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Let $X$ be a bounded metric space. Let $Y$ be a subspace of $X$. Prove that $Y$ is bounded and that $\operatorname{diam}(Y) \le \operatorname{diam}(X)$.

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Is this (homework)? What did you try? – martini Apr 20 '12 at 16:03
Hint: If there is a number $r$, such that $d(x,y) \le r$ for all $x$ and $y$ in $X$, then $d(x,y) \le r$ for all $x$ and $y$ in $Y$. – David Mitra Apr 20 '12 at 16:04
This is the sort of question that textbooks ask only to check that you have read and understood the definitions. You should (re)read the definitions and think about what they mean, and then the answer to the question should be obvious. The point of this remark is that we cannot do your learning for you; you have to do that yourself. – Harald Hanche-Olsen Apr 20 '12 at 16:55
What I had as my answer was the following: If x,y %\in% X then d(x,y) < diam(X) and – Nicky Apr 20 '12 at 18:45
What I had as my answer was the following: "If X is bounded then $d(x,y) \le K$ for some $K \ge 0$ and $sup[d(x,y)] = diam(X)$ so $ d(x,y) \le diam(X)$. Since Y is a subset of X $sup[d(x,y)] \ge sup[d(a,b)]$ for some $a,b \in Y$ and hence Y is bounded by diam(X). Then, be definition, $diam(X) \ge diam(Y)$." I am just unsure whether or not this is correct since it is not homework, I am studying for my exam, but there are no answers available to me. – Nicky Apr 20 '12 at 19:38
up vote 0 down vote accepted


  1. Recall what the diameter of a bounded metric space is. This should involve the notion of $\sup$ (supremum), so recall what that is.

  2. As you said in a comment, conclude that for any pair of points in $Y$ the distance between them is bounded by the diameter of $X$.

  3. Once again, use the definition of the diameter, and conclude that the $diam(Y)$, as the supremum of pairwise distances in $Y$, is bounded by the diameter of $X$.

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