## The Image of a Compact Space

One of the nice things about connectedness is that it’s preserved under continuous maps. It turns out that compactness is the same way — the image of a compact space under a continuous map is compact.

Let’s take an open cover of the image . Since is continuous, we can take the preimage of each of these open sets to get a bunch of open sets in . Clearly every point of is the preimage of *some* point of , so the form an open cover of . Then we can take a finite subcover by compactness of , picking out some finite collection of indices. Then looking back at the corresponding to these indices (instead of their preimages) we get a finite subcover of . Thus any open cover of the image has a finite subcover, and the image is compact.

Advertisements

## 6 Comments »

### Leave a Reply

Advertisements

I’ve been lurking and reading your topology posts. Please keep it up!

Comment by Tony | January 16, 2008 |

iam in seargh on a filters and topology have you any information?

Comment by bushi | February 4, 2008 |

[…] subset of the disk of radius . Now the function is a continuous, real-valued function on , and the image of a compact space is compact, so takes some maximum value on […]

Pingback by Uniform Convergence of Power Series « The Unapologetic Mathematician | September 10, 2008 |

I stumbled upon your blog (nice!) while searching for something. I have a question related to the statement here. Consider the open disk in $R^2$, and polar coordinates $(r, \phi)$. Now consider a map like

$f(r) = r exp(1/1-r)$

$0 \mapsto 0$ and any distance r is stretched more and more as you move closer to 1. This looks like it maps the disk to $R^2$ and also looks continuous everywhere inside the open disk. Then a compact space is mapped to a non-compact space by a continuous mapping. Where am I going wrong?

Comment by Amitabha | January 19, 2011 |

The open disk isn’t compact!

Comment by John Armstrong | January 19, 2011 |

Oops!

Comment by Amitabha | January 20, 2011 |