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Jul 14, 2016 at 23:21 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Jun 14, 2016 at 22:47 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
May 15, 2016 at 22:19 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Sep 11, 2015 at 2:14 answer added Henry.L timeline score: 2
Sep 11, 2015 at 0:17 comment added Henry.L @ChristianRemling I have already tried to write up a proof following Mr.Edgar's reference, I think I can answer my own question later. Thanks for you kindness.
Sep 10, 2015 at 22:50 comment added Christian Remling I think you'll have to read up a little on the basics of ordinals and cardinal arithmetic, then (2) becomes essentially trivial. The hint (your box) gives the claimed representation of $\sigma(A)$, and the claim on the cardinality follows from $|A_{\alpha}|=c$ (by induction) and $c^{\omega}=c$.
Sep 10, 2015 at 2:38 comment added Nate Eldredge Please add a link to the Math.SE question.
Sep 10, 2015 at 0:55 comment added Gerald Edgar I first saw this in Hewitt & Stromberg, Real and Abstract Analysis.
Sep 10, 2015 at 0:52 comment added Henry.L @GeraldEdgar I am very confused about author's words so do you know any text that addresses the same problem more clearly? Thanks.
Sep 10, 2015 at 0:47 comment added Gerald Edgar I think "a countable sequence of these operations" is in this notation $\mathcal A_\omega$. But a countable ordinal is "a countable ordered set" so that part refers to $\mathcal A_\alpha$ for a countable ordinal.
Sep 10, 2015 at 0:25 history asked Henry.L CC BY-SA 3.0