Past exam of the topology course of the University of Oxford / 2023 / 1

Give the nonempty subset $A\subseteq X$ its subspace topology. It is compact if every open cover admits a finite subcover. Equivalently, whenever $A\subseteq {\bigcup }_{j\in J}{U}_j$ with each $U_j$ open in $X$, there are $j_1,\dots ,j_m$ such that $A\subseteq U_{j_1}\cup \cdots \cup U_{j_m}$.

It is connected if it has no separation of a topological space: there are no disjoint nonempty open subsets $B,C$ of $A$ with $A=B\cup C$. Here openness is relative to the subspace topology; equivalently, the only clopen sets of $A$ are $\varnothing$ and $A$.

Let $(V_j{)}_{j\in J}$ be an open cover of $f(K)$ by open subsets of $Y$. The continuity of $f$ makes each preimage $f^{-1}(V_j)$ open in $X$, and these preimages cover $K$. By compactness, choose $j_1,\dots ,j_m$ withApplying $f$ gives $f(K)\subseteq {\bigcup }_{r=1}^m{V}_{j_r}$. Thus every open cover of $f(K)$ has a finite subcover, proving the continuous image of a compact space theorem. A cover by sets open in $f(K)$ gives the same argument after expressing them as intersections with ambient open sets.

Let $K\subseteq X$ be compact, with $X$ a Hausdorff space. The empty set is closed, so suppose $K\ne \varnothing$. Fix $x\in X\setminus K$. For each $y\in K$, the Hausdorff property supplies disjoint open neighbourhoods $U_y$ of $y$ and $V_y$ of $x$.

The $U_y$ form an open cover of $K$. Choose a finite subcover $U_{y_1},\dots ,U_{y_m}$ and put $V={\bigcap }_{r=1}^m{V}_{y_r}$. This is an open neighbourhood of $x$. If a point of $V$ belonged to $K$, it would belong to some $U_{y_r}$ as well as $V_{y_r}$, a contradiction. Hence $V\subseteq X\setminus K$. Every point of the complement has such an open neighbourhood, so $X\setminus K$ is open and $K$ is closed.

Use the Sierpiński space $X=\{0,1\}$ with topology $\{\varnothing ,\{1\},X\}$. The singleton $A=\{1\}$ is compact: from any open cover, choose one member containing $1$. But $X\setminus A=\{0\}$ is not open, so $A$ is not closed. This also exhibits why the Hausdorff hypothesis in the preceding proof matters.

The intended hypothesis is that $A$ is not connected. The original PDF has the same grammatical ambiguity as the transcription. If only $X$ were assumed not connected, the assertion would fail for the discrete space $X=\{0,1\}$ with its usual metric and the closed subset $A=\{0\}$: two disjoint sets cannot both meet that singleton.

Under the intended reading, choose a separation of a topological space $A=B\bigsqcup C$. Both $B$ and $C$ are nonempty and closed in $A$, because each is the complement of the other relatively open part. Since $A$ is closed in $X$, both are closed in $X$.

For a nonempty subset $F$ of the metric space $(X,d)$, the distance to a set satisfiesIndeed, the triangle inequality gives $d(x,F)\le d(x,z)+d(z,F)$, and swapping $x,z$ proves the bound. Thus the two distance to a set functions below are continuous. SetThese are disjoint open sets. For $b\in B$, we have $d(b,B)=0$ and $d(b,C)>0$, since $C$ is closed and $b\notin C$. Hence $B\subseteq U$. Similarly $C\subseteq V$. Therefore $A\subseteq U\cup V$ and both $U$ and $V$ meet $A$, as required. The construction is a concrete instance of the normality of every metric space.

Suppose $K\setminus \{a\}$ were compact. By the compact subset of a Hausdorff space theorem, it would be closed in $X$, hence closed in $K$. The singleton $\{a\}$ is also closed in the Hausdorff space $X$: each other point has an open neighbourhood avoiding $a$.

Consequently both $\{a\}$ and $K\setminus \{a\}$ are clopen in $K$. They are disjoint, nonempty, and cover $K$, giving a separation of a topological space in contradiction to the connectedness of $K$. Thus $K\setminus \{a\}$ is not compact.

Removing $1/2$ from the closed interval $I=[0,1]$ leaves the separation of a topological spaceThus $1/2$ is a cut point of $I$.

In contrast, removing any point $p$ from the square $Q=[0,1{]}^2$ leaves a path-connected space. To prove this, take distinct $u,v\in Q\setminus \{p\}$. Choose an interior point $w$ outside the two lines through $(u,p)$ and $(v,p)$. Such a choice exists: a small circle in the interior meets each of those lines in at most two points. Since $Q$ is a convex set, the line segments $[u,w]$ and $[w,v]$ stay in $Q$; our choice of $w$ ensures that both avoid $p$. Their concatenation is a continuous path from $u$ to $v$ in $Q\setminus \{p\}$.

The product topology on $I\times I$ agrees with its usual subspace topology in the Euclidean plane. If $h:Q\to I$ were a homeomorphism, its restriction would give a homeomorphismThe domain is connected, since path connectedness implies connectedness, while the target is not connected. This contradicts the preservation of connectedness by a homeomorphism.

The answer is no, because the number of non-isolated points differs. In $X$, every $1/n$ is an isolated point: a sufficiently small open interval around it contains no other member of $X$. The only non-isolated point is $0$, since every neighbourhood of $0$ contains some $1/n$.

In the product topology, $(x,y)\in X\times X$ is an isolated point exactly when both coordinates are isolated points of $X$. One direction follows because the product of two open singletons is open. For the other, any basic open rectangle containing $(x,y)$ contains another point if either coordinate has no open singleton. Hence the set of non-isolated points of $X\times X$ iswhich is infinite. A homeomorphism takes open singletons to open singletons, so it induces a bijection between the sets of non-isolated points. A space with just one such point cannot be homeomorphic to a space with infinitely many.