Let $(X,d)$ be a metric space. If every subset of $X$ is bounded, does it mean that the space itself is bounded?
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Perhaps you meant every that every proper subset of $X$ is bounded. In that case, take any $A, B \subsetneq X$ such that $A \cup B = X$. Hence, $X$ is bounded as a finite union of two bounded sets.
Namely $$\operatorname{diam} X = \operatorname{diam}(A\cup B) \le \operatorname{diam} A + d(A, B) + \operatorname{diam} B < +\infty$$
mechanodroid
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Technically, this argument does not work if $X$ has just one point. – tomasz Dec 13 '17 at 01:54
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Maybe we can add: for finite $X = {x_1, \dots, x_n}$, we have $\operatorname{diam} X = \max {d(x_i, x_j) \mid i, j = 1, \dots, n}$ hence bounded. – Alex Vong Dec 13 '17 at 02:26
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@tomasz : In fact, it fails if $A$ or $B$ are empty. – Eric Towers Dec 13 '17 at 03:36
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@EricTowers: If $A$ or $B$ is empty and both are proper subsets of $X$, then $A\cup B\neq X$. What I meant however, is that you cannot choose such $A$ and $B$ in the first place if $X$ has just one point. – tomasz Dec 13 '17 at 17:51
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Of course. After all, $X$ is a subset of itself. On the other hand, if $X$ is bounded, every subset of $X$ is bounded too.
José Carlos Santos
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