We started off by reviewing what the basis of a topology is and how to verify it. Then we covered the first problem in the fifth written assignment.

Next, we attemted to prove a false theorem that in any topological space, if it is separable and first-countable, then it is second-countable. We discussed a proof that started off convincing but ran into a snag. Indeed, the lower limit topology provides a counterexample to this (we will discuss it in more detail next Wednesday).

Next, we discussed the cofinite topology on \(\mathbb{R}\) and saw that it is separable but not first-countable (and thus not second-countable either).

I promised to cover a solution to the last problem in the fifth written assignment. If \(x \in A\) then we have \(d(x, A) = 0\) so we can take \(a = x\), so assume \(x \not\in A\). Let \(c = d(x, A)\) and for each \(\epsilon > 0\) let \(U_\epsilon = \{y \in X \mid d(y, x) > c + \epsilon\}\) It’s straightforward to check that each \(U_\epsilon\) is open (using triangle inequality). If we assume for contradiction there is no \(a \in A\) with \(d(x, a) = c\), then we have that \(\mathcal{O} = \{U_\epsilon \mid \epsilon > 0\}\) forms an open cover for \(A\). Since \(A\) is compact, \(\mathcal{O}\) has a finite subcover of \(A\), thus in particular there is \(\epsilon_0 > 0\) small enough such that \(A \subseteq U_{\epsilon_0}\). However, since \(d(x, A) = c\), there must be some \(b \in A\) with \(d(x, b) < c + \epsilon_0\), which is a contradiction.