In this lecture we did a grab-bag of exercises and propositions to hopefully solidify our understanding of the basic concepts we will be building on in this course.

First, we proved that in a Hausdorff space, finite sets are closed. This is simply because that in a Hausdorff space, singletons are closed (discussed last time) and finite unions of closed sets are closed.

Next, we showed that if \(X\) and \(Y\) are topological spaces with \(Y\) Hausdorff, and \(f : X \rightarrow Y\) is continuous, then the graph of \(f\), defined \(\textrm{graph}(f) := \{(x, f(x)) \mid x \in X\}\) is closed as a subset of \(X \times Y\). This proof ended up being more challenging than I expected, so I include the argument here. Suppose \((x, y)\) is an arbitrary element of \(X \times Y\) not in the graph of \(f\). Then we have \(y \not= f(x)\). So because \(Y\) is Hausdorff, we can find disjoint open \(V_1, V_2\) with \(y \in V_1\) and \(f(x) \in V_2\). Let \(U := f^{-1}(V_2)\) which is open. Then we can see \(U \times V_1\) is an open neighborhood of \((x,y)\). To check that it is disjoint from the graph of \(f\), fix some \(x' \in X\) and suppose for contradiction that \((x', f(x')) \in U \times V_1\). But by definition of \(U\) we have \(f(x') \in V_2\). So we have \(f(x') \in V_1 \cap V_2\) which contradicts that they are disjoint.

We also gave another proof that the graph is closed. First, we proved a lemma which states that if \(X_1, X_2, Y_1, Y_2\) are topological spaces and \(f : X_1 \rightarrow Y_1\) and \(g : X_2 \rightarrow Y_2\) are continuous functions then the function \(h : X_1 \times X_2 \rightarrow Y_1 \times Y_2\) defined by \(h(x_1, x_2) := (f(x_1), g(x_2))\) is continuous. Second, we proved that for any continuous function, preimages of closed sets are always closed. With these lemmas, we can provide another proof of the previous result. We can define the function \(h : X \times Y \rightarrow Y \times Y\) by \(h(x, y) = (f(x), y)\). Then the preimage under \(h\) of the diagonal on \(Y\) (which we proved to be closed last time) is precisely the graph of \(f\), which is thus closed.

Finally, saw that if \(X\) is a topological space and \(x \in X\) and if \(X\) has a countable basis, then we can construct a decreasing sequence \((U_n)\) of open neighborhoods of \(x\) such that for any other open neighborhood \(V \ni x\) we can find some \(N\) such that for all \(n \ge N\) we have \(U_n \subseteq V\). Then if we construct a sequence \((a_n)\) with \(a_n \in U_n\) we have \(a_n \rightarrow x\). In particular, if \(A \subseteq X\) is closed and \(x\) is a limit point of \(A\) then we can find a sequence \((a_n)\) from \(A\) such that \(a_n \rightarrow x\). But it is very important to keep in mind we used the assumption that there is a countable basis of \(X\). We cannot do this in general.