Find an example of a metric space \(X\) and a subset \(E \subseteq X\) such that \(E\) is closed and bounded but not compact.
Solution
1. Example: \(X = (0, 1)\) with the usual metric \(d(x,y) = |x - y|\), and \(E = (0, 1) = X\) itself.
explicit space and subset.
2. \(E\) is closed in \(X\).
\(E = X\) is the whole space, and the whole space is always closed (its complement \(\varnothing\) is open).
3. \(E\) is bounded.
\(E \subseteq B(x_0, 1)\) for any \(x_0 \in X\) (the diameter of \((0,1)\) is \(1\); e.g. \(E \subseteq B(1/2, 1)\)).
4. \(E\) is not compact.
the sequence \(x_k = 1/(k+1) \in E\) has no convergent subsequence in \(X\): it converges to \(0 \notin X\), and any subsequence also converges to \(0\), so no subsequence converges to a point of \(X\).
(Compactness of metric spaces is equivalent to sequential compactness.)
5. Q.E.D.
<1>2–<1>4 show closed + bounded \(\not\Rightarrow\) compact in general metric spaces; compactness needs completeness (or the ambient space must be \(\RR^n\)). Alternative example: \(X = \QQ\), \(E = \QQ \cap [0,1]\) (closed and bounded in \(\QQ\), not compact).