Let \(X\) be a metric space.
Suppose that for some \(\epsilon > 0\), every \(\epsilon\)-ball in \(X\) has compact closure. Show that \(X\) is complete.
Suppose that for each \(x \in X\) there is an \(\epsilon > 0\) such that the ball \(B(x, \epsilon)\) has compact closure. Show by means of an example that \(X\) need not be complete.
Solution
Let \((x_n)\) be a Cauchy sequence. Choose \(N\) such that \[ d(x_n,x_N)<\varepsilon/2\qquad(n\ge N). \] Then the tail of the sequence lies in \(B(x_N,\varepsilon)\), whose closure is compact by hypothesis. Hence the tail has a convergent subsequence \(x_{n_k}\to x\). Since the original sequence is Cauchy, the whole sequence converges to \(x\): given \(\eta>0\), choose \(N_1\) so \(d(x_m,x_n)<\eta/2\) for \(m,n\ge N_1\), then choose \(k\) with \(n_k\ge N_1\) and \(d(x_{n_k},x)<\eta/2\). For \(n\ge N_1\), \[ d(x_n,x)<\eta. \] Thus every Cauchy sequence converges and \(X\) is complete.
Take \[ X=(0,1) \] with the usual metric. For each \(x\in(0,1)\), choose \[ 0<r<\min\{x,1-x\}. \] Then the closure in \(X\) of \(B(x,r)\) is the ordinary closed interval \([x-r,x+r]\), which is compact. Thus every point has a ball with compact closure.
Nevertheless \(X\) is not complete: the sequence \(x_n=1/n\) is Cauchy in \(X\) but has no limit in \(X\). Hence the local hypothesis in (b) does not imply completeness.