0
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Finding a relationship between $Y_{\varepsilon + \delta}$ and $(Y_\varepsilon)_\delta$ helps. $\endgroup$ – Daniel Fischer Jul 5 '15 at 19:38
  • 1
    $\begingroup$ I have attempted to transcribe your images into the "text" format allowed here. Please see this page for further information about this. Also, please ensure that I didn't unintentionally alter the meaning of your question. $\endgroup$ – user642796 Jul 5 '15 at 19:43
  • $\begingroup$ FYI, this metric is known as the Hausdorff metric. $\endgroup$ – Theo Bendit Jul 5 '15 at 20:07
3
$\begingroup$

HINT: Use the fact that $d$ satisfies the triangle inequality to show that if $A\subseteq B_\epsilon$, and $B\subseteq C_\delta$, then $A\subseteq C_{\delta+\epsilon}$.

$\endgroup$
  • $\begingroup$ Ok I can prove this result, but I don't see how it helps towards the required result. $\endgroup$ – Extremal Jul 5 '15 at 20:09
  • 1
    $\begingroup$ @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. $\endgroup$ – Brian M. Scott Jul 5 '15 at 20:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.