Bounded and closed set, but not compact

Question

What kind of set if bounded and closed but not compact?

Edit for re-openers: re-tagged as big-list.

Answer 1

If we are in a topological vector space (say, Banach), it has to be infinite-dimensional. In subsets Euclidean spaces, closedness and boundedness are together equivalent to compactness.

Answer 2

By the Heine Borel theorem, you will need an example of a topological space that isn't simply a closed subset of $\Bbb R^n$ with the usual topology.

One example is $X = (0,1)$ with the topology inherited from $\Bbb R$. In this example, $X$ is itself closed and bounded within $X$, but $X$ is not a compact space.

Another option is $\Bbb R$ with the discrete metric. Any subset is closed, but any infinite subset fails to be compact.

Another classic example is $\Bbb R$ with the metric $$ d(x,y) = \min\{|x - y|,1\} $$ In this case, every subset is bounded, but not every subset is compact.

Answer 3

Look at the sequence space $\ell^\infty$ of bounded sequences with the norm $$\|\|x\|\|_\infty = \sup_n |x_n|.$$

Let $e_n$ be the element of $\ell^\infty$ that is $1$ at $n$ and $0$ otherwise. The set of these is bounded. It is not compact because it is discrete and infinite.

Answer 4

Take $\mathbb R$ with the discrete topology. It is a metrizable space with the discrete metric induced from the topology. Then the set $\mathbb R$ is a closed subset (since the whole space is always closed). This set is also bounded because the distance between every 2 points in it is never more than 1. Take the open covering of $\{\{x\},x\in \mathbb R\}$. Can you find a finite sub covering of it?

Answer 5

Take any metric space $X$ that is bounded but not compact, e.g. $(0,1)$ with the metric of $\mathbb R$. Then $X$ itself is closed (in $X$) and bounded but not compact.

Alternatively, take any metric space $X$ that is not complete. A Cauchy sequence in $X$ that does not converge is a set that is bounded and closed but not compact.

Answer 6

The set of rational numbers $\mathbb Q$ with metric $d(p,q)=|p-q|$ has a subset $A=${$q\in \mathbb Q: 2\le q^2\le 4$}, which is closed and bounded but not compact due to the non-completeness of $\mathbb Q$.