# the topologist’s sine curve has the fixed point property

The typical example of a connected space that is not path connected (the topologist’s sine curve) has the fixed point property.

Let ${X}_{1}=\{0\}\times [-1,1]$ and $$, and $X={X}_{1}\cup {X}_{2}$.

If $f:X\to X$ is a continuous map^{}, then since ${X}_{1}$ and ${X}_{2}$ are both path connected, the image of each one of them must be entirely contained in another of them.

If $f({X}_{1})\subset {X}_{1}$, then $f$ has a fixed point because the interval^{} has the fixed point property.
If $f({X}_{2})\subset {X}_{1}$, then $f(X)=f(cl({X}_{2}))\subset cl(f({X}_{2}))\subset {X}_{1}$, and in particular
$f({X}_{1})\subset {X}_{1}$and again $f$ has a fixed point.

So the only case that remains is that $f(X)\subset {X}_{2}$. And since $X$ is compact^{}, its projection to the first coordinate is also compact so that it must be an interval $[a,b]$ with $a>0$. Thus $f(X)$ is contained in
$S=\{(x,\mathrm{sin}(1/x)):x\in [a,b]\}$. But $S$ is homeomorphic to a closed interval, so that it has the fixed point property, and
the restriction of $f$ to $S$ is a continuous map $S\to S$, so that it has a fixed point.

This proof is due to Koro.

Title | the topologist’s sine curve has the fixed point property |
---|---|

Canonical name | TheTopologistsSineCurveHasTheFixedPointProperty |

Date of creation | 2013-03-22 16:59:37 |

Last modified on | 2013-03-22 16:59:37 |

Owner | Mathprof (13753) |

Last modified by | Mathprof (13753) |

Numerical id | 6 |

Author | Mathprof (13753) |

Entry type | Proof |

Classification | msc 55M20 |

Classification | msc 54H25 |

Classification | msc 47H10 |