# 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

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

${X}_{2}$ | $=\left\{(x,\mathrm{sin}{\displaystyle \frac{1}{x}})\mid x>0\right\}$ | |||

$X$ | $={X}_{1}\cup {X}_{2}$ |

${X}_{2}$ 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.

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 |