A space is totally disconnected if its only connected subspaces are one-point sets. Show that if \(X\) has the discrete topology, then \(X\) is totally disconnected. Does the converse hold?
Solution
1. A discrete space \(X\) is totally disconnected.
Let \(C\subseteq X\) contain distinct points \(x\) and \(y\). Every subset of \(C\) is open in \(C\), so \(\theset{x}\) and \(C\sm\theset{x}\) are disjoint nonempty open subsets of \(C\) with union \(C\), and \(C\) is not connected.
2. The converse is false: \(\QQ\) with the subspace topology from \(\RR\) is totally disconnected but not discrete.
Let \(C\subseteq\QQ\) contain rationals \(p<q\), and choose an irrational \(r\) with \(p<r<q\). Then \(C\cap(-\infty,r)\) and \(C\cap(r,\infty)\) are disjoint nonempty open subsets of \(C\) with union \(C\), so \(C\) is not connected. No singleton \(\theset{q}\) is open in \(\QQ\), since every open interval about \(q\) contains other rationals.
3. Q.E.D.
Step 1 proves the statement, and step 2 answers the question about the converse.