Question on metric space

Question

Let $(X,d)$ be a metric space, $A\subset X$ be closed and pick $y\in X-A$, then we define $d(y,A)=\inf\{d(x,y):x\in A\}$. Dumb question, how do we know the inf here is defined? Closed set doesn't necessarily mean bounded, doesn't that mean the inf does not necessarily exist?

Answer 1

You're right that a closed set is not necessarily bounded, of course. But in this case the relevant set to look at is not $A\subset X$, and you're right that $\operatorname{inf}A$ would not exist (actually, there isn't even a sensible way to say what this even means, since infimum is not something that can be defined in this context) but what is relevant is that $\operatorname{inf}\{d(x,y)\mid x\in A\}$ exists. The set $\{d(x,y)\mid x\in A\}$ is, in fact, bounded below by zero simply by definition of a metric (it measures "distance" and is always nonnegative). Any subset of $\mathbb R$ bounded below has an infimum.