Distance between a point and a set

Question

The problem I'm trying to solve is

Prove that $d(a, B \cup C)$ is the smaller of $d(a,B)$ and $d(a,C)$ for a point $a$ and subsets $B, C$ of a metric space.

So I think what I need to show is that $d(a, B\cup C) = \min(d(a,B), d(a,C))$

It's obvious that $d(a, B\cup C) \le d(a,B), d(a,C)) $

So I suppose without loss of generality that $d(a, B)$ is less than $d(a, C)$ and try to show that $d(a, B) \le d(a,B\cup C)$ . This way I can show that the two are equal and I'm done.

I'm trying to prove this using the definition of infimum but have been unsuccessful so far. Any advice on the approach I should be taking? Maybe this isn't the right approach in the first place.

Answer 1

Hint: Try to prove $\inf (X\cup Y) = \min(\inf X, \inf Y)$ for $X,Y\subseteq\mathbb R_{\ge 0}$ first.

Answer 2

$\forall x \in B \quad d(a,B \cup C) \leq d(a,x)$ (to see this note that the definition of the distance on the LHS is the infimum $d(a,y)$ over $y \in B$ and note that $B \subseteq B \cup C$. Taking infima over $x \in B$ gives $d(a,B \cup C) \leq d(a,B)$.

Similarly $d(a,B \cup C) \leq d(a,C)$.

So we have $d(a,B \cup C) \leq \min \quad (1)$, where $\min$ stands for the minimum you mention above in your question.

Without loss suppose that $\min = d(a,C)$. Suppose for a contradiction that we have a strict inequality in $(1)$.Then either:

I $\quad$ for some $c \in C$ we have $d(a,c) < d(a,C)$, an obvious contradiction; or

II $\quad$ for some $b \in B$ we have $d(a,b) < d(a,C)$, contradicting our assumption that $\min = d(a,C)$.

So in either case we arrive at a contradiction, and the inequality must be an equality.