Past exam of the Probability, statistics and options II course of the University of Oxford / 2008 / E (Topology)

The distance function $x\mapsto d(x,c)$ is continuous because the triangle inequality gives $|d(x,c)-d(y,c)|\le d(x,y)$. Hence $\overline{B}(c,r)=\{x:d(x,c)\le r\}$ is the inverse image of the closed interval $(-\infty ,r]$, so is closed. Since it contains $B(c,r)$, it also contains the closure of that open ball:

The condition is $\max (|x-1/2|,|y|)\le 1/2$. Therefore the ball is the closed squareIts four vertices are $(0,\pm 1/2)$ and $(1,\pm 1/2)$.

The induced-metric balls are intersections with the corresponding ambient balls. On the horizontal arm the distance from the centre is $|x-1/2|$; on the vertical arm it is $\max (1/2,y)$. ConsequentlyThe closure relative to $E$ of the open ball is $[0,1]\times \{0\}$. Every point $(0,y)$ with $y>0$ has positive distance $y$ from that horizontal arm, so none of those points is in the closure. In particular $(0,1/4)$ lies in the closed ball but not in the closure of the open ball, proving strict inclusion.

Let the radius be $r>0$. One inclusion follows from (a). For the reverse, if $\Vert x-c\Vert \le r$, set $x_n=c+(1-1/n)(x-c)$ for $n\ge 2$. Then $\Vert x_n-c\Vert <r$ and $x_n\to x$, so every closed-ball point is in the closure of the open ball. HenceThe positive-radius convention matters: if radius zero is allowed, the open ball is empty whereas the closed ball is $\{c\}$, so the asserted equality fails at that degenerate radius.

In the usual real line take $\boxed{F_n=[n,\infty )}$. Each set is nonempty and closed, and $F_{n+1}\subseteq F_n$. No real number lies in them all, because the positive integers are unbounded. Thus their intersection is empty. These sets are not compact.

Compact subsets of a Hausdorff space are closed. Regard every $K_n$ as a closed subset of the compact set $K_1$. Any finite intersection equals the member with largest index and is therefore nonempty. The finite intersection property for closed subsets of a compact space impliesEquivalently, an empty intersection would give an open cover of $K_1$ by their complements, with a finite subcover contradicting nested nonemptiness.

Suppose no such $n_0$ exists. Nestedness then makes every $K_n\setminus U$ nonempty. These are nested compact sets: $X\setminus U$ is closed and $K_n$ is compact. By (i) their intersection is nonempty. But that intersection is $({\bigcap }_n{K}_n)\setminus U$, contrary to the hypothesis. Thus $K_{n_0}\subseteq U$ for some $n_0$, and nestedness gives $\boxed{K_n\subseteq U\text{ for all }n\ge n_0}$ immediately.

The inclusion $f({\bigcap }_n{K}_n)\subseteq {\bigcap }_{n}f(K_n)$ holds for every map. For the reverse choose $y\in {\bigcap }_{n}f(K_n)$ and let $L_n=K_n\cap f^{-1}(\{y\})$. These sets are nonempty and nested. A Hausdorff space has closed singletons, and continuity makes the fibre closed, so each $L_n$ is compact. Part (b)(i) gives a point $x\in {\bigcap }_n{L}_n$. Then $x\in {\bigcap }_n{K}_n$ and $f(x)=y$. Hence