I read the following sentence in Wikipedia. It is the second one in the paragraph.

... a $T_0$ topological space which is homeomorphic to itself and exhibits pointwise convergence...

Isn't every space homeomorphic to itself? The identity mapping is a homeomorphism, to point one.

  • $\begingroup$ I too am confused. The identity mapping is always continuous. $\endgroup$ Dec 13, 2013 at 19:35
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    $\begingroup$ Sounds like we should change Wikipedia. $\endgroup$
    – Ian Coley
    Dec 13, 2013 at 19:36
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    $\begingroup$ So somebody wrote something odd on wikipedia. Disaster! xkcd.com/386 $\endgroup$ Dec 13, 2013 at 19:37
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    $\begingroup$ Guess: the topological space is homeomorphic to a proper subset of itself. Another possibility is that there is a nontrivial homeomorphism from the space to itself. $\endgroup$
    – dfeuer
    Dec 13, 2013 at 19:48
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    $\begingroup$ … or that homeomorphisms act transitively on the space? Or that the identity is the only homeomorphism of the space with itself? The possibilities are endless. $\endgroup$ Dec 13, 2013 at 19:53

1 Answer 1


This seems to be an extremely badly worded sentence. On p.157 of Encyclopedia of General Topology (Hart, Nagata, Vaughan), they describe Scott topology, and mention that Scott constructed "continuous lattices" that are homeomorphic to their space of self maps $[L \to L]$. Here's the relevant excerpt (I believe fair use allows me to quote a sentence from a book):

[...] The category of domains and Scott continuous functions is Cartesian closed, and Scott produced a canonical construction of continuous lattices $L$ which have a natural homeomorphism onto the space $[L \to L]$ of continuous self-maps; in such a $T_0$-space as "universe", every "element" is at the same time a "function" whence expressions, like $f(f)$, are perfectly consistent. [...]

This is also the main point of Scott's paper Continuous lattices (which you can find online):

The main result of the paper is a proof that every topological space can be embedded in a continuous lattice which is homeomorphic (and isomorphic) to its own function space.

To address the self homeomorphism thing, yes, of course, the identity is always a self homeomorphism. But what can happen is that you have two different topologies on the same set, and then the identity will not be a homeomorphism. But this is unrelated to what's happening in this article.

  • $\begingroup$ I'd say that extremely badly worded sentence is an understatement; extremely badly worded article is, in my opinion, nearer the mark. +1 for still finding what was probably meant by this sentence. $\endgroup$ Dec 13, 2013 at 20:14
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    $\begingroup$ The article is a mess. $\endgroup$ Dec 13, 2013 at 20:44
  • $\begingroup$ Yeah, it's not very well organized. It's too bad, the original writer obviously put a lot of effort in it. $\endgroup$ Dec 13, 2013 at 21:04

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