# 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. --prime mover (talk) 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)
Update

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 Metric Space is Weakly Countably Compact iff Countably Compact. 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 These old superseded pages may be relegated to a separate category "deprecated" or something.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)

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)

The statement and proof should be okay now (though I'm not sure if the title is since it sounds more general than the statement, but maybe that's okay). The issue was in getting the open set for each point which contains only finitely many terms. What justifies doing this that's missing in general topological spaces is kind of tricky and was glazed over in the comment that was previously there along the lines of "the contrary of this is that a subsequence converges". One can only demonstrate that statement when you can find some kind of decreasing infinite basis of open sets at each of the points in the space (i.e. when the space is first-countable). Otherwise there's a problem with making sure the sequence you try to construct actually gets "trapped" in the open sets around the point, if you get what I mean.

There's some commentary on this issue here: http://math.stackexchange.com/questions/44907/whats-going-on-with-compact-implies-sequentially-compact

-- Qedetc 17:00, 12 June 2011 (CDT)

Title certainly is not all right. The title reflected the incorrect nature of the original page (which I'm now wishing I'd just bloody well deleted in the first place because it's turned into a complete bleeding nuisance). I'll get to it, but I can't be arsed to fill in all the missing links for a start. --prime mover 17:21, 12 June 2011 (CDT)

The definition of term of a sequence in this wiki is unclear. More importantly, the use of the notion of term in this proof is highly confusing.

In the proof, something similar to the following is stated: $U$ contains only finitely many terms of the sequence.

I now believe this is expected within the proof to mean that (1) $\left\{n\in\mathbb{N} : x_n\in U\right\}$ is finite; and not that (2) the set $\left\{x_n : n\in\mathbb{N}\right\}\cap U$ is finite (since then the proof would be incorrect/incomplete).

On the other hand, one cannot plausibly mean the number $n$ when referring to the n-th term of the sequence. Therefore, I can only infer that the n-th term of the sequence must refer to the element $x_n\in X$. Thus to say that $U$ contains only finitely many terms of the sequence can only be rigorously interpreted as (2), instead of the needed (1).

I think the proof must be rewritten to make this clear, avoiding the confusion between interpretations (1) (which we want) and (2) (which is insufficient for the proof, yet corresponds to standard practice, and moreover seems to be the only possible rigorous interpretation).

Yeah, sorry. I hadn't looked at the definition of "term" offered in Definition:Sequence, and was using it as though the index distinguished them. I agree it might not be clear to someone coming across the page, and I've changed the proof to hopefully avoid the confusion.
I don't think that (2) above is really the standard though. I'm saying this partly based off my expectation that if I walked into a math building, wrote the sequence $(1,2,1,2,1,2,1,2,...)$ on a chalk board, and asked how many terms were even, there would not be a lot of people who say "exactly one of the terms is even." I also suspect you'd hear many of them say describe $(1,1,1,1,0,0,0,...)$ as "the sequence whose first four terms are $1$ and remaining terms are $0$". I've got at least one analysis text (tbh I haven't checked the others, I just grabbed the one on the floor) that never formally defines "term" but makes a point of distinguishing between "number of terms" and the size of the range, so that, in their example, the sequence defined by $a_n = (-1)^n$ has an infinite number of terms while it only takes on two values.
My guess is "term" is usually thought of in a way that would make sense if it were formally defined to be a pair $(n, x_n)$ but informally used to mean the value $x_n$ when it is used like a value for convenience. Thinking of it as a pair like this isn't a completely ridiculous notion when you remember that sequences are usually formalized as functions $\mathbb N \to X$.
For anyone who is interested, I think the proof works regardless of which interpretation you use, though you need to make the extra observation that if $(x_n)$ had infinitely many repetitions of the same value, then the subsequence consisting of only that value would be convergent.
-- Qedetc 12:18, 13 June 2011 (CDT)
That is a very good point. I had never thought about the possibility that the n-th term could stand for $(n,x_n)$. In fact, I don't think I had ever used it in that way, which would actually be convenient. I also agree with your example of how most mathematicians would respond to the question regarding the even terms of $(1,2,1,2,1,...)$.
Nevertheless, this certainly tripped me up in the previous version of the proof. Furthermore, without a rigorous definition of term, I think it had the potential to mislead a lot of other people who immediately translate term to $x_n$ for some $n$. One solution is to rewrite the proof, like you did. Another solution is to explicitly define n-th term to mean $(n,x_n)$ (for example in the definition of sequence), and make that clear within the proof.
Finally, I agree that the proof would be essentially correct even with interpretation (2), as long as one added the remark that the sequence would then attain only finitely many values, and therefore would have a constant subsequence.