# 2.04 Connectedness, path-connectedness

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

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# 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:

*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.

*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

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

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**Compactness**. - Index of all lectures.