Past exam of the Probability, statistics and options II course of the University of Oxford / 2010 / E / E4

A subset $K$ is compact if every cover of $K$ by sets open in the ambient space has a finite subcover. Equivalently, every relatively open cover of the subspace $K$ has a finite subcover. These agree because relatively open sets are intersections with ambient open sets.

A subset $B$ of a metric space is bounded if it lies in a ball of finite radius: there exist a point $x_0$ and $R<\infty$ such that $d(x,x_0)\le R$ for every $x\in B$. Equivalently its diameter ${\sup }_{x,y\in B}d(x,y)$ is finite, with the empty set bounded by convention.

In any topological space a closed subset of a compact set is compact: add its open complement to an open cover and take a finite subcover of the compact set. However, compact subsets need not be closed in a general space. In the indiscrete two-point space a singleton is compact but not closed. Conversely a closed subset need not be compact: $\mathbb{R}$ itself is closed in $\mathbb{R}$ but the cover $\{(-n,n):n\ge 1\}$ has no finite subcover.

In a Hausdorff space every compact subset $K$ is closed. Given $x\notin K$, separate $x$ and each $k\in K$ by disjoint neighborhoods; compactness selects finitely many neighborhoods covering $K$, and the intersection of their companion neighborhoods of $x$ misses $K$. Thus the complement is open. Closedness still need not imply compactness, as the same real-line example shows. It does imply compactness if the ambient space is compact.

In a metric space compactness implies boundedness: finitely many radius-one balls cover the set, so their centers have a finite maximal distance from any fixed center and the triangle inequality gives a global bound. The converse is false. An infinite set with discrete metric $d(x,y)=1$ for $x\ne y$ is bounded, but its singleton cover has no finite subcover. It is even closed and complete, so closedness plus boundedness alone does not suffice in general metric spaces.

By the Heine–Borel theorem a subset of $\mathbb{R}$ is compact exactly when it is closed and bounded. Consequently the nonempty compact intervals are exactly $\boxed{[a,b](-\infty <a\le b<\infty )}$. This includes the degenerate interval $[a,a]$. The empty interval is compact as well; nonclosed bounded intervals and all unbounded intervals fail the criterion.

The image $f(K)$ is compact: pull any open cover back by the continuous map, use a finite subcover of $K$, then map forward. For a general topological target, closedness is not guaranteed; a continuous image into the indiscrete two-point space can be a nonclosed singleton. No connectedness assertion follows merely from compactness of $K$.

If $Y$ is metric, $f(K)$ is compact, therefore closed in $Y$, bounded and totally bounded. It is also complete in its induced metric and sequentially compact. For the latter properties, compact metric spaces have a convergent subsequence of every sequence; a Cauchy sequence with one convergent subsequence converges to that same limit. Total boundedness follows by selecting finitely many balls of each prescribed radius. These are properties of the image itself, irrespective of whether $Y$ is complete.

A sequence is Cauchy ifThe comparison is between sufficiently late terms, without presupposing a limiting point in the space.

A metric space is complete if every Cauchy sequence of its points converges to a point of that same space. The requirement on the location of the limit distinguishes, for example, $\mathbb{R}$ from its incomplete subspace $\mathbb{Q}$.

Yes. A Cauchy sequence of continuous functions in the supremum norm is Cauchy in the given complete space of bounded functions, so it has a uniform limit there. A uniform limit of continuous functions is continuous: at a point, use two uniform errors smaller than $\epsilon /3$ and continuity of one sufficiently late term for the remaining error. Thus the limit is in $C([0,1])$, proving completeness. Equivalently this is a closed subspace of the given complete space.

Continuity and nonvanishing on the compact interval implyHence $m\Vert f-g{\Vert }_{\infty }\le d_{\alpha }(f,g)\le M\Vert f-g{\Vert }_{\infty }$. A $d_{\alpha }$-Cauchy sequence is therefore uniformly Cauchy, so (i) gives a continuous uniform limit; the upper bound then gives convergence in $d_{\alpha }$. Thus $\boxed{(C([0,1]),d_{\alpha })\text{ is complete}}$. The positive lower bound is what makes the argument work.