Prove if A is compact then it is closed and bounded using definition that A is compact if every sequence in A contains a convergent subsequence. Suppose A is contained in the metric space (M,d). Prove that if A is compact, then it is closed and bounded. I was wondering if there was a way of proving this using the definition that A is compact if every sequence ${x_n}$ in A contains a subsequence that converges to an element of A. I do not want to use the finite subcover definition of compactness if this is possible.
 A: Let's get all the definitions on the table:


*

*If $(M,d)$ is a metric space, then $A \subset M$ is a closed subset of $M$ iff every sequence that lies in $A$, and converges in $M$, converges in $A$. That is, if $a_{n} \in A$ for all $n$ and $a_{n} \to a$, then $a \in A$.

*$A \subset M$ is bounded if there exists a positive real number $R$ and an "origin" $x \in M$ such that for all $a \in A$, $d(x,a)<R$. That is, $A$ is contained in the ball of radius $R$ centred at $x$.  

*A metric space $M$ is compact iff every sequence in $M$, ($a_{n} \in M$ for all $n$) has a subsequence $a_{n_{k}}$ which converges in $M$. A subset $A \subset M$ is compact iff the metric subspace $(A,d)$ (where the metric $d$ is restricted to $A \times A$) is compact. This is equivalent to your characterisation.


Alright, so let $(M,d)$ be a metric space and $A \subset M$ a compact subset. Let's try to show $A$ is a closed subset of $M$. Well, let $a_{n}$ be a sequence in $A$, which converges in $M$. Say $a_{n} \to a \in M$. Since we don't know $A$ is closed, we can only guarantee that the limit is in $M$ (since that is the definition of convergence in $M$). What else do we know? Well, since $A$ is compact, we can take a subsequence $a_{n_{k}}$ which also converges, say to some limit $b \in A$. Now, $a_{n_{k}}$ tends to both $a$ and $b$, but limits of sequences in metric spaces are unique. So $b=a$. Since we know $b \in A$, we also have that $a \in A$. So $A$ is closed.  
In order to show $A$ is bounded, let's suppose that it isn't. What does that mean? It means that for every $x \in M$ and every $R > 0$,  there exists an $a \in A$ such that $d(x,a)>R$. This is supposed to contradict the compactness of $A$, so let's try to build a sequence in $A$ with no convergent subsequence. Let $a_{1} \in A$ be anything you like (we can certainly pick something, as otherwise $A$ would be vacuously bounded!). Since $A$ isn't bounded, we know there have to be elements of $A$ that are at least $1$ away from $a_{1}$, otherwise the ball of radius $1$ with centre $a_{1}$ would contain $A$. So let $a_{2}$ be any such element. Now, it's tempting to say "let $a_{3}$ be at least $1$ away from $a_{2}$", but that need not imply it is far away from $a_{1}$! Indeed, we could then choose precisely $a_{1}$, and we would not get a contradiction at all. Instead, let's choose $a_{3}$ far away from both of them. Let $a_{3}$ be at least $d(a_{1},a_{2})$ away from $a_{1}$. Then it's certainly at least $1$ away from $a_{1}$, and by the triangle inequality
$$d(a_{3},a_{2})+d(a_{2},a_{1}) \ge d(a_{3},a_{1}) \ge d(a_{2},a_{1})+1$$
So that $d(a_{3},a_{2}) \ge 1$. Similarly, choose a sequence $a_{n}$ in $A$ by always choosing $a_{n+1}$ to be at least $\max_{i}d(a_{1},a_{i})+1$ away from $a_{1}$. Then, for all $n \in \mathbb{N}$, and $m<n$, we have $d(a_{n},a_{m})>1$. So there is no hope of $a_{n}$ having a convergent subsequence, and therefore $A$ is not compact, a contradiction. 
