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

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.