example of a connected space that is not path-connected

This standard example shows that a connected topological space need not be path-connected (the converse is true, however).

Consider the topological spaces

	$\displaystyle X_{1}$	$\displaystyle=\left\{(0,y)\mid y\in[-1,1]\right\}$
	$\displaystyle X_{2}$	$\displaystyle=\left\{(x,\sin\frac{1}{x})\mid x>0\right\}$
	$\displaystyle X$	$\displaystyle=X_{1}\cup X_{2}$

with the topology induced from $\mathbb{R}^{2}$ .

$X_{2}$ is often called the “topologist’s sine curve”, and $X$ is its closure.

$X$ is not path-connected. Indeed, assume to the contrary that there exists a path (http://planetmath.org/PathConnected) $\gamma\colon[0,1]\to X$ with $\gamma(0)=(\frac{1}{\pi},0)$ and $\gamma(1)=(0,0)$ . Let

c=\inf\left\{t\in[0,1]\mid\gamma(t)\in X_{1}\right\}.

Then $\gamma([0,c])$ contains at most one point of $X_{1}$ , while $\overline{\gamma([0,c])}$ contains all of $X_{1}$ . So $\gamma([0,c])$ is not closed, and therefore not compact. But $\gamma$ is continuous and $[0,c]$ is compact, so $\gamma([0,c])$ must be compact (as a continuous image of a compact set is compact), which is a contradiction.

But $X$ is connected. Since both “parts” of the topologist’s sine curve are themselves connected, neither can be partitioned into two open sets. And any open set which contains points of the line segment $X_{1}$ must contain points of $X_{2}$ . So $X$ is not the disjoint union of two nonempty open sets, and is therefore connected.

Title	example of a connected space that is not path-connected
Canonical name	ExampleOfAConnectedSpaceThatIsNotPathconnected
Date of creation	2013-03-22 12:46:33
Last modified on	2013-03-22 12:46:33
Owner	yark (2760)
Last modified by	yark (2760)
Numerical id	16
Author	yark (2760)
Entry type	Example
Classification	msc 54D05
Related topic	ConnectedSpace
Related topic	PathConnected
Defines	topologist’s sine curve