= Solution
Removing $1/2$ from the <closed interval> $I=[0,1]$ leaves the <separation of a topological space>
$$
I\setminus\{1/2\}=[0,1/2)\sqcup(1/2,1].
$$
Thus $1/2$ 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 <homeomorphism>
$$
Q\setminus\{h^{-1}(1/2)\}\longrightarrow I\setminus\{1/2\}.
$$
The domain is <connected>, since <path connectedness> implies <connectedness>, while the target is not <connected>. This contradicts the preservation of <connectedness> by a <homeomorphism>.
Back to article page
