Past exam of the topology course of the University of Oxford / 2014 / 2 / b

The interval $[0,1]$ is compact by the Heine-Borel theorem, so its square is compact by part (a). The half-open interval $[0,1)$ is not compact because it is not closed in the real line.

Each of the other two products has a continuous surjective coordinate projection onto $[0,1)$. If either product were compact, that image would be compact, a contradiction. Compactness is preserved by homeomorphisms, so neither product is homeomorphic to the closed square.

A boundary reparameterization gives an explicit homeomorphism. Let $Q=[0,1{]}^2$, with centre $c=(1/2,1/2)$, and parametrize its boundary once byHere $b(0)=b(4)$. The omitted top edge of $[0,1]\times [0,1)$ is $b([0,1])$. The omitted top and right edges of $[0,1{)}^2$ are $b([0,2])$.

Use the increasing boundary parameter mapIt takes $[0,4]$ bijectively onto itself, respects the identified endpoints, and maps the first omitted arc exactly onto the second. Extend it radially:Every noncentral point has a unique boundary ray and a unique radial parameter $r=2\Vert q-c{\Vert }_{\infty }$; the only duplicate boundary parameters are zero and four, whose images agree. The inverse uses ${\varphi }^{-1}$ by the same construction. Both maps are continuous off the centre, and at the centre their radial distances tend to zero. Thus $H$ is a homeomorphism of the whole closed square.

It carries the first omitted edge precisely onto the second omitted pair of edges, so restriction givesThis is the radial extension of a square boundary homeomorphism; it avoids any need to treat corners as topologically distinguished points.