\displaystyle \mathcal{P}(A)\) is bigger, a contradiction. To answer the question, \(\displaystyle |\mathcal{P}(\mathbb{R})|>|\mathbb{R}|\), so yes there is such an \(\displaystyle S\).

And as for your second question, the fact that \(\displaystyle (0,1)\) and \(\displaystyle \mathbb{R}\) are numerically equivalent doesn't mean that any infinite subset of \(\displaystyle \mathbb{R}\) is numerically equivalent to \(\displaystyle \mathbb{R}\) because \(\displaystyle \mathbb{N}\subset\mathbb{R}\) yet \(\displaystyle \mathbb{R}\) is not countable. The purpose of the theorem is anybody's guess, you might as well ask what the meaning of life is!
 
  • Like
Reactions: novice
Feb 2010
470
154
The fact that \(\displaystyle |A| < |\mathcal{P}(A)|\) means that we can match every element of \(\displaystyle A\) to an element of \(\displaystyle \mathcal{P}(A)\) like students to chairs, with some elements in \(\displaystyle \mathcal{P}(A)\) to spare, which intuitively means that the set \(\displaystyle \mathcal{P}(A)\) is 'bigger' than \(\displaystyle A\). The fact that there is no biggest set means that there is no set which cannot be 'fit inside' another set in this way: no matter how many students we have, we can always find a classroom to sit them all with some chairs left over.
That might be misleading. In general

|S| < |T| if and only if there is an injection from S into T but there is no bijection from S onto T.

That there is an injection from S into a proper subset of T (the fact you've mentioned) does not in itself entail that |S| < |T|.
 
  • Like
Reactions: nimon