Non-compact subsets of a metric space $(X,d)$.

Question

I'm trying to come up with an example of a metric space $(X,d)$ such that a subset $A \subset X$ is not compact, but is closed and bounded.

Essentially I want to find an example that shows that a subset of a metric space is compact iff it is closed and totally bounded.

Answer 1

$X=(0,1)$ with Euclidean metric. Then $X$ itself is closed and bounded yet not compact.

Answer 2

HINT: Start with any non-compact metric space, and define a new metric $d'$ by setting $d'(x,y)=\min\{d(x,y),1\}$ for $x,y\in X$; $d'$ generates the same topology as $d$.

Answer 3

Consider $X = \mathbb{Z}$ with a metric $d$ such that $d(m, n) = 1$, if $m \neq n$ and $d(n, n) = 0$. $d$ is a metric.

Take any infinite subset $A$ (for example, $\mathbb{N}$). Then $A$ is open (and closed) and it is also bounded (because the maximum distance is 1).

Moreover, $A$ is not compact, because the family of sets defined by $\{\{x\} \mid x \in A\}$ is an open cover of $A$, but you there isn't finite subcover, because $A$ is infinite.