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Given a metric space $(X,d)$, an element $x_0 \in X$ an a non-empty subset $A \subset X$, we define the distance from $x_0$ to $A$ as $$d(x_0,A) := \inf\{d(x_0,a) : a \in A\}.$$

Now, in $\mathbb{R}$ with the usual distance we have $A := (1,2]$ and $x_0 := 1$, then $d(1,A) = 0.$

My question is, how you can take $a = 1 \notin A$ to conclude $d(1,1)=0$?

Is the reason that you consider $1$ is the lower limit of $A$? So you should define $d(x_0,A) := d(x_0,\inf A)$, right?

Andrés E. Caicedo
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user183002
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    No. Consider $A = {0}\cup{3}$ and $x_0=2$. Then $d(x_0,A) = 1$, but $d(x_0,\inf A) = d(2,0) = 2$. – amsmath Oct 09 '19 at 18:09

1 Answers1

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You take the infimum of the set $\{d(1,a) : a \in A\} = (0,1]$, which is of course $0$.

Klaus
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