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)