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Sep
23
comment How do I show $|\frac{i\overline{z}}{2} - \frac{i}{2}|=|z - 1|?$
@Timbuc I didn't. And thanks for your help.
Sep
23
comment How do I show $|\frac{i\overline{z}}{2} - \frac{i}{2}|=|z - 1|?$
@Timbuc No, only that $|Z| < 1$
Sep
23
comment How do I show $|\frac{i\overline{z}}{2} - \frac{i}{2}|=|z - 1|?$
So, it's a mistake in the text?
Sep
23
comment How do I show $|\frac{i\overline{z}}{2} - \frac{i}{2}|=|z - 1|?$
Of course I tried that before asking here. This isn't my homework, btw.
Sep
23
asked How do I show $|\frac{i\overline{z}}{2} - \frac{i}{2}|=|z - 1|?$
Jul
2
awarded  Curious
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.
May
17
asked Connected spaces: is there a mistake in the example below from James Munkres' Topology?
Feb
22
awarded  Teacher
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
revised How do I show that $\kappa^+ \le 2^\kappa$ for every cardinal $\kappa$?
added 45 characters in body
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
accepted How do I show that $\kappa^+ \le 2^\kappa$ for every cardinal $\kappa$?
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.
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 $\kappa$?