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In a metric space $(X,\Omega)$ given that $A$ is a subset of $X$: if every sequence of $A$ contains a subsequence that converges to some point in $X$, then, $A$ is compact relative to $\Omega$.

Proof:

Suppose that every sequence $x_i ∈ A$ has a convergent subsequence and let $F$ be a cover of $A$. Find a countable open cover $F_0 = \{U_i ⊆ X |U_i \text{ is open} , i ∈ N\}$ with the property that for every $U_i ∈ F_0$ there is some $V ∈ F$ with $U_i ⊆ V$.

Claim: $F_0$ has a finite subcover. To prove the claim, suppose the opposite. Then by picking an arbitrary $x_n ∈ A − \bigcup_{i-1}^{n-1}U_i$ we obtain a sequence. By the assumption, there must be a subsequence $y_i = x_{n_k}$ with converges to $y_0$. Since $F_0$ is a cover,there is some $m ∈ N$ with $y_0 ∈ U_m$. But then $y_j \notin U_m$ for all $j ≥ m$ which is a contradiction. This concludes the proof of the claim.

Thus let $\{U_1, .., U_N \}$ be a finite cover of $A$ and let $V_i ∈ F$ be such that $U_i ⊆ V_i$ for all $i = 1, ...., N$. The $\{V_1, ..., V_n\}$ is also a finite cover of $X$.

Could the step of contradiction in the proof be explained?

tryst with freedom
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H Cruz
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1 Answers1

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The sequence $y_i$ is a convergent subsequence of $x_n$. The thing it converges to is $y_0$. We defined $x_n$ to be a sequence of things which are not members of any finite subcover $U_i$ by excluding the union of the $U_i$'s up to $n-1$ in the definition of $x_n$. But $y_0$ is the limit of the subsequence $y_i$ and since $y_0\in A$, and since $U_i$ is a cover of $A$, this contradicts that $x_n \notin U_i$.

In other words the limit of the convergent subsequence of the $x_n$s, a thing we constructed by saying that it is not in any finite cover, has to be in one or the other of the elements of the finite cover. We label this element $U_m$. However, nothing which converges to $y_0$ (the convergent subsequence $y_i$) can be in $U_m$.

Suzu Hirose
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