Mark
Reputation
363
Next privilege 500 Rep.
Access review queues
 Oct21 comment How do I show that $\kappa^+ \le 2^\kappa$ for every cardinal $\kappa$? @bof Yes I know Cantor's theorem. And all the cardinals are comparable only when you assume AC. I haven't seen a proof of $\lambda < \kappa^+$ implies $\lambda \le \kappa$. I'll look into that. I think it's best to assume I know everything, and answer me. I'd ask if I didn't understand something. Oct21 revised How do I show that $\kappa^+ \le 2^\kappa$ for every cardinal $\kappa$? edited body; edited title Oct21 asked How do I show that $\kappa^+ \le 2^\kappa$ for every cardinal $\kappa$? Aug13 awarded Fanatic Aug11 awarded Yearling May9 awarded Nice Question May6 accepted How do I prove that an uncountable subset $S$ of $\mathbb{R}$ has the in-between property? May6 comment How do I prove that an uncountable subset $S$ of $\mathbb{R}$ has the in-between property? @BrianM.Scott: I see your point. I didn't thought about that either. May6 comment How do I prove that an uncountable subset $S$ of $\mathbb{R}$ has the in-between property? @BrianM.Scott Gee I didn't knew the proof of the NI theorem depends on what I'm trying to use it to prove. And yes rays and open/closed intervals is all I care about. The reason I asked for a 'more rigorous'proof is that the text I remember containing this problem didn't seemed like it would ask for a proof if this was such a trivial problem. May6 comment How do I prove that an uncountable subset $S$ of $\mathbb{R}$ has the in-between property? @BrianM.Scott: I understand that would work, but I was wondering if there was a more rigorous proof. Also, you can edit the title if you'd like. That's the best I could come up with. May6 asked How do I prove that an uncountable subset $S$ of $\mathbb{R}$ has the in-between property? Dec30 awarded Enthusiast Dec9 comment What's the difference between saying that there is no cardinal between $\aleph_0$ and $\aleph_1$ as opposed to saying that… You wrote "it could also be the case that there are smaller cardinalities but none which are strictly between $\aleph_0$ and the countable of such set." It's not clear to me what those cardinalities are smaller compared to. And also, is the last line a typo and you were going for $\aleph_1$ in place of 'the countable of such set'? Dec9 accepted What's the difference between saying that there is no cardinal between $\aleph_0$ and $\aleph_1$ as opposed to saying that… Dec6 comment What's the difference between saying that there is no cardinal between $\aleph_0$ and $\aleph_1$ as opposed to saying that… @Arthur Fischer: In Set Theory by Thomas Jech, the author writes that "one cannot prove without the Axiom of Choice that $\omega_1$ is not a countable union of countable sets." That is to say, without AC, the uncountability of $\omega_1$ is unprovable. It negates what you wrote in the second paragraph that nowhere was AC used, doesn't it? Dec5 awarded Commentator Dec5 asked What's the difference between saying that there is no cardinal between $\aleph_0$ and $\aleph_1$ as opposed to saying that… Dec5 revised Why is $\omega$ the smallest $\infty$? deleted 76 characters in body Dec5 revised Why is $\omega$ the smallest $\infty$? deleted 350 characters in body Dec5 comment Why is $\omega$ the smallest $\infty$? @Marvis I was under the impression that part of the question was already resolved. If it isn't still, I've edited my post explaining how I think $\aleph_0$ is the least of infinite cardinals.