We first observed that given two topological spaces \(X\) and \(Y\), if those spaces are actually metric spaces with metrics \(d_X\) and \(d_Y\) respectively, then we can characterize a function \(f : X \rightarrow Y\) being continuous if and only if for every \(x \in X\) and \(\epsilon > 0\) there is some \(\delta > 0\) such that for any \(x' \in X\), if \(d_X(x, x') < \delta\) then \(d_y(f(x), f(x')) < \epsilon\).

We then used this definition to show that given any two continuous functions \(f, g : \mathbb{R} \rightarrow \mathbb{R}\) and scalars \(c, d \in \mathbb{R}\) we have the linear combination \(cf + dg\) defined by \((cf + dg)(x) = cf(x) + dg(x)\) is continuous. This tells us that the continuous functions from \(\mathbb{R}\) to \(\mathbb{R}\)

We reviewed the notion of homemorphism. Recall that a function \(f : X \rightarrow Y\) between topological spaces is a homeomorphism iff it is bijective (and thus has an inverse \(f^{-1}\)) and moreover both \(f\) and \(f^{-1}\) are continuous. In this case we say \(X\) and \(Y\) are homeomorphic, often denoted \(X \cong Y\). We can view two homeomorphic spaces as being ``the same”. In particular, if \(X\) is connected and \(X \cong Y\), then so is \(Y\). If \(X\) is Hausdorff then so is \(Y\), etc. This allows us conversely to say that two spaces are not homemorphic if there is some topological property that one space exhibits but the other doesn’t. For example, \(\mathbb{R}\) and \(\mathbb{R}\setminus\{0\}\) are not homeomorphic because \(\mathbb{R}\) is connected while the other is not. We can also say \(\mathbb{R}\) and \(\mathbb{R}^2\) are not homeomorphic because for the first, removing any one point makes it disconnected, while that is not true for the second.

We started discussing the next main topic: compactness.

We say a topological space \(X\) is compact iff every open cover of \(X\) has a finite subcover. What is an open cover? An open cover of \(X\) is a family \(\{U_{i \in I}\}\) of open sets in \(X\) such that the union \(\bigcup_{i \in I} U_i\) is equal to \(X\). A subcover is simply a subfamily \((U_i)_{i \in I_0}\) for some \(I_0 \subseteq I\) whose union is still \(X\). So in other words, \(X\) is compact if and only if for every collection \(\{U_i\}_{i \in I}\) of open sets that unions up to \(X\), there is a finite \(I_0 \subseteq I\) such that \(\{U_i\}_{i \in I_0}\) still unions up to \(X\).

The real line \(\mathbb{R}\) is not compact, since it has a cover \(\{(z, z+2) \mid z \in \mathbb{Z}\}\) which does not have a finite subcover.

We started but didn’t finish a proof that \([0, 1]\) is compact. More generally, we will prove that a subset of \(\mathbb{R}\) is compact if and only if it is closed and bounded.