# Question on metric space

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?

• Infs always exist. (For negative reals you may need $-\infty$ to be an allowable value of the $\inf$ but that isn't an issue here.) Oct 22, 2022 at 17:41
• its bounded below by 0, since X is a metric space Oct 22, 2022 at 17:41

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.