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seen Aug 24 at 22:15

May
17
comment Connected spaces: is there a mistake in the example below from James Munkres' Topology?
Thank you for your answer.
May
17
comment Connected spaces: is there a mistake in the example below from James Munkres' Topology?
Yes, you're right. I was skipping over the word 'contains'. And, no it doesn't say that the subspace is connected. Lots of confusion in my head, is all. Thanks for your help.
Oct
22
comment How do I show that $\kappa^+ \le 2^\kappa$ for every cardinal $\kappa$?
Right. So thanks again for all your help. Cheers. :)
Oct
22
comment How do I show that $\kappa^+ \le 2^\kappa$ for every cardinal $\kappa$?
I see we're talking just about the ordinals. Only when $|A \times A|=|A|$ for all infinite sets then that imply AC. And I know if it's a theorem of ZF that makes it a theorem of ZFC also. I didn't phrase my comment very well. What I meant to say earlier was that your first theorem looked to be that of ZFC and not ZF.
Oct
22
comment How do I show that $\kappa^+ \le 2^\kappa$ for every cardinal $\kappa$?
By Hessenberg sums for ordinals do you mean the canonical well-ordering of ordinal pairs? Used to prove that $\aleph_\alpha \times \aleph_\alpha = \aleph_\alpha$? And does not $|\kappa \times \kappa|=|\kappa|$ imply AC? Your first theorem seems also to be that of ZFC to me. Correct me if I'm wrong.
Oct
21
comment How do I show that $\kappa^+ \le 2^\kappa$ for every cardinal $\kappa$?
Thank you so very much for your answer.
Oct
21
comment How do I show that $\kappa^+ \le 2^\kappa$ for every cardinal $\kappa$?
@bof Yes, assuming AC.
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.
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.
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
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
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.
Dec
1
comment Is there a number that's right in the middle of this interval $(0, 1)$?
OK, so because my partitioning isn't countable, what I did wasn't right.
Dec
1
comment Is there a number that's right in the middle of this interval $(0, 1)$?
I don't quite follow your post. Can't the entire interval [0, 1] be partitioned into infinitely-many singleton sets, with each adding length zero and thus $[0, 1] = \{0\} \cup \dots \cup \{1\} = 0$? What am I missing?
Aug
24
comment If $\kappa$ is a cardinal, $\aleph$ is any $\aleph$-number, and if $\kappa\leq\aleph$ then $\kappa$ can be well ordered as well.
Thank you very much for your help.
Aug
24
comment If $\kappa$ is a cardinal, $\aleph$ is any $\aleph$-number, and if $\kappa\leq\aleph$ then $\kappa$ can be well ordered as well.
@Arthur: I don't really know. I'm getting this from here: math.stackexchange.com/questions/56466/… in the proof we're not assuming the AC holds so I guess working without it.
Aug
19
comment If $a, p$ are cardinals satisfying $2 p = p$ and $a+p=2^p,$ then $a \ge 2^p.$
I don't know why I didn't think of it that way. Thanks.
Aug
19
comment If $a, p$ are cardinals satisfying $2 p = p$ and $a+p=2^p,$ then $a \ge 2^p.$
@Asaf Karagila: I'm new... will try to format my questions better next time. And thanks for the link.