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It's well known that metric spaces like $\mathbb{R}^n$ possess a property that every bounded closed subsets are compact whenever $n<\infty$. I wondered whether there is a certain metric spaces where every bounded closed subsets are compact. Or, furthermore, whether every such spaces can be embedded into some $\mathbb{R}^n$ (isometrically)?

Jon Snow
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    What do you mean by "I wondered whether there is a certain metric spaces where every bounded closed subsets are compact." ? $\mathbb R^n$ is an example of this(as you yourself noted). – math-physicist Jul 12 '22 at 03:12
  • @sadman-ncc I think it's clear that the OP is asking for other examples besides those (and besides subsets of $\mathbf R^n$). – KCd Jul 12 '22 at 03:16
  • Do you already know that every closed subset of a compact metric space is compact? – Moishe Kohan Jul 12 '22 at 03:25
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    The space of analytic functions on a region in $\mathbb C$ is an example of metric space (with a suitable metric that metrizes uniform convergence on compact subsets) in which closed bounded sets are compact. (Ref. Montel's Theorem). There are no normed linear spaces of infinite dimension which have this property. – geetha290krm Jul 12 '22 at 05:03

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In every metric space, compact subsets are closed and bounded, so you're asking about metric spaces where compactness is equivalent to closedness + boundedness.

That compact = closed and bounded in each $\mathbf R^n$ is true essentially because it is true in $\mathbf R$. Another example of a metric space where compact = closed and bounded is $\mathbf Q_p$, the field of $p$-adic numbers for a prime $p$. This, like $\mathbf R$, is a completion of $\mathbf Q$ for some absolute value and it is locally compact. In the product space $\mathbf Q_p^n$, compact = closed and bounded, so it is similar to $\mathbf R^n$. You can't isometrically embed $\mathbf Q_p$ into $\mathbf R$ in any useful way. (As a field, $\mathbf Q_p$ does not embed into $\mathbf R$ because some negative integers are perfect squares in $\mathbf Q_p$.)

Boundedness of a metric is a less robust property than it may appear: for every metric space $(X,d)$, we can change the metric to a bounded metric without changing the notion of convergent sequences: the function $d'(x,y) = \min(d(x,y),1)$ is a metric on $X$ and $d'(x,y) = d(x,y)$ when $d(x,y)$ is small, so the notions of convergent sequence, open subset, closed subset, and compact subset is the same on $X$ using either metric $d$ or $d'$. When a metric $d$ is bounded, compact = closed + bounded exactly when compact = closed, and that's true exactly when the metric space itself is compact (not very interesting).

A good analogue for general metric spaces of the property "compact = closed + bounded" for subsets of $\mathbf R^n$ is that for subsets of a general metric space, "compact = complete + totally bounded". (You can look up the definition of a totally bounded metric space on Wikipedia. A subset of a metric space is totally bounded when it is totally bounded as a metric space in its own right. Or look at a related MSE page here.) In a complete metric space, a subset is complete if and only if it is closed, so for subsets of a complete metric space, "compact = closed + totally bounded".

KCd
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