Let $X$ be a compact metric space, show that a continuous function $f:X\rightarrow\mathbb{R}$ attains a maximum and a minimum value on $X$.
Attempt: So the important thing is that I have previously shown that such a function is bounded and that for compact $X$, $f(X)$ is compact given $f$ continuous. In $\mathbb{R}$, compact $\implies$ closed and bounded. So $f(X)$ is closed and contains its accumulation points, and it is bounded so $\exists \sup(A),\inf(A)$ and since closed $\implies \sup(A)\in A, \inf(A)\in A$.
Did I miss anything/make an unwarranted leap of logic?
Best Answer
Here’s an example of how the same argument could be written up nicely.