Question:
Let $X$ be a bounded metric space. Let $Y$ be a subspace of $X$. Prove that $Y$ is bounded and that $\operatorname{diam}(Y) \le \operatorname{diam}(X)$.
Bounded subspaces and diameters.
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metric-spaces
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0What 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. – 2012-04-20
1 Answers
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Steps:
Recall what the diameter of a bounded metric space is. This should involve the notion of $\sup$ (supremum), so recall what that is.
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 $X$.
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 $X$.