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

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.