The Unapologetic Mathematician

Mathematics for the interested outsider

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 X under a continuous map f:X\rightarrow Y is compact.

Let’s take an open cover \{U_i\} of the image f(X). Since f is continuous, we can take the preimage of each of these open sets \{f^{-1}(U_i)\} to get a bunch of open sets in X. Clearly every point of X is the preimage of some point of f(X), so the f^{-1}(U_i) form an open cover of X. Then we can take a finite subcover by compactness of X, picking out some finite collection of indices. Then looking back at the U_i corresponding to these indices (instead of their preimages) we get a finite subcover of f(X). Thus any open cover of the image has a finite subcover, and the image is compact.


January 16, 2008 - Posted by | Point-Set Topology, Topology


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

    Comment by Tony | January 16, 2008 | Reply

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

    Comment by bushi | February 4, 2008 | Reply

  3. […] 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 | Reply

  4. 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 | Reply

    • The open disk isn’t compact!

      Comment by John Armstrong | January 19, 2011 | Reply

      • Oops!

        Comment by Amitabha | January 20, 2011 | Reply

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