Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 4 / 12E (Analysis II)

Let $(x_n)$ be a Cauchy sequence in a compact metric space $X$. By sequential compactness of a compact metric space, it has a convergent subsequence $x_{n_j}\to x\in X$. Given $\epsilon >0$, choose $N$ such that $d(x_m,x_n)<\epsilon /2$ whenever $m,n\ge N$, and then choose $j$ such that $n_j\ge N$ and $d(x_{n_j},x)<\epsilon /2$. For every $n\ge N$, the triangle inequality givesThus the whole sequence converges to $x$. Every Cauchy sequence converges in $X$, so $X$ is a complete metric space.

No. Give an infinite set $X$ the discrete metricThis space is bounded. Every Cauchy sequence is eventually constant, so it is complete. However, the open cover by the singleton balls $B_{1/2}(x)=\{x\}$ has no finite subcover. Hence $X$ is not compact.

For any $\epsilon >0$, the balls $\{B_{\epsilon }(x):x\in X\}$ form an open cover of the compact metric space $X$. Compactness supplies a finite subcoverThis is exactly total boundedness.

Let $(x_n)$ be any sequence in a complete, totally bounded metric space $X$. Cover $X$ by finitely many balls of radius $2^{-1}$; one contains infinitely many terms. From those terms choose an infinite subsequence lying in one ball of radius $2^{-2}$, and continue. The diagonal argument produces a subsequence $(x_{n_j})$ whose tail from its $j$th term onward lies in a ball of radius $2^{-j}$. Hencewhenever $k,\ell \ge j$, so the subsequence is Cauchy. Completeness makes it converge in $X$.

Every sequence in $X$ therefore has a convergent subsequence, so $X$ is sequentially compact. Sequential compactness is equivalent to compactness for metric spaces, and thus $X$ is compact. This proves that a complete totally bounded metric space is compact.

The space is not compact because it is not complete. For $n\ge 2$, define the continuous functionThese functions converge in the L1 norm to the step function $h=1_{(1/2,1]}$, because $f_n-h$ is supported on an interval of length $2/n$ and bounded in absolute value by one. They are consequently Cauchy for the metric $d$, since for sufficiently close pairs the outer minimum with $1$ does nothing.

If $(f_n)$ converged in this metric to some $f\in C[0,1]$, it would also converge to $f$ in $L^1$. Uniqueness of an $L^1$ limit would give $f=h$ almost everywhere. Continuity would then force $f=0$ on $[0,1/2)$ and $f=1$ on $(1/2,1]$, which is impossible at $1/2$. Thus this Cauchy sequence has no limit in $C[0,1]$, while every compact metric space is complete.