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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?

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    $\begingroup$ 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.) $\endgroup$ Oct 22, 2022 at 17:41
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    $\begingroup$ its bounded below by 0, since X is a metric space $\endgroup$
    – Chris
    Oct 22, 2022 at 17:41

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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.

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