Talk:Compact First-Countable Space is Sequentially Compact

The notation $\{x_n\}_{n \in \N}$ is, I believe, not recommended, as (using the "set brace" notation that it does), it carries the implication that the order of terms does not matter.

I've been fairly consistently using $\left \langle {x_n} \right \rangle$ for a general sequence, but $\left({x_n} \right)$ is frequently seen. I prefer the former because round brackets are arguably overused.

Hope you don't mind (in the interests of consistency) my changing it to $\left \langle {x_n} \right \rangle$ as appropriate. --Matt Westwood 12:25, 8 March 2009 (UTC)

WARNING

The statement in this proof is WRONG without any other assumption on the space $(X, \vartheta)$. There exist compact spaces which are not sequentially compact; for example $[0,1]^{[0,1]}$ is one of them. You may want to add hypothesis of 2nd countability to $X$ and, in this case, even relax compactness to Lindelöf. Reading more carefully the proof you will be convinced that it is, in fact, wrong (look at the last part!).


 * (whew) this one wasn't one of mine, thank goodness. Okay, I'll put a qmark by it - feel free to do the dirty. Why not set yourself up a username while you're about it, if you're up for it ... your comments are appreciated. --prime mover 13:36, 12 March 2011 (CST)

An amendment has been made by an anonymous user changing "topological space" to "metric space". Fair enough, but now the page is redundant, as there already exist Countably Compact Metric Space is Compact (the statement includes an "iff") and Countably Compact Metric Space is Sequentially Compact (also including an "iff"). Both of these make use of the results that Metric Space is First-Countable and (indirectly) that a countably compact metric space is second-countable. There are a few other redundant pages round here from before this area was approached systematically. These old superseded pages may be relegated to a separate category "deprecated" or something. Haven't thought what yet, the idea has only just occurred to me (we don't want to delete these pages as there may be external links, but OTOH we don't want a lot of redundancy cluttering up what needs to be clearer. Comments appreciated. --prime mover 16:16, 22 May 2011 (CDT)
 * Update


 * Further update:
 * Another amendment has been made by another anonymous user who has added further constraints on the conditions under which a compact space is sequentially compact. This is a good direction to go in, but now the proof does not match the theorem stated (and in fact nor does the title any more).


 * I expect this will end up being rewritten as something like: Let $T$ be a first-countable T1 space. Then $T$ is compact iff it is sequentially compact. Or there may be a stronger way of wording it. Also the proof itself will need to be amended to take into account the actual T1-ness and first-countableness of the set. At the moment this is not the case, and the proof remains flawed. --prime mover 10:35, 11 June 2011 (CDT)