Compactness Proof I have a doubt respect this theorem
A compact subset $M$ of a metric space is closed and bounded.
Proof by my lecture:
For every $x\in \bar{M}$ there is a sequence $(x_n)$ in $M$ such that $x_n\longrightarrow x \cdots$. My question is Why this?. I know that if $x$ is a boundary point then exists $x_n\longrightarrow x$. But How I will be able to prove when $x$ is not boundary point?
 A: There will be a sequence $( x_n )$ such that $x_n \to x$ simply because if $x$ is not a boundary point, then it is in $M$, and we have the trivial sequence $x_n = x$.
If $x \notin M$, then $x$ is a boundary point.  Then we can select $x_n \ne x$ such that $d(x_n,x) < \frac{1}{n}$.  That this sequence should converge to $x$ is a consequence of the definition of convergence and the Archimedean property of the reals.
A: A metric space is always Hausdorff, and compact sets in Hausdorff spaces are closed. Alternatively, devise a proof using the equivalent sequential compactness: every sequence in $K$ has a convergent subsequence that converges to an element in $K$. Relate this to limit points in metric spaces, which are always limits of sequences in the said set.
Pick $x\in K$, $K$ a compact set. Note that the collection $$\mathscr B=\{B(x;\epsilon):\epsilon >0\}$$ is an open cover. By compactness, there exists a finite cover, $B(x,\epsilon_1),\dots,B(x;\epsilon_r)$. But then $$K\subseteq B(x,\epsilon)$$ where $\epsilon=\max\limits_{1\leq i\leq r} \{\epsilon_i\}$.
Alternatively, using the equivalent sequential compactness and assuming unboundedness you can find a sequence such that $d(x_k,x_{n+1})\geq n$ for $k=1,2\dots,n$ whence no convergent subsequence will exist.
A: Let $X$ be a topological space. The question raised above can be stated more generally as the following lemma.
On limit points:
Let $A\subset X$. If there is a sequence $\{x_n\} \subset A$ that converges to $x$, then $x$ is a limit point of $A$. The converse holds if $X$ has a countable basis at $x$, that is a countable collection of nbhd $\{U_n\}$ of $x$ such that any nbhd $U$ of $x$ contains at least one $U_n$. For then we may form the sequence $\{x_n\}$ by taking a point $x_n \in U_1\cap ...\cap U_n$, that is easily verified to converge to $x$.
On continuous functions:
Let $f: X\rightarrow Y$. If $f$ is continuous, then for every convergent sequence $x_n\rightarrow x$ in $X$, the sequence of images $f(x_n)\rightarrow f(x)$. The converse holds if $X$ has a countable basis at every point.
This result goes by the name of the sequence lemma (coined by James Munkres). A space that has a countable basis at each point is called first-countable. Sequences are adequate to detect limit points and continuity of functions in a first-countable space (which is not true in general). A metric space is always first countable for $\{B(x,1/n)\}$ is a countable basis at x. Metric spaces are very special and this is a particular example of the many properties they enjoy.
