Prove compact subsets of metric spaces are closed
Note, this question is more of analyzing an incorrect proof of mine rather than supplying a correct proof.
My Attempted Proof
Suppose $X$ is a metric space. Let $A \subset X$ be a compact subset of $X$ and let $\{V_{\alpha}\}$ be an open cover of $A$. Then there are finitely many indices $\alpha_{i}$ such that $A \subset V_{\alpha_{1}} \cup \ … \ \cup V_{\alpha_{n}}$.
Now let $x$ be a limit point of $A$. Assume $x \not\in A$. If $x \not\in A$ put $\delta = \inf \ \{\ d(x, y) \ | \ y \in A\}$. Take $\epsilon = \frac{\delta}{2}$, then $B_d(x, \epsilon) \cap A = \emptyset$ so that a neighbourhood of $x$ does not intersect $A$ asserting that $x$ cannot be a limit point of $A$, hence $x \in A$ so that $A$ is closed. $\square$.
Now there must be something critically wrong in my proof, as I don't even use the condition that $A$ is compact anywhere in the contradiction that I establish. The above proof would assert that every subset of a metric space is closed.
I think my error must be in the following argument : $\delta = \inf \ \{\ d(x, y) \ | \ y \in A\}$. For if we take $X = \mathbb{R}$ and $A = (0, 1) \subset \mathbb{R}$, then $\delta = 0$ if $x = 1$ or $x = 0$.
Am I correct in analyzing this aspect of my proof?
Best Answer
Suppose $A$ is compact. Then let $x$ be in $X \setminus A$. We must find a neighbourhood of $x$ that is disjoint from $A$ to show closedness of $A$.
For every $a \in A$ let $U_a = B(a, \frac{d(a,x)}{2})$ and $V_a =B(x, \frac{d(a,x)}{2})$. Then $U_a$ and $V_a$ are disjoint open neighbourhoods of $a$ and $x$ respectively (disjoint follows from the triangle inequality, check this).
Then the $U_a ,a \in A$ together form a cover of $A$, and now we use compactness of $A$ to get finitely many neighborhoods $U_{a_1}, \ldots, U_{a_n}$, that also cover $A$. But then $V_{a_1} \cap \ldots V_{a_n}$ is an open neighbourhood of $x$ that misses $A$ (since a point in $A$ cannot be in all of $V_{a_1}, \ldots, V_{a_n}$ because it belongs to some $U_{a_i}$ which is disjoint with $V_{a_i}$.).
So $A$ is closed.
An alternative with sequences: suppose $x \in \overline{A}$. Then being a point in the closure we find a sequence in $A$, $(a_n)$ that converges to $x$. The sequence $(a_n)$ from $A$ has a convergent subsequence in $A$ by compactness of $A$, so there is some $a \in A$ and some subsequence $a_{n_k} \rightarrow a$. But also $a_{n_k} \rightarrow x$, and as limits of sequences in metric spaces are unique: $a =x$, but then $x \in A$ as required: $\overline{A} = A$.