A continuous function does not always map open sets to open sets, but a continuous function will map compact sets to compact sets. One could make list of such preservations of topological properties by a continuous function $f$: $$ f( \mathrm{open} ) \neq \mathrm{open} \;,$$ $$ f( \mathrm{closed} ) \neq \mathrm{closed} \;,$$ $$ f( \mathrm{compact} ) = \mathrm{compact} \;,$$ $$ f( \mathrm{convergent \; sequence} ) = \mathrm{convergent \; sequence} \;.$$ Could you please help in extending this list? (And correct the above if I've erred!)

Edit. Thanks for the several comments and answers extending my list. I was hoping that I could see some common theme among the properties preserved by a continuous mapping, separating those that are not preserved. But I don't see such a pattern. If anyone does, I'd appreciate a remark. Thanks!

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    $\begingroup$ Wouldn't it be better to start with inverse images rather than images? $\endgroup$ – anon Aug 26 '10 at 14:20
  • $\begingroup$ There are a couple here: en.wikipedia.org/wiki/… - Connected and Path-Connected $\endgroup$ – anon Aug 26 '10 at 14:22
  • $\begingroup$ $f(\text{Hausdorff}) \neq \ \text{Hausdorff}$. $\endgroup$ – Agustí Roig Aug 26 '10 at 14:32
  • $\begingroup$ $f(\text{locally (path) connected}) \neq \ \text{locally (path) connected}$ (but quotients of locally (path) connected spaces are locally (path) connected). $f(\text{locally compact}) \neq \ \text{locally compact}$. $\endgroup$ – Agustí Roig Aug 26 '10 at 14:44
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    $\begingroup$ I just saw this question by coincidence: on page 510 of Engelking's General Topology there's a rather extensive table called Invariants and inverse invariants of mappings. You'll find topological properties with indication of whether they are preserved by (various kinds of) continuous maps or not (such as open maps, closed maps, quotient maps, perfect maps, etc.). For mere continuous most things have been mentioned: simple covering properties (variations on compactness, connectedness, Lindelöf) and separability. More complicated covering properties such as paracompactness aren't preserved. $\endgroup$ – t.b. Jun 6 '12 at 7:56
  1. Connectedness and path connectedness
  2. If $f$ is a local homeomorphism, then $f$ is an open and closed map.
  3. If $f$ is onto and the domain is normal (can separate closed sets) or the map has compact fibers, then the image will be Hausdorff if the domain is.
  4. Second-countability is preserved under open maps.
  5. The image of a simply connected space need not be simply connected.

That's off the top of my head.. there are many more. Also, this should probably be community wiki.

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    $\begingroup$ A local homeomorphism is not necessarily a closed map. Let S=Z+ times (0,1) with the dictionary topology, and let f(n,p)=p. Then f is a local homeomorphism, but the set of ordered pairs {(n,1/2+1/(3n))|n is in Z+} is closed in S, but its image under f is not closed in (0,1). $\endgroup$ – dfeuer Nov 4 '11 at 22:18

These examples may be silly, but just to add to your list:

1) Sequential compactness (i.e. every sequence has a convergent subsequence)

2) Countable compactness (i.e. every countable open cover has a finite subcover)

3) $\sigma$-compactness (i.e. the space is a countable union of compact sets)

Less trivially, as mentioned in the Wikipedia article, the Lindelof property and separability are both preserved.


Continues image of a connected set is connected. Continues image of a complete set is not complete. Continuity does not preserve Cauchy sequences, unless it's a uniform continuity.


Following from muad's comment one could say that

1) $A \subseteq X_2$ is open, then $f^{-1}(A) \subseteq X_1$ is open
2) same thing for closed sets

which I think are more simple conditions for characterizing a continuous function as opposed to those in your list, since the concept of open and closed sets are "simpler" than compact of connected sets.

  • $\begingroup$ While I agree with what you're saying, I'm not sure it really answers the OP's question. $\endgroup$ – Jesse Madnick Sep 30 '10 at 6:56
  • $\begingroup$ I see hwat you're saying: since I'm considering pre-images and you can't tell if the image of an open set will be open, this isn't that appropriate, right? $\endgroup$ – Andy Sep 30 '10 at 8:24

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