subsets of convergence

Question

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For every collection of open sets {Uγ : γ ∈ Γ} with S ⊆ Uγ∈Γ Uγ, there is a finite collection {Uγ : γ ∈ F} such that S ⊆ Uγ∈F Uγ. Then prove that every sequence in S has a convergent subsequence that converges to a point of S.***

This is a question that I've been trying to figure out for ages.

I know that I need to show that if all subsequences of a sequence (xn)∞ n=1 diverge, then for all s ∈ S and all ε > 0, there are only finitely many terms of the sequence (xn)∞ n=1 that belong to the set (s − ε, s + ε) and use this fact to find a cover for S

But I don't understand how to do this

Answer 1

Background

Say you are working in a metric space $(X,d)$ with $S \subseteq X$. The assumed property of $S$ is called compact.

$S$ is compact iff each of its open covers has a finite subcover. That is, for any collection $\{U_\gamma\}_{\gamma \in \Gamma}$ of open subsets of $X$ such that $S \subseteq \cup_{\gamma \in \Gamma}\ U_\gamma$, there is a finite subset $F$ of $\Gamma$ such that $S \subseteq \cup_{\gamma \in F}\ U_\gamma$.

This property is particularly useful in analysis since every interval $[a,b]$ is compact. As a consequence, you can have Bolzano-Weierstrass Theorem as pointed out in another answer. Extreme value theorem is another direct consequence of the compactness of an interval.

Back to business, assuming $S$ is compact, the question is asking you to show that $S$ is sequentially compact:

S is sequentially compact iff every sequence $\{x_n\}$ in $S$ has a convergent subsequence whose limit is in $S$.

But I would like to proceed by proving that $S$ is limit point compact:

S is limit point compact iff every infinite subset $E$ of $S$ has a limit point in $S$.

The sequentially compact property and the limit point compact property are actually equivalent to compactness if you are working in a metric space. As you will learn later, compactness is equivalent to totally boundedness + completeness in a uniform space.

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Compactness implies limit point compactness

Let $E$ be any infinite subset of $S$. Suppose $E$ had no limit point in $S$. Then each point $s$ in $S$ has the following property:

$$\exists \ r(s) > 0 \text{ s.t. }\forall \ p\in E,\ [s = p \vee d(s,p) > r(s)]$$

For each $s \in S$, let $U_s = B_{r(s)}(s)$. If $s \notin E$, then this neighborhood does not contain any point of $E$. If $s \in E$, then $U_s$ contains $s$ as the only point of $E$.

Now $\mathscr{C} = \{U_s\}_{s \in S}$ is an open cover of $S$. But for any $e \in E$, $U_e$ is the only open set in $\mathscr{C}$ that contains $e$. Therefore, each subcover $\mathscr{D}$ must have $\{U_s\}_{s \in E}$ as a subset of $\mathscr{D}$. That is, $\mathscr{D}$ must be infinite. This contradicts the compactness of $S$.

Now each infinite subset $E$ of $S$ must have a limit point in $S$.

Limit point compactness implies Sequential compactness

Let $E$ be the range of $\{x_n\}$.

Suppose $E$ is finite. Then there is some constant subsequence which is obviously convergent. Let $x_n = (-1)^n$ as an example, the sequence $x_{2n} = 1$. You might need to do mathematical induction on the cardinality of range of $\{x_n\}$ to prove the existence of such a constant subsequence for a general sequence with finite range.

Suppose $E$ is infinite, then $E$ has a limit point in $S$. For each $k$, $B_{1/k}(p)$ contains infinitely many points of $E$. Take $x_{n_1} = x_1$. Choose $x_{n_k} \in B_{1/k}(p) \cap E$ such that $n_k > n_{k-1}$. It follows that $x_{n_k} \rightarrow p$.

In summary, compactness implies limit point compactness implies sequential compactness. QED.

Answer 2

If every open cover of a set has a finite subcover,then the set will be compact set. By the given conditions , S is a compact set. Now by Bolzano-Weierstrass therom, Every bounded sequence has a convergent subsequence. Thus S has a convergent subsequence. Again since S is closed, every sequence of S that converge ,must converge to a point of S .