Imagine you remove a point of the real line \(\mathbb{R}\). How many “pieces” are you left with? Most would say “2”. Now imagine you remove a point from the real plane \(\mathbb{R} \times \mathbb{R}\). Most would say it remains in one piece but with a “hole”. Thus suggests a way to formally prove that \(\mathbb{R}\) and \(\mathbb{R} \times \mathbb{R}\) with the standard topologies are not the same topological space. Of course, this is intuitively obvious, but if you reflect for a moment you may become convinced that we would need to pursure something along these lines to actually prove it. It’s clear that they’re different spaces when you consider that \(\mathbb{R}\) has one dimension while \(\mathbb{R} \times \mathbb{R}\) has two, but it’s not clear that after boiling everything down to just the topology we still have a way of making sense of the “dimension” of a space.

The notion of connectedness allows us to make formal the idea of a topological space having “just one piece”. If you’re curiuos, the number of “holes” of a topological space is formalized by homotopy and the fundamenal group, which is taught in a course on algebraic topology (which is usually taken directly after a course on topology).

Given a topological space \(X\), we say that \(X\) is connected if the only clopen sets are \(\emptyset\) and \(X\).

It turns out in practice working with the notion of being disconnected (i.e. not connected) is a often a bit easier, and there are several equivalent ways to characterize it. In particular, the following are all equivalent:

  1. \(X\) is disconnected;

  2. There are non-empty disjoint open \(U_1, U_2 \subseteq X\) such that \(U_1 \cup U_2 = X\);

  3. There are non-empty disjoint closed \(C_1, C_2 \subseteq X\) such that \(C_1 \cup C_2 = X\).

Given a set \(A \subseteq X\) is connected if it is connected in the relative topology as a subspace of \(X\). If we recall how the relative topology is defined, we can restate this as saying that \(A \subseteq X\) is connected if there is no pair of open sets \(U_1, U_2 \subseteq X\) that both have nonempty intersection with \(A\) that are disjoint within \(A\) (i.e. \(U_1 \cap U_2 \cap A = \emptyset\)) and cover \(A\) (i.e. \(A \subseteq U_1 \cap U_2\)).

We proved that the closed interval \([0, 1]\) is connected. This is actually quite important so we labeled it as a theorem. We presented two proofs, the first of which was similar but slightly simpler to the one in the book, which I reproduce here. Considering equivalent statement (3) above and the observation in the previous paragraph, as well as the fact that \([0, 1]\) is closed, it is sufficient to show that any two nonempty closed sets \(C_1, C_2 \subseteq [0, 1]\) with \(C_1 \cup C_2 = [0, 1]\) must intersect. Indeed, assume without loss of generality \(0 \in C_1\). Let \(b\) be the infimum of \(C_2\) (i.e. the greatest lower bound). It’s straightforward to check that \(b\) must be a limit point of \(C_2\) and thus an element of \(C_2\). Thus if \(b = 0\) we have that \(0 \in C_1 \cap C_2\), so assume that \(b > 0\). But in order for \([0, 1]\) to be a subset of \(C_1 \cap C_2\) we must then have that \([0, b) \subseteq C_1\). But then \(b\) is also a limit point of \(C_1\) and thus in \(C_1\) and so we have \(b \in C_1 \cap C_2\).

Next, we discussed several examples of topological spaces and decided whether they are connected or disconnected:

  1. \((0, 1) \cup \{2\}\) with the standard topology

  2. \((0, 1) \cup \{2\}\) with the upper topology

  3. \(\mathbb{R}\) with the lower limit topology

  4. \(\mathbb{R}\) with the discrete topology

  5. \(\mathbb{R}\) with the cofinite topology

  6. \((\mathbb{R} \times \mathbb{R}) \setminus (\{(x, 0) \mid x \in \mathbb{R}\} \cup \{(0, x) \mid x \in \mathbb{R}\})\) with the standard product topology

The spaces (1), (3), (4) and (6) are disconnected while (2) and (5) are connected.

Finally, we proved that if \(X\) is connected and \(f : X \rightarrow Y\) is continuous then the image \(f(X)\) is connected. This is also an important result and we labeled it as a theorem.