I have a continuous function $f$ that is defined on a compact set. And $f(x_0)$ is its minimum. If I have a sequence $x_n$ such that $f(x_n)\to f(x_0)$, how can I show that $x_n\to x_0$? I tried proving by contradiction but I got lost. Anyone could help me? Thank you very much.
Update:$f$ only achieves minimum at $x_0$.
With the additional assumption the answer is yes. Assume the contrary, that $x_n\not\to x_0$. Since the space is compact, $x_n$ must have a convergent subsequence, say $x_{n_k}\to a$ as $k\to\infty$, and as we assume $x_n$ is not convergent, it must have another convergent subsequence, say $x_{n_j}\to b\not=a$ as $j\to\infty$. By continuity we will have $f(a)=f(x_0)=f(b)$ that is the minimum is achieved in both points $a$ and $b$, contradicting that $f$ only achieves its minimum at a single point.
Edit. I assumed that the compact domain is a subset of Euclidean space, and hence the Bolzano Weierstrass theorem applies (the link does not properly format, you may need to google Bolzano Weierstrass), namely every bounded sequence in $\mathbb R^n$ has a convergent subsequence. In other words I assume the domain is sequentially compact. There are compact spaces that are not sequentially compact. At any rate, a compact space is countably compact and limit point compact, that is, every sequence has a limit point. (In general this is not the same as the limit of a sequence: A sequence may have more than one limit point.) Hence the sequence $x_n$ (as above) must have a limit point $a$. Then $f(a)$ is the minimum, and since $f$ only achieves minimum at $x_0$ we have $a=x_0$. So the sequence $x_n$ has a unique limit point $x_0$ and it follows that $x_n\to x_0$. This may look like the same argument I wrote first, but in this version I need not pick convergent subsequences. At any rate, I felt the original version using convergent subsequences was more appropriate (even formally incomplete), and in addition you may follow the above links to find out about different versions of compactness.
