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seen Nov 27 at 10:32

Oct
21
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.
Oct
21
revised How do I show that $\kappa^+ \le 2^\kappa$ for every cardinal $\kappa$?
edited body; edited title
Oct
21
asked How do I show that $\kappa^+ \le 2^\kappa$ for every cardinal $\kappa$?
Aug
13
awarded  Fanatic
Aug
11
awarded  Yearling
May
9
awarded  Nice Question
May
6
accepted How do I prove that an uncountable subset $S$ of $\mathbb{R}$ has the in-between property?
May
6
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.
May
6
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.
May
6
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.
May
6
asked How do I prove that an uncountable subset $S$ of $\mathbb{R}$ has the in-between property?
Dec
30
awarded  Enthusiast
Dec
9
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'?
Dec
9
accepted What's the difference between saying that there is no cardinal between $\aleph_0$ and $\aleph_1$ as opposed to saying that…
Dec
6
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?
Dec
5
awarded  Commentator
Dec
5
asked What's the difference between saying that there is no cardinal between $\aleph_0$ and $\aleph_1$ as opposed to saying that…
Dec
5
revised Why is $\omega$ the smallest $\infty$?
deleted 76 characters in body
Dec
5
revised Why is $\omega$ the smallest $\infty$?
deleted 350 characters in body
Dec
5
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.