A is a subset of a metric space under given conditions. Prove there is a $y \in A$ that minimizes the distance from $x$ to elements of $A$ Say I have a metric space $(M, d)$ and know that $A$ is a non-empty sequentially compact subset of $M$.
Let $x \in M$. Then I'm trying to prove that there must exist a point $y \in A$ such that $d(x, y)$ = inf$\{d(x,z) : z \in A\}$.
(Basically, that there is a $y \in A$ that minimizes the distance from $x$ to elements of $A$)
Sequential compactness of the subset of a metric space implies that A is also compact.
We know that a compact subset of a metric space must be closed and bounded.
Hence the locations of all points in $A$ can be found. So we know that $y\in A$ will also be definable within the known boundaries. Still not sure how to prove that this $y$ actually exists, though.
Thoughts on a formal proof for this?
 A: Set $M = \inf\{d(x,z):z\in A \}$
By definition there exists $y_{j}\in A$ so that
$$M\leq d(x,y_{j}) \leq M+\frac{1}{j}$$
There is a subsequence $y_{n_{1}},y_{n_{2}}...$ converging to some $y \in A$ due to sequential compactness (here $n_{1}<n_{2}<...$). W.L.O.G assume that (if not pick another subsequence) $d(y_{n_{j}},y)\leq \frac{1}{j}$ Now note that
$$d(x,y_{n_{j}}) \leq d(x,y)+d(y,y_{n_{j}})$$
Thus for all $j \in \mathbb{N}$
$$d(x,y) \geq M-\frac{1}{j}$$
Thus by taking $j \rightarrow \infty$
$d(x,y) \geq M$. Note also that
$d(x,y) \leq d(x, y_{n_{j}})+d(y_{n_{j}},y) \leq M +\frac{1}{n_{j}}+\frac{1}{j}$.
Thus by taking $j \rightarrow \infty$
$$d(x,y) \leq M.$$
Combining the two inequalities we have $d(x,y) = M$.
A: Take $l=\inf\{d(x,z):z\in A\}$ and $n\in \mathbb{N}$. Then $l+{1 \over n}$ is not a lower bound of $\{d(x,z):z\in A\}$. So there is some $z_n\in A$ such that $l\leq d(x,z_n)< l+{1 \over n}$. Since $A$ is sequentially compact, there is some subsequence $\{z_{n_k}\}$ of $\{z_n\}$ that converges to some point $z^*\in A$. Now $$l\leq d(x,z^*)\leq d(x,z_{n_k})+d(z_{n_k},z^*)\leq l+\frac{1}{n_k}+d(z_{n_k},z^*)$$ Taking $k \longrightarrow \infty$ yields $l\leq d(x,z^*)\leq l$ so $d(x,z^*)=l$ and we're done.
