0

So I did this but I'm not sure if it is an actual proof because it is based on a hypothesis, and also I'm not sure if I understand the Continuum Hypothesis correctly.

Because $\boldsymbol{I} \subseteq \boldsymbol{R}$, l can have either the same or less cardinality than $\boldsymbol{R}$. By the Continuum Hypothesis, l can not have a lower cardinality than $\boldsymbol{R}$, so it must have the same cardinality.

Is this correct, and/or is there a more rigorous method?

Also, can I use the notation $\operatorname{card} A < \operatorname{card} B$?

mle
  • 2,247
  • 1
  • 18
  • 33
Dude
  • 13
  • 5
  • it isn't proven, and it could be a larger cardinality. – Jacob Claassen May 03 '17 at 02:16
  • how can a subset of a set have a larger cardinality than the set? – Dude May 03 '17 at 02:17
  • You still need to show that the cardinality isn't lower than that of $\mathbb{R}$. That is, why can't the irrationals be countable? – AspiringMathematician May 03 '17 at 02:20
  • Oh yes I already did show that both R and I are uncountable I just didn't think I should put the whole proof up here – Dude May 03 '17 at 02:21
  • They are not countable. Doesn't mean they are exactly C – user76568 May 03 '17 at 02:21
  • @Dude Then I think you're done. – AspiringMathematician May 03 '17 at 02:22
  • Idk man, I'm not throwing any possibilities in math out after I learned about 1+2+3+...=-1/12 – Jacob Claassen May 03 '17 at 02:30
  • Why does CH mean it can't have lower cardinality? If you mean cardinality of I and R areuncountable, you haven't shown that that R isn't a higher uncountable cardinality. The standard thing you have to do is show removing a countable subset from an uncountable maintains the same cardinality. – fleablood May 03 '17 at 02:57
  • @fleablood CH means in particular that every uncountable set of reals has size continuum - since $I$ is a set of reals, we're done. – Noah Schweber May 03 '17 at 03:18
  • Well, if you have can rely upon *that* you are done, but there is little point in worrying about proving the cardinality of I. – fleablood May 03 '17 at 03:21
  • @fleablood I was just responding to your question "Why does CH mean it can't have lower cardinality?". Of course there's no value to using CH here. – Noah Schweber May 03 '17 at 04:46
  • No, that does explain the the statement. I was assuming the op hadn't considered that uncountable sets may have different cardinalities. The CH is very powerful. More powerful than the impression I got from the post. – fleablood May 03 '17 at 05:00
  • Possibile duplicate of https://math.stackexchange.com/questions/265451/the-cardinality-of-mathbbr-mathbb-q?noredirect=1&lq=1 and https://math.stackexchange.com/questions/732/proof-that-the-irrational-numbers-are-uncountable – mle May 03 '17 at 07:38

1 Answers1

2

Use the fact that for infinite cardinals $$\kappa+\lambda=\max\{\kappa,\lambda\}$$

Now $$\mathbb{R}=\mathbb{Q}\cup I$$ Thus we have $$|\mathbb{R}|=\max\{|\mathbb{Q}|,|I|\}$$ and it follows $$|\mathbb{R}|=|I|$$

Rene Schipperus
  • 38,213
  • 2
  • 28
  • 74
  • Axiom of choice. But yeah :P – user76568 May 03 '17 at 02:25
  • I didn't know the first fact. What is it called, so I can look more into it? – Dude May 03 '17 at 02:28
  • @Dror I wasn't aware I had to avoid it, in that case $\mathfrak{b}+\aleph_0=\mathfrak{b}$, if $\aleph_0\leq \mathfrak{b}$ is trivial and this hypothesis clearly holds for the irrationals. – Rene Schipperus May 03 '17 at 02:30
  • The first fact is quite easy and is in most books on set theory. It follows from $\lambda+\lambda=\lambda$ for infinite $\lambda$, which is just an interweaving argument. – Rene Schipperus May 03 '17 at 02:31
  • @ReneSchipperus You don't have to avoid it ofc! But it is still interesting to add to an answer – user76568 May 03 '17 at 02:34