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

Consider the topological spaces \begin{eqnarray*} X_1 &= \left\{(0,y)\mid y\in[-1,1]\right\}\\ X_2 &= \left\{(x,\sin\frac{1}{x})\mid x>0\right\}\\ X &= X_1 \cup X_2\\ \end{eqnarray*}with the topology induced from $\Reals^2$ .

$X_2$ 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 $\gamma\colon[0,1]\to X$ 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.