Past exam of the topology course of the University of Oxford / 2015 / 1

A topological space is connected if it has no separation: two disjoint nonempty open sets whose union is the whole space. Equivalently, its only subsets that are both open and closed are the empty set and the whole space.

A subset is connected when its subspace topology has this property. Thus the sets in a proposed separation of a subset need only be relatively open, rather than open in the ambient space.

Suppose $A\cup B=U\cup V$ were a separation. By connectedness, all of $A$ lies on one side, say $U$: otherwise $A\cap U,A\cap V$ would separate $A$. A point of $A\cap B$ then lies in $B\cap U$, so connectedness of $B$ forces all of $B$ into $U$ as well. This makes $V$ empty, a contradiction.

Therefore the union of two intersecting connected subsets is connected. The argument also proves the connected union with a connected hub rule for arbitrarily many connected pieces meeting one connected subset.

If $f(A)=U\cup V$ were a separation, thenwould be disjoint nonempty relatively open sets covering $A$. They are relatively open by continuity of the restriction $f{|}_A:A\to f(A)$. This contradicts connectedness of $A$.This is the continuous image of a connected space theorem; no separation property of the target is required.

Use the basis of a topology consisting of open rectangles:It covers $X\times Y$, since $X\times Y\in \mathcal{B}$. Whenever a point lies in two basis rectangles, their intersection is another basis rectangle containing it:These are precisely the two basis of a topology axioms. Arbitrary unions of the rectangles therefore form a topology, the product topology. To check the finite-intersection axiom directly, distribute the intersection of two unions over all pairs of rectangles and use the displayed formula. The empty union gives the empty set.

If both factors are Hausdorff, two distinct product points differ in at least one coordinate. Disjoint open neighbourhoods in that coordinate, multiplied by the other whole factor, give disjoint open neighbourhoods of the product points. Thus the product is Hausdorff.

Conversely, assume the product is Hausdorff. Fix $y_0\in Y$, using nonemptiness. For $x\ne x^{\prime }$, choose disjoint open neighbourhoods of $(x,y_0)$ and $(x^{\prime },y_0)$ and refine them to basis rectangles $U\times V$ and $U^{\prime }\times V^{\prime }$. Since $y_0\in V\cap V^{\prime }$, their disjointness forces $U\cap U^{\prime }=\varnothing$. These separate $x,x^{\prime }$, proving that $X$ is Hausdorff. Fixing a point of $X$ proves the same for $Y$.The nonempty hypothesis is essential: a product with the empty set is empty regardless of the other factor. This is the Hausdorff property of a nonempty product.

Choose $x_0\in X\setminus A$ and $y_0\in Y\setminus B$, possible because the two subsets are proper. The crosslies in the indicated complement. Its two arms are homeomorphic to the connected factors and meet at $(x_0,y_0)$, so $H$ is connected by part (a)(ii).

For any remaining point $(x,y)$, either $x\notin A$ or $y\notin B$. In the first case the whole connected subset $\{x\}\times Y$ lies in the complement and meets $H$ at $(x,y_0)$. In the second case use $X\times \{y\}$, which meets $H$ at $(x_0,y)$. These connected fibres together with $H$ cover the complement. The connected union with a connected hub argument therefore givesNeither proper subset needs to be open, closed, or connected.

Every topological subspace $S\subseteq X$ again has the cofinite topology: its nonempty open subsets have finite complements in $S$, and every finite exclusion is obtained by intersecting with a cofinite open subset of $X$. As usual, the empty set is also open; this exception is implicit in the paper's shorthand definition.

If $S$ is infinite, any two nonempty relatively open sets intersect, since their complements are finite. Hence $S$ is hyperconnected space and therefore connected. If $S$ is finite, its cofinite topology is discrete topology. It is then connected exactly when it has at most one point. ThusThere are also infinite disconnected subsets of the product. Choose distinct $a,b\in X$ and put $D=\{a,b\}\times X$. This is infinite. The traces on $D$ of the ambient open setsare respectively $\{a\}\times X$ and $\{b\}\times X$. They are disjoint, nonempty and cover $D$, giving a separation. This answers the infinite-subset question, rather than only producing two isolated product points.