# 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 ExampleOfAConnectedSpaceThatIsNotPathconnected 2013-03-22 12:46:33 2013-03-22 12:46:33 yark (2760) yark (2760) 16 yark (2760) Example msc 54D05 ConnectedSpace PathConnected topologist’s sine curve