example of a connected space that is not path-connectedExampleOfAConnectedSpaceWhichIsNotPathConnected2002-06-10 09:17:012008-05-28 14:39:15Examplepathtopologist's sine curve\usepackage{amssymb}
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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 ``\emph{topologist's sine curve}'', and $X$ is its closure.
$X$ is not path-connected.
Indeed, assume to the contrary that there exists a \PMlinkname{path}{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.