Past exam of the options II course of the University of Oxford / 2010 / E / E4

A subset $K$ of a topological space is compact if every cover of $K$ by open subsets of the ambient space has a finite subcover. This is equivalent to compactness of $K$ with its subspace topology.

A subset $B$ of a metric space is bounded if it lies in some ball of finite radius: there are $x_0$ in the space and $R<\infty$ with $d(x,x_0)\le R$ for every $x\in B$. For a nonempty subset, this is equivalent to finite diameter, ${\sup }_{x,y\in B}d(x,y)<\infty$. The empty subset is bounded by convention.

In an arbitrary topological space, compact need not imply closed. In a two-point indiscrete space, a singleton is compact but not closed. Closed need not imply compact either: $\mathbb{R}$ is closed in itself but the cover $\{(-n,n):n\ge 1\}$ has no finite subcover. One general theorem that does hold is that a closed subset $F$ of a compact space $K$ is compact: adjoin $K\setminus F$ to any relative open cover of $F$, take a finite cover of $K$, and discard that additional set.

In a Hausdorff space, every compact subset $K$ is closed. For $y\notin K$, separate $y$ from each $x\in K$ by disjoint open sets $V_x$ about $y$ and $U_x$ about $x$. Finitely many $U_x$ cover $K$; the intersection of their corresponding $V_x$ is a neighborhood of $y$ disjoint from $K$. Thus the complement of $K$ is open. The converse still fails, as the real-line example shows.

In every metric space, compact subsets are bounded. Cover a nonempty $K$ by unit balls centered on $K$ and take a finite subcover; the triangle inequality bounds all points relative to any one of the finitely many centers. The converse fails even for closed bounded subsets of general metric spaces. An infinite set with discrete metric $d(x,y)=1$ for distinct points is closed and bounded in itself, but its cover by singleton balls has no finite subcover. Thus compact implies closed and bounded in a metric space, while closed and bounded need not imply compact.

By the Heine-Borel theorem, a subset of $\mathbb{R}$ is compact exactly when it is closed and bounded. Consequently the compact intervals areincluding singleton intervals. The empty interval is compact too, if included in the interval convention. Nonempty open, proper half-open, and unbounded intervals are not compact.

For any topological target $Y$, $f(K)$ is compact. Pull an open cover of $f(K)$ back by $f$, take a finite subcover of $K$, and map the covering statement forward. Without a separation assumption on $Y$, the image need not be closed, as a constant map into a nonclosed singleton of an indiscrete space shows.

If $Y$ is a metric space, $f(K)$ is also closed in $Y$, bounded, totally bounded, complete with its inherited metric, and sequentially compact. Closedness uses the Hausdorff theorem from part (iii); total boundedness follows by taking finite subcovers of the covers by balls of any fixed radius. Compact metric spaces are sequentially compact, and a Cauchy sequence with a convergent subsequence converges to the same limit, proving completeness. These properties are asserted for the image subset, without requiring the whole ambient metric space to be complete.

A sequence $(x_n)$ in a metric space is a Cauchy sequence ifThis compares sufficiently late terms with each other; it does not assume the existence of a limit in the space.

A metric space is complete if every Cauchy sequence in it converges to a point of that same space. A limit in a larger ambient completion is insufficient for this definition.

Yes. A Cauchy sequence in the supremum norm converges in the given complete space $B([0,1])$ to a bounded function $f$. Supremum-norm convergence is uniform convergence. The uniform limit of continuous functions is continuous: for a point $x_0$ and tolerance $ϵ$, choose a continuous $f_n$ within $ϵ/3$ of $f$ uniformly, then use its continuity at $x_0$ and the triangle inequality. Hence $f\in C([0,1])$. Thus $C([0,1])$ is a closed, complete subspace of $B([0,1])$.

Yes. Continuity and nonvanishing on the compact interval give constantsConsequentlyA $d_{\alpha }$-Cauchy sequence is therefore supremum-norm Cauchy and has a continuous uniform limit $f$ by part (i). The upper bound then makes it converge to $f$ also in $d_{\alpha }$. Equivalently, multiplication by $\alpha$ is an isometry from this weighted space onto ordinary $C([0,1])$, since division by the continuous nonvanishing $\alpha$ gives its inverse.

This is completeness of a continuous weighted uniform norm.