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Is every subset of a metric space a metric subspace? A simple proof does justify that all are subspaces, still, wanted to know if I missed something.

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To be completely precise, all subsets of a metric space equipped with the induced metric are metric subspaces. You could also equip the subsets with other metrics, and then they wouldn't be metric subspaces.

(Strictly speaking you could also not equip them with a metric at all, and then they aren't metric subspaces because they aren't even metric spaces).

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  • $\begingroup$ Yes, exactly what I meant! Thanks a lot! I'm trying to find if there's any criteria to find if a normal space has a (not- closed) subspace which is not normal, now I can rule out metric spaces. $\endgroup$ – Jesse P Francis Oct 25 '12 at 14:29
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All the definition of a metric subspace of $(X,d)$ is is a subset $Y\subset X$ together with the metric $d|_{Y\times Y}$. So the subsets themselves aren't metric subspaces, but they can be made into subspaces by giving them a restriction of $d$ as a metric.

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    $\begingroup$ Technically it's $d|_{Y\times Y}$ rather than $d|_Y$. $\endgroup$ – kahen Oct 25 '12 at 14:30
  • $\begingroup$ Yeah sorry, that's what I meant! $\endgroup$ – user123123 Oct 25 '12 at 14:31
  • $\begingroup$ @user123123 you answer, is for me the useful one. Thanks! $\endgroup$ – R. Ferreira Jul 21 '17 at 2:39
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Let $(x,p)$ be metric space and letting $Y$ be a non empty subset of $X$. Define the function $§$ on $Y$. $Y$ by $§(x,y)=p(x,y)$ for all $x,y$ in $Y$.Then $(Y,§)$ is also a metric space called a subspace of the metric space $(x,p)$.

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