TITLE: limits of a sequence in a compact metric space
QUESTION [0 upvotes]: I have a compact metric space $Y$, continuous real function $g$, and a point $\rho$ such that $g(\rho)$ = inf $g(y)$,  $ \forall y \in Y$.
If I have a sequence {$y_{n}$} $\in Y$  and  am given that $\lim_{n\to\infty} g(y_{n})$ = $g(\rho)$, how can I show that:
$\lim_{n\to\infty} y_{n}$ = $\rho$?

REPLY [2 votes]: It is not true, consider a constant function on a closed interval.
If $\rho$ is unique such point, then you have a sequnce of closed subsets $A_n=g^{-1}([g(\rho)-1/n, g(\rho)+1/n]$) and the uniqueness is the same as
$$
\bigcap A_n=\{\rho\}.
$$
For large $N$ for all $k>N$ $y_k$ lies in $A_n$, so the only possible limit point is $\rho$. As it is compact you always have a subsequence, convergeing to $\rho$ and so the whole sequence converges.

REPLY [1 votes]: No, suppose that $g$ is the constant function.
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