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Show that if a set $A$ in a metric space is bounded, so is each subset $B \subseteq A$.

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First of all you put "hw" in the title, so why didn't you tag it with "homework"? And second of all it's very impolite to post in the imperative - you're not writing a textbook nor are you a teacher assigning homework – kahen Feb 27 '11 at 5:25
What does it mean for $A$ to be bounded? I.e., how would you write that out explicitly? – Jonas Meyer Feb 27 '11 at 5:26

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If $A$ is bounded then there exists $x \in M$ and $r>0$ such that for all $a \in A$, we have $d(x,a) <r$. So how would you show that each subset $B \subseteq A$ is bounded?

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Definition: A metric space $M$ is called bounded if there exists some number $r>0$, such that $d(x,y) \leq r$ for all $x, y \in M$.

Since you are assuming $A$ to be bounded then there is some number $k$ such that $d(x,a) \leq k$ for all $x,a\in A$. Since $B \subset A$ and this condition happens for all elements in $A$, therefore it should happen for elements in $B$ as well.

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