# 2.04 Connectedness, path-connectedness

## Video

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

## Notes

### Connectedness

(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 $$V=\emptyset$$.

### Path-connectedness

(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.

### Examples

(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.