Metric spaces where bounded closed subsets are compact 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)?
 A: 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".
