The classical theorem states that an Hausdorff space admits a Stone-Cech compactification if and only if it is completely regular or Tychnoff (i.e. such that any point $x$ and closed set $C$ with $x\not\in C$ can be separated by a continuous function with values in $[0;1]$ with $f(x)=0$ and $f\restriction C=1$). The following proof


gives the construction of the Stone-Cech compactification of any topological space. The proof shows that $X$ can be mapped (injectively if $X$ is Hausdorff) in a compact space with the universal property given by its Stone-Cech compactification (if $X$ has one). Otherwise the map most likely is not a topological embedding: it is certainly not injective if $X$ is not Hausdorff, and most likely it should not be an homeomorphism with the image if $X$ is not Tychonoff. I do not see in the proof how the assumption that X is Tychonoff is necessary and sufficient to grant that the embedding of $X$ into its Stone-Cech compactification is a homeomorphism with the image; nor I see any gap in the proof. Is there a simple example of a non-Tychonoff Hausdorff space which does not admit a Stone-Cech compactification? Thanks for the help.

  • $\begingroup$ The "injectively if Hausdorff" part is not in that proof. $\endgroup$ – Henno Brandsma Apr 13 '18 at 10:40
  • $\begingroup$ In fact there is no guarantee that the "canonical map" from $X$ to $\beta(X)$ is every injective, let alone an embedding! The proof is silent on that. $\endgroup$ – Henno Brandsma Apr 13 '18 at 10:44
  • $\begingroup$ I added some stuff to my answer that might clarify things. $\endgroup$ – Henno Brandsma Apr 13 '18 at 11:07
  • $\begingroup$ thanks now I Think I understand the issue properly, thanks for your clarifications. $\endgroup$ – curious on mathematics Apr 13 '18 at 11:51

If $i: X \to Y$ is an embedding of $X$ into a compact Hausdorff space $Y$, then this implies that $X$ is Tychonoff:

$Y$, being compact and $T_2$, is $T_4$ and thus Tychonoff by Urysohn's lemma and $T_1$-ness. And $i[X]$ is a subspace of $Y$ and subspaces of Tychonoff spaces are Tychonoff. So $i[X]$ is Tychonoff and so $X$ too (as it's homeomorphic to $i[X]$).

So any space that is not Tychonoff a fortiori has no Stone-Cech compactification.

Note that in the linked to proof on stacks, it is never shown that the canonical map of $X$ to $\beta(X)$ is even injective (it need not be) let alone an embedding. What is constructed is a continuous map with a universal property, not an embedding.

The classical theorem you stated at the beginning is for the classical definition of a compactification of $X$: a pair $(i,Y)$ such that

  • $Y$ is compact Hausdorff.
  • $i: X \to Y$ is an embedding, i.e. $i: X \to i[X]$ is a homeomorphism where $i[X]$ has the subspace topology w.r.t. $Y$ of course.
  • $i[X]$ is dense in $Y$.

And $(i,Y)$ is called a Čech-Stone compactification when moreover for every compact Hausdorff $Z$ and every continuous $f: X \to Z$ there exists a continuous $\beta(f): Y \to Z$ such that $\beta(f) \circ i = f$. If we identify $i[X]$ with $X$, this just says that we can extend all continuous maps from $X$ to its superset $Y$ when we have a compact Hausdorff codomain.

Then indeed $X$ has a Čech-Stone compactification iff $X$ is Tychonoff. And it's essentially unique as well.

The definition you linked to is more general but less powerful: this is a continuous $\beta: X \to \beta(X)$ where $\beta(X)$ is compact Hausdorff, such that for any $f: X \to Y$ where $Y$ is compact Hausdorff there exists a map $\beta(f): \beta(X) \to Y$ with $\beta(f) \circ \beta = f$.

This is a purely category theoretical definition in terms of morphisms; it does not mention embeddings or dense-ness. The proof in your link shows only a "construction" for $\beta,\beta(X)$ for any $X$ and this $\beta(X)$ could be very degenerate for some spaces, e.g. not a nice extension space of $X$, etc.

So that proof is not "wrong", it shows the existence of another type of object.

  • $\begingroup$ thanks, are there simple examples of non-Tychonoff Hausdorff spaces for which it is "transparent" that a compactification cannot exist? $\endgroup$ – curious on mathematics Apr 13 '18 at 9:56
  • $\begingroup$ @curiousonoperatoralgebras not being Tychonoff is transparent (to me). $\endgroup$ – Henno Brandsma Apr 13 '18 at 10:08
  • $\begingroup$ yes thanks now I understand the issue correctly. $\endgroup$ – curious on mathematics Apr 13 '18 at 11:50

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