2.04 Connectedness, path-connectedness

(2.54) This is essentially the intermediate value theorem. Suppose there are open sets \(U,V\subset [0,1]\) which are disjoint and such that \([0,1]=U\cup V\). We want to prove that one of them is empty.

(4.06) Define a function \(F\colon[0,1]\to\mathbf{R}\) by \[F(x)=\begin{cases}0&\mbox{ if }x\in U\\1&\mbox{ if }x\in V.\end{cases}\] This function is well-defined because every point is in either \(U\) or \(V\) but not both. It is continuous because the preimage of any open set is one of \(U\), \(V\), \([0,1]\) or \(\emptyset\) (depending on whether it contains \(0\), \(1\), both or neither) which are all open.

(6.18) If \(F\) takes on both values \(0\) and \(1\) then, by the intermediate value theorem, it takes on all the intermediate values. This function doesn't: it only takes on the values \(0\) and \(1\). That means \(F\) does not take on both values, hence either \(U=\emptyset\) or \(V=\emptyset\).

(9.57) Let \(X\) be a path-connected space and let \(U,V\subset X\) be disjoint open sets such that \(U\cup V=X\). If they are both nonempty then we can pick a point \(x\in U\) and \(y\in V\). By path-connectedness, there is a continuous path \(\gamma\) from \(x\) to \(y\). Now \(\gamma^{-1}(U)\) and \(\gamma^{-1}(V)\) are disjoint open sets in \([0,1]\) whose union is \([0,1]\) and they are both nonempty because \(0\in\gamma^{-1}(U)\) and \(1\in\gamma^{-1}(V)\). This contradicts the previous theorem, so either \(U=\emptyset\) or \(V=\emptyset\), and we deduce that \(X\) is connected.

(16.51) If there is a homeomorphism \(F\colon\mathbf{R}\to\mathbf{R}^2\) then \(F|_{\mathbf{R}\setminus\{0\}}\colon\mathbf{R}\setminus\{0\}\to\mathbf{R}^2\setminus\{F(0)\}\) is also a homomorphism. We may as well take \(F(0)=(0,0)\) (by composing \(F\) with a translation, for example). But we know that \(\mathbf{R}\setminus\{0\}\) and \(\mathbf{R}^2\setminus\{(0,0)\}\) are not homeomorphic: one is connected, one is disconnected.