A simple proof that compactness implies closeness and boundedness in metric space

Question

I'm trying to prove this well-known result by using net. The proof is surprisingly simple. Could you have a check on my attempt?

Let $X$ be a metric space, $\tau$ the induced topology on $X$, and $A \subseteq X$ be compact (in the subspace topology $\tau_A$). Then $A$ is closed and bounded.

Lemma: Let $a\in A$ and $(x_d)_{d \in D}$ be a net in $A$. Then $x_d \to a$ in $\tau$ if and only if $x_d \to a$ in $\tau_A$. [A proof is provided here]

My attempt:

Let $a\in X$ and$(x_d)_{d\in D}$ be a net $A$ such that $x_d \to a$ in $\tau$. By this characterization of closed set, it suffices to prove $a\in A$. Because $A$ is compact, there exist $a' \in A$ and a subnet $(x_{\varphi(s)})_{s\in S}$ of $(x_d)_{d\in D}$ such that $x_{\varphi(s)} \to a'$ in $\tau_A$. By our lemma, $x_{\varphi(s)} \to a'$ in $\tau$. Any convergent subnet of $(x_d)_{d\in D}$ converges to $a$. In particular, $x_{\varphi(s)} \to a$ in $\tau$. Because $(X, \tau)$ is Hausdorff, $a=a'$ and thus $a\in A$. Hence $A$ is closed.

Let $A_x := \mathbb B_A(x, 1) := \{y \in A \mid d(y, x) < 1\}$. Then $A_x \in \tau_A$, i.e., $A_x$ is open in $A$ for all $x\in A$. Then $(A_x)_{x\in A}$ is an open cover of $A$. Then there is a finite subset $I \subseteq A$ such that $(A_x)_{x\in I}$ covers $A$. Pick $x_o \in I$ and Let $d := \max \{d(x, x_0) \mid x\in I\}\}$. Then $d(x_0, y) \le d(x_0, x) + d(y, x)$ where $y \in A_x$ for some $x\in I$. It follows that $d(x_0, y) \le d+1$ for all $y\in A$.

Answer 1

Your boundedness proof does not use nets at all and can also be simplified: let $A$ be compact and let $a_0 \in A$. Consider the open cover $\{\mathbb{B}(a_0,n)\mid n \in \Bbb N\}$ of $A$: it has a finite subcover and the one with the largest radius in it contains all other ones (the balls are increasing) so for one $N$ we have $A \subseteq \Bbb{B}(a_0,N)$ and we're done. (no triangle inequality etc.)

The closedness argument is fine: Let $(x_d)_{d \in D}$ be a net in $A$ converging to some $x \in X$ in $(X,\tau)$ and we want to show $x \in A$. By compactness of $(A,\tau_A)$ for some $a_0 \in A$ and some subnet $(x_{\phi(i)})_{i \in I}$ we have $x_{\phi(i)} \to a_0$ in $(A,\tau_A)$ and so also (lemma) $x_{\phi(i)} \to a_0$ in $(X,\tau)$. As subnets of a convergent net have the same limits, $x_{\phi(i)} \to x$ in $(X,\tau)$ as well and by $T_2$ ness of $X$, $a_0=x$ and so $x \in A$ as required.

This proof is not noticably simpler than the standard proof using a cover and for a beginning student the latter proof is probably more insightful.

Answer 2

To me it is ok, but since you are in a metric space you could use sequences as well to characterize the closure of your set.

If instead you wanted to do it because you wanted to exercise with nets (whose convergence is sometimes called Moore-Smith convergence), then all you have to do is to mimick the sequence proof as you did.