example of a connected space that is not path-connected

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

Consider the topological spaces

$X_1$$=\{(0,y)\mid y\in [-1,1]\}$

$X_2$$=\left\{(x,\sin {\displaystyle \frac{1}{x}})\mid x>0\right\}$

$X$$=X_1\cup X_2$

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

$X_2$ 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)$\gamma :[0,1]\to X$ with$\gamma (0)=(\frac{1}{\pi },0)$ and $\gamma (1)=(0,0)$.Let

$$c=inf\{t\in [0,1]\mid \gamma (t)\in X_1\}.$$

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.