A topological space $X$ has a compactification if and only if $X$ is a Tychonoff space

Question

Following a reference from "General Topology" by Ryszard Engelking

Lemma

Let be $(X,\mathcal{T})$ a not compact topological space and let be $\infty\notin X$; thus on $X^\infty=X\cup\{\infty\}$ we consider the topology $$ \mathcal{T}^\infty:= \{U \subseteq X^\infty\mid U \cap X \in \mathcal{T} \land (\infty \in U \implies X \setminus U \mathrm{\ compact)}\} $$ and the function $i:X\rightarrow X^\infty$ defined as $$ i(x)=x $$ So the pair $(i,X^\infty)$ is a compactification of the space $X$, that we name Alexandroff compactification of the space $X$.

Proof. Here one can see the proof that $\mathcal{T}^\infty$ is a topology on $X^\infty$. So we have only to prove that $(i,X^\infty)$ is a compactification of $X$. First of all we observe that the function $i$ is an embedding of $X$ in $X^\infty$: indeed if $x,y\in X:x\neq y$ then clearly $i(x)\neq i(y)$ and so $i$ is injective; then for any open $U$ of $X$ it results that $i(U)$ is open in $X^\infty$ since $\mathcal{T}\preccurlyeq\mathcal{T}^\infty$ and so $i$ is open; finally we observe that for any $V\in\mathcal{T}^\infty$ it results that $i^{-1}(V)=V\cap X$ that is open in $X$ by the definition of $\mathcal{T}^\infty$ and so we can claim that $i$ is an embedding. So now we prove that $X^\infty$ is compact, using the topology $\mathcal{T}^\infty$: indeed if $\mathcal{U}$ is an open cover of $X^\infty$ we pick $U_0\in\mathcal{U}$ such that $\infty\in\mathcal{U_0}$; then by the definition of $\mathcal{T}^\infty$ we know that $X\setminus U_0$ is compact in $X$ and so in $X^\infty$ too, since $X\subseteq X^\infty$, and so there exist $U_1,...,U_n\in\mathcal{U}$ such that $X\setminus U_0\subseteq U_1\cup...\cup U_n$, since indeed $X\setminus U_0\subseteq X^\infty\subseteq\bigcup\mathcal{U}$ and so $\mathcal{U}$ cover $X\setminus U_0$ too, and so $\{U_0,U_1,...,U_n\}$ is a finite subcover of $\mathcal{U}$, form which we can claim that $X^\infty$ is compact.

Now we observe that if $X$ is a Tychonoff space so we know that through some embedding $h$ it is embeddable in $[0,1]^k$ that is compact and so the pair $(h,h[X])$ is a compactification of $X$. However if $X$ is not completely regular for what we proved in the previous lemma we can claim that there exist the compactification $(i,X^\infty)$ and so it seems that the theorem $3.5.1$ in the image is false. Perhaps is it the lemma I proved flase? then if the statement of the lemma is true, is my proof correct? Moreover I'd like that one show that if $\mathcal{U}$ is an open cover of $X^\infty$ then there exsist a its finite subcover: indeed I doubt that this passage of my proof is wrong. Could someone help me, please?

Answer 1

If $X$ has a compactification $(c,Y)$ then $Y$ is compact Hausdorff (as Engelking assumes Hausdorff as part of the definition of compactness, which is sometimes confusing when you use it as a reference compared to another text which does not) and thus normal (standard theorem) and hence Tychonoff, and so all subspaces are Tychonoff, including $c[X]$ which is homeomorphic to $X$. So $X$ is Tychonoff.

The fact that a Tychonoff (including $T_1$ !) space $X$ has a compatification follows from the Tychonoff embedding theorem, where we embed $X$ into some $[0,1]^I$ space and take the closure of its image there.

The Alexandroff compactification won't be Tychonoff in general, not even Hausdorff for $X$ not locally compact, so then it is not even a compactification in Engelking's definition (!). It's just an extension of $X$ to a quasi-compact space.

Answer 2

You should check the definitions and conventions made in the book.

As far as I know, Tikhonov space implies Hausdorff.¹ and most likely also compact spaces are assumed to be Hausdorff by Engelking, judging from the wording of the theorem and the previous considerations.

The Aleksandrov compactification² that you describe is not Hausdorff in general. A necessary condition is, of course, that $X$ is Hausdorff. Suppose $X\cup\{\infty\}$ is Hausdorff and take $x\in X$; then there are a neighborhood $U$ of $x$ and a neighborhood $(X\setminus K)\cup\{\infty\}$ of $\infty$, with $K$ a compact subset of $X$, such that $$ (X\setminus K)\cap U=\emptyset $$ This means $U\subseteq K$, so $x$ has a compact neighborhood, so it is locally compact. The converse is obvious.

Note that for some authors locally compact means “every point has a basis of neighborhoods consisting of compact sets”. The two definitions agree on (completely) regular spaces.

Thus your proposed counterexample is invalid, because it doesn't provide a Hausdorff compactification. It does when the space is Tikhonov and locally compact, but then it is not a counterexample.

If $X$ is a Tikhonov space that is not locally compact, the Aleksandrov compactification is not Hausdorff, so it is not a compactification in Engelking's sense. But this doesn't disprove the theorem, because you can find other compactification, for instance the Stone-Čech one.

Outline of the proof. If $X$ has a compactification, then it is (homeomorphic to) a subspace of a compact space, in particular it is Tikhonov, because this is a hereditary property.

Conversely, suppose $X$ is Tikhonov. Then one can prove that, for every closed subset $C$ and every $x\notin C$, there exists a continuous map $f\colon X\to[0,1]$ such that $f(c)=0$ for every $c\in C$ and $f(x)=1$. This can be used to provide an embedding of $X$ into $[0,1]^\Lambda$, where $\Lambda$ is the set of continuous maps $X\to[0,1]$.

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Footnotes.

¹ The name is Андрей Николаевич Тихонов, that can be transliterated either as Tichonov or Tikhonov. The latter style is more common in English texts. There's no reason whatsoever for using “y” in the name. The final ff is probably of German origin.

² The name is Павел Сергеевич Александров, that can be transliterated Aleksandrov (unfortunately the silly spelling Alexandroff is frequent).