# Distance from an element to a subset

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?

• 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 9 at 18:09

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