# A metric on the set of closed bounded subsets of a metric space

Define the distance from a point $$p$$ in a metric space $$(X,d)$$ to a subset $$Y \subset X$$ by $$d(p,Y) := \inf \{ d(p,y) : y \in Y \}.$$ For any $$\varepsilon > 0$$, define $$Y_\varepsilon := \{ x \in X : d(x,Y) \leq \varepsilon \}.$$

[G]iven any two bounded sets $$A,B \subset X$$, define $$d_S (A,B) = \inf \{ \varepsilon > 0 : A \subset B_\varepsilon\text{ and }B \subset A_\varepsilon \}.$$

1. Show that $$d_S$$ yields a metric on the set of closed bounded subsets of $$X$$.
2. Show that $$d_S$$ fails to do so on the set of bounded subsets of $$X$$.

Regarding part (1), I proved the first 2 properties of a metric, but failed to prove the third, which is the triangle inequality. Can somebody help me by giving a hint?

• Finding a relationship between $Y_{\varepsilon + \delta}$ and $(Y_\varepsilon)_\delta$ helps. – Daniel Fischer Jul 5 '15 at 19:38
HINT: Use the fact that $d$ satisfies the triangle inequality to show that if $A\subseteq B_\epsilon$, and $B\subseteq C_\delta$, then $A\subseteq C_{\delta+\epsilon}$.
• @EpsilonDelta: It implies that if $d_S(A,B)<\epsilon$ and $d_S(B,C)<\delta$, then $d_S(A,C)<\epsilon+\delta$, from which it’s a fairly small step to the desired result. – Brian M. Scott Jul 5 '15 at 20:14