The first half of this lecture addressed a single result: that the product of two compact spaces is compact. We briefly used this result to prove a corollary: a subset of \(\mathbb{R} \times \mathbb{R}\) which is closed and bounded is compact (the other direction is also true and requires a proof similar to the proof we saw for subsets of \(\mathbb{R}\)). The proof in the textbook is quite different from the one we did in lecture, so I sketch it here.

Let \(X\) and \(Y\) be two compact spaces. We show that \(X \times Y\) is compact. To do this, let \(\mathcal{O}\) be any open cover of \(X \times Y\). Our goal is to show that \(\mathcal{O}\) has a finite subcover.

By the result from the previous lecture, since the set \(\{U \times V \mid U \subseteq X \text{ open}, V \subseteq Y \text{ open}\}\) of “open rectangles” forms a basis for the topology on \(X \times Y\), we can assume without loss of generality that \(\mathcal{O}\) is of the form \(\{U_i \times V_i \mid i \in I\}\).

For each \(y \in Y\), define \(I^y := \{i \in I \mid y \in V_i\}\) and \(\mathcal{O}^y := \{U_i \mid i \in I^y.\}\). For each \(y \in Y\) we argued in lecture that \(\mathcal{O}^y\) is an open cover of \(X\). Thus, by compactness of \(X\), there is a finite \(I^y_0 \subseteq I^y\) such that \(\mathcal{O}^y_0 := \{U_i \mid i \in I^y_0\}\) is a cover of \(X\).

Now, for each \(y \in Y\), define \(W_y := \bigcap \{V_i \mid i \in I^y_0\}\), which is open since \(I^y_0\) is finite, and moreover contains \(y\) by definition of \(I^y\). Thus \(\mathcal{O}^* := \{W_y \mid y \in Y\}\) is an open cover of \(Y\). By compactness of $Y$, there is finite \(Y_0 \subseteq Y\) such that \(\mathcal{O}^*_0 := \{W_y \mid y \in Y_0\}\) is a cover of \(Y\). Now let \(I_0 = \bigcup_{y \in Y_0} I^y_0\) and \(\mathcal{O}_0 := \{U_i \times V_i \mid i \in I_0\}\). We claim that \(\mathcal{O}_0\) is our finite subcover of \(X \times Y\). We know it is finite because \(Y_0\) is finite and each \(I^y_0\) is finite for \(y \in Y_0\), so we proceed to show that it is a cover.

To this end, let \((\hat{x}, \hat{y}) \in X \times Y\) be arbitrary. Since \(\mathcal{O}^*_0\) is a cover of \(Y\), there is some \(y^* \in Y_0\) such that \(\hat{y} \in W_{y^*}\). Since \(\mathcal{O}^{y^*}_0\) is a cover of \(X\), there is some \(\hat{i} \in I^{y^*}_0\) with \(\hat{x} \in U_{\hat{i}}\). Then \((\hat{x}, \hat{y}) \in U_{\hat{i}} \times V_{\hat{i}}\) and \(\hat{i} \in I_0\) so indeed the point \((\hat{x}, \hat{y})\) is covered by \(\mathcal{O}_0\).

It’s a difficult proof! It will take some time to fully digest.

In the second half, we started to set up a deeper discussion of sequences. We defined three properties of a topological space \(X\):

  1. \(X\) is second-countable if the topology on \(X\) has a countable basis;

  2. \(X\) is first-countable if every point in \(X\) has a countable local basis; and

  3. \(X\) is separable if there is a countable dense subset.

Recall that given \(x \in X\) a local basis of \(x\) is a collection \(\mathcal{B}_x\) of open neighborhoods of \(x\) with the property that for any open neighborhood \(U \ni x\) of \(x\), there exists some \(V \in \mathcal{B}_x\) with \(V \subseteq U\).

For \(\mathbb{R}\) with the standard topology, and \(\mathbb{R}\) with the discrete topology, I asked the class to determine if these properties hold. The standard topology, is second-countable, first-countable, and separable. The discrete topology is not secound-countable, nor is it separable, but it is first-countable.

We will discuss implications between these properties and additional examples in the next lecture.