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While trying to prove that a set is sequentially compact, I was suggested to prove by contradiction -- this is how it went, at least part of it:

Definition. We say that a set $A$ is sequentially compact if every sequence $(x^m)$ with $\{x^m\}\subseteq A$ has a sequence $(x^{m_k})$ that $x^{m_k}\rightarrow x\in A$.

Theorem. Being $(M,d)$ a metric space, a set $A\subseteq M$ is compact only and only if it is also sequentially compact.

part of what you should assume for the demonstration. Suppose initially that $A$ is a compact set. Suppose too that the sequence $(x^m)$ is such that $\{x^m\}\subseteq A$ and $(x^m)$ does not possess any sequence that converges to any element of $A$. As we know that $A$ is a closed set, that implies that $(x^m)$ does not possess any convergent sequence in $(M,d)$.

Actual Demonstration (part of it [the part I cannot understand]). Begin trying to define $\{x^m\}$ as a closed set. Take $y\in M-\{x^m\}$ and suppose $\{x^m\}$ as a not closed set (implying $M-\{x^m\}$ as a not open set). Suppose that for every ε>0, and for every N, $B(y,ε)∩({(x^m):m≥N})≠∅$. For $k=1$, take $x^{m_1}\in\{x^m\}\cap B(y,1)\neq\emptyset$. Take $x^{m_2}\in\{x^m\}\cap B,(y,1/2)\neq\emptyset$, such that $m_2>m_1$. To see that this is possible, define $\delta:=\min(x^m,y)$. By hypothesis, $\{x^m\}\cap B,(y,\delta)\cap B(y,1/2)\neq\emptyset$; so we can take $x^{m_2}\in\{x^m\}\cap B,(y,\delta)\cap B(y,1/2)\neq\emptyset$. We can guarantee that $m_2>m_1$. Proceeding inductively in such manner, we can obtain a subsequence $(x^{m_k})$ of $(x^m)$ such that, for every $k$, $d(x^{m_k}, y)<1/k$. So $x^{m_k}\rightarrow y$. That is a contradiction with the statement $(x^m)$ does not have any converging subsequences. We conclude that $\{x^m\}$ is a closed set.

Here it is what I haven't figured out: I don't know if I understood the whole demonstration correctly (or copied correctly from the board in my class), but how can we prove that $\{x^m\}$ is closed if a subsequence $(x^{m_k})$ of $(x^m)$ converges to a point outside of $\{x^m\}$, since $y\in M-\{x^m\}$? The trick was in assuming that $\{x^m\}$ didn't have any converging sequences in (M,d) and proving a contradiction in the demonstration. But how does that make $\{x^m\}$ a closed set? I still cannot see what I am missing. Can someone point if I am understanding something wrong or if there is any correction to be made in this demonstration?

Notation: I used $\{x^m\}$ as the set of the sequence $(x^m)$. If there are any questions about notation, please ask.

Observations (edited): I'm having some trouble in writing in this system -- I welcome any help. My question changed a bit now because of further development of the discussion of this demonstration with friends.

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  • $\begingroup$ 1) Is $x^m$ supposed to be an element of $A$ ? Then you soulf write $x^m \in A$ instead of $x^m \subset A$. 2) What do you mean by 'define $x^m$ as a closed set' ? 3) What is $M\{x^m\}$ ? $\endgroup$ – user10676 May 5 '13 at 9:34
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    $\begingroup$ There is a big issue with what you have copied. The sentence 'Suppose that for every $\varepsilon>0$...' should be replaced by something like 'Claim: $\forall \varepsilon>0, \forall N, B(y,\varepsilon) \cap \{ x^m: m \geq N \} \neq \emptyset$' (I let you figure out why it is true). $\endgroup$ – user10676 May 5 '13 at 12:47
  • $\begingroup$ I am sorry! something about the notation came out wrong. I think the system here took away the "\" between {x^m} and M. Writing in another way: M-{x^m}. So y belongs to complement of {x^m}. For the other questions: x^m is an element of A, but I was trying to define the set of the elements of the sequence (x^m) to be contained in A. Hence the notation {x^m}. As for {x^m} being a closed set, the demonstrations tries to prove that. And I think you're right, user10676, we have to define an element N for which the statement starts to be true -- I will edit it right now. $\endgroup$ – John Doe May 5 '13 at 15:57
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I think $M\{x^m\}$ is supposed to be the margin of $\{x^m\}$, i.e. $$ M(X) := \overline{X} \setminus X \text{.} $$ where $\overline{A}$ is the closure of $A$.

The proof then shows that for every $y \in M(\{x^m\})$ there's a sub-sequence of $(x^m)$ which converges to $y$. Since $(x^m)$ is not supposed to have any convergent sub-sequences, it follows that $M\{x^m\}$ is empty, and thus that $\overline{\{x^m\}} = \{x^m\}$, i.e. that ${x^m}$ is closed.

You do the rest with Heine-Borel. If $A$ is compact then $A$ is closed and bounded, hence $\{x^m\}$ is bounded. But by the above$\{x^m\}$ is also closed, hence also compact. But if $\{x^m\}$ contains no convergent subsequence, you can create a covering of $\{x^m\}$ with open balls (centers as the $x^m$, you have pick a suitable $\epsilon$) with no open subcover, which is a contradiction to $\{x^m\}$ compact.

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  • $\begingroup$ fgp, there was a notation error. I meant to write M-{x^m} or the complement of {x^m}. My trouble is with understanding this bit: why can we have $(x^{m_k}\)→y$ when y belongs to the complement of {x^m}? But thank you for your help anyhow. $\endgroup$ – John Doe May 5 '13 at 16:38

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