# Distance from a proper subspace of a metric space.

Let $(X,d)$ be a metric space. And $A \subset X$ be a proper closed subspace. Proper here implies every closed and bounded ball is compact. Define $d_A(x)= inf_{a \in A} d(x,a)$. Show that this infimum is actually a minimun that is there is a $a_1\in A$ such that $d_A(x)=d(x,a_1)$.

Please give some idea for A being compact it is clear. If the statement is wrong please give a counter example.

Let $\epsilon >0$ and $a\in A$ so that $d(x, a) \leq d(x, A) + \epsilon$. In particular,

$$\tilde A= A \cap B_x(d(x, A) + \epsilon)$$

is nonempty, closed and bounded. Also $d(x, A) = d(x, \tilde A)$.

• Sorry I think I got the point that $d(x, A) = d(x, \tilde A)$ are equal thanks.
– GGT
Commented Nov 11, 2014 at 13:45

Suppose $A\subseteq X$, with $A$ compact. We wish to show that for fixed $x$, there exists $a_0\in A$, such that $d(x,a_0)=\inf_{a\in A} d(x,a)$.

We know that $f:A \to\mathbb{R}$ such that $f(a)=d(x,a)$ is continuous. By the compactness of $A$ it has an infimum which is attained at some point.

Suppose $A$ is proper. Let $\inf_{a_\in A} d(x,a)=r$. Let $B_t(x)$ be the ball of radius $t$ centred at $x$. We know that $C_i=B_{r+\frac{1}n}(x) \cap A$ is non empty. Choose any $a_i\in C_i$. Clearly $\overline{\{a_i\}}$ is compact and $\inf_{i}d(x,a_i)=r$, allowing us to repeat the arguments for the compact case.

• I mentioned for A being compact it is very clear. I want the proof for A being proper.
– GGT
Commented Nov 11, 2014 at 10:27
• Well you have to show that $d(x,C_i)=d(x,A)$ ?
– GGT
Commented Nov 11, 2014 at 13:29