2.04 Connectedness, path-connectedness


Below the video you will find accompanying notes and some pre-class questions.



(00.23) Let \(X\) be a topological space. We say that \(X\) is disconnected if there exist open sets \(U,V\subset X\) such that \(U\cap V=\emptyset\), \(U\neq\emptyset\), \(V\neq\emptyset\) and \(U\cup V=X\). We say that \(X\) is connected if it is not disconnected.

(2.09) The interval \([0,1]\), equipped with the subspace topology inherited from \(\mathbf{R}\), is connected.

(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 \(V=\emptyset\).


(8.08) We can use the fact that \([0,1]\) is connected to prove that lots of other spaces are connected:

A space \(X\) is path-connected if for all points \(x,y\in X\) there exists a path from \(x\) to \(y\), that is a continuous map \(\gamma\colon[0,1]\to X\) such that \(\gamma(0)=x\) and \(\gamma(1)=y\).

(9.16) A path-connected space is connected. (The converse fails.)

(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.


(13.00) The space \(X=\mathbf{R}^2\setminus\{(0,0)\}\) is path-connected. Given \(x,y\in X\), if the straight line \(\overline{xy}\) misses \((0,0)\) then it gives a path in \(X\) connecting \(x\) and \(y\). Otherwise, first follow a semicircle centred at \((0,0)\) passing through \(x\), then follow a straight path to \(y\). Therefore any two points are connected by a path.

(14.56) By contrast, \(\mathbf{R}\setminus\{0\}=(-\infty,0)\cup(0,\infty)\) is a disjoint union of nonempty open sets, so is disconnected.

The spaces \(\mathbf{R}\) and \(\mathbf{R}^2\) are not homeomorphic, in other words there is no continuous bijection \(F\colon\mathbf{R}\to\mathbf{R}^2\) such that \(F^{-1}\) is also continuous (for more about homeomorphisms, see this section).

(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.

Pre-class questions

  1. True or false? Every indiscrete space is connected.
  2. True or false? Every indiscrete space is path-connected.
  3. True or false? A subspace of a connected space is connected.


CC-BY-SA, Jonny Evans 2017