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A recent comment under this old answer claims there's some issue with the following proof.

Prop. A compact metric space $(M,d)$ is sequentially compact.

The "proof":

Suppose a sequence $(x_n)\subset M$ has no convergent subsequence in a compact metric space $M$. Then, for every $y\in M$, there is an open neighborhood $U_y\ni y$ s.t. the set $\{n\in\mathbb{N}: x_n\in U_y\}$ is finite. Then the collection $\{U_y: y\in M\}$ is an open cover of $M$. By compactness, it admits a finite subcover $\{U_{y_1},\cdots, U_{y_k}\}$. But then the index set $\{n\in \mathbb{N}: x_n\in M\}=\cup_{i=1}^k\{n\in \mathbb{N}: x_n\in U_{y_i}\}$ is a finite set. Contradiction.

The comment claims nothing in the above proof uses the fact $M$ is a metric space. We can fix this issue by using open ball $B_y(r_y)$ for some $r_y>0$, instead of using arbitrary open neighborhood $U_y$.

However, if dropping the assumption that $M$ is metric, compactness does NOT imply sequential compactness. But, in that context, I can't seem to find the falsehood in the above false proof. Every step still makes sense to me. So I'm confused. Where does the above "proof" break down in general compact spaces? Thanks in advance.

Update. As the comments below, one mistake I didn't realize is the statement "$(x_n)$ admits a subsequence converging to some point $y$" is NOT equivalent to "$(x_n)$ is frequently in every neighborhood of $y$". Because the latter one cannot guarantee a fixed sub-index $\{n_1,\cdots, n_k,\cdots\}$ making $(x_{n_k})$ eventually in all the neighborhoods of $y$. So point $y$ not being the limit of any subsequence is not equivalent to the requirement that some neighborhood of $y$ must only contains finitely many terms in $(x_n)$.

It's a different type of confusion, comparing with the suggested duplicated question.

user760
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    @ArturoMagidin compactness does not imply sequential compactness- a counterexample is $[0,1]^{[0,1]}$. Neither implies the other in general – Lavender Sep 25 '24 at 14:55
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    I suspect this "proof" might implicitly assume first-countability to deduce the existence of $U_y$? – Lavender Sep 25 '24 at 14:57
  • @Lavender But isn't the existence of such a $U_y$ just due to negating the definition of "there's a subsequence converging to $y$"? – user760 Sep 25 '24 at 15:01
  • @user760 No, the argument is that if there were no such neighborhood, you could construct a convergent subsequence by choosing "smaller and smaller" neighborhoods containing some point in the sequence. But arranging these "smaller and smaller" neighborhoods in such a way that suffices for convergence of the subsequence uses first countability... you take a countable local base $U_n$ and the smaller and smaller neighborhoods are $U_0$, $U_0\cap U_1,$ $U_0\cap U_1\cap U_2,$ etc. – spaceisdarkgreen Sep 25 '24 at 15:04
  • @Lavender Thanks for the link. But what's wrong with this reasoning: A subsequence of $(x_n)$ converges to $y$ iff the original sequence is frequent in every open neighborhood of $y$. That is, for every open $U\ni y$, and every $n\in \mathbb{N}$, there is some $n'>n$ s.t. $x_{n'}\in U$. So negating this would mean there exist a $U_y\ni y$ and some $n_0\in \mathbb{N}$ s.t. for any $n>n_0$, $x_n\notin U_y$. Then this $U_y$ would suffice. No? This seems different to the route in the linked question. – user760 Sep 25 '24 at 15:27
  • @user760 I'm not convinced by this 'frequent' notion. $x_\bullet\to y$ if every neighbourhood eventually contains the sequence i.e. there is an $N$ associated to $U$ such that $x_n\in U$ for all $n>N$. This is different to your definition involving 'frequent', at first glance anyway. And its negation does not entail there is some $U$ with only finitely many $x_\bullet$ contained within; yours does however. Thus I conclude your definition of convergence is wrong – FShrike Sep 25 '24 at 15:30
  • @FShrike If the original sequence is eventually in every neighborhood of $y$, then the sequence itself converges to $y$. But I'm only claiming a subsequence does, not the original sequence. – user760 Sep 25 '24 at 15:32
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    In standard language, if some subsequence (which we hold fixed!) of $x_\bullet$ is eventually in any neighbourhood of $y$, $x_\bullet$ converges to $y$ along a subsequence. Your definition reads: if for every neighbourhood of $y$, some subsequence (possibly this subsequence depends on the neighbourhood) is eventually in that neighbourhood. These aren't the same – FShrike Sep 25 '24 at 15:33
  • You should have to reverse quantifiers, my friend. You should choose the $(x_{n'})$ first; in your definition, as written, the $n'$ may depend on $U$ and this is where the problem lies – FShrike Sep 25 '24 at 15:34
  • @FShrike It seems both "eventually in" and "frequently in" are standard, which mean different things. https://math.stackexchange.com/questions/2849974/sequences-eventually-and-frequently-in-a-set. – user760 Sep 25 '24 at 15:38
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    aye, I've edited to clarify - I misread what you were initially saying as a definition of convergence, rather than subsequential convergence. My point is your quantifier order is reasonable but unfortunate, subtly leaving open the possibility that the choices of $n'>n$ (for any given $n$) depend on $U$. – FShrike Sep 25 '24 at 15:40
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    @FShrike Ah, so "$(x_n)$ is frequently in every neighborhood of $y$" is not equivalent to "$(x_n)$ admits a subsequence converging to $y$". I think I got it. Thanks – user760 Sep 25 '24 at 15:45
  • @Lavender Remembered that belatedly; thanks. – Arturo Magidin Sep 25 '24 at 15:50

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