There is not the set of all compactification.

Question

Definition (0)

Let be $X$ a topological space. So a pair $(h,K)$ is a compactification of $X$ if $K$ is a compact space and if $h:X\rightarrow K$ is an embedding of $X$ in $K$ such that $h[X]$ is dense in $K$. Moreover a compactification $(h,K)$ of $X$ is a $T_2$ compactification if $K$ is a Hausdorf -compact- space.

Achtung (1)

In the previoius definition many authors require that $K$ must be Hausdorff!

Statement (2)

There is not the set of all compactification

Proof. Using the previous definition it's possible to show that any topological space has at least one compactification -its Alexandrov compactification: here the proof. So if there exist the set of all compactification it would contain the set of all set, since infact any set $X$ can organise as a topological space and $X\subseteq X^\infty$.

Well I have this trouble: using the definition of my text -see the lemma (4) here- we can use the Alexandrov compactification only on not compact spaces and so the proof of the previous statement would be valid iff for any set $X$ there exist a not compact topology. Furthermore if in the definition (0) I require that $K$ must be $T_2$, then the statement (2) it is still true?

Could someone help me, please?

You need to be precise here. If $X$ is Hausdorff and locally compact and non-compact, $\alpha(X)$, the Alexandroff compactification of $X$, is a Hausdorff compactification of $X$, in fact the smallest one.
If $X$ is not locally compact, or not Hausdorff, then $\alpha(X)$ is not a Hausdorff compactification at all. So do you want to consider a set of compactifications, or a set of Hausdorff compactifications? — Henno Brandsma, Mar 27 '20 at 10:35
@HennoBrandsma Primarily I consider the set of compactification, but anyway I ask to me if the statement is true or false when I consider Hausdorff compactification. — Antonio Maria Di Mauro, Mar 27 '20 at 10:39
Clearly every set can be given a well order and each well order has a Hausdorff compactification. Thus the set of all compactiifications would allow the set of all sets. This cannot happen in ZFC. — William Elliot, Mar 27 '20 at 10:50
@WilliamElliot That is a nonsense argument because the well-order is not related to the topology that $X$ has. — Henno Brandsma, Mar 27 '20 at 10:52

Henno Brandsma · Accepted Answer · 2020-03-27T10:56:44.420

If $X$ has one compactification $(h,K)$, only demanding $h[X]$ dense in $Y$ and $h$ an embedding into $Y$, and $Y$ compact (without Hausdorff), then there is no set of compactifications of $X$, because for any set $Z$ such that $|Y| = |Z|$, we can transport the topology and $h$ to $Z$ too and get (an essentially the same) compactification of $X$: if $f$ is a bijection from $Y$ to $Z$, define a topology on $Z$ by $\{f[O]: O \in \mathcal{T}_Y\}$ and define $h' = f \circ h$ and then $(h',Z)$ is as required. And it is well-known that for any non-empty set, there is no set of all sets with the same cardinality. That's why cardinal numbers were invented, essentially.

Any any space $X$ has a trivial compactification $Y=X\cup \{\infty\}$ where the topology on $Y$ is given by $\mathcal{T}_X \cup \{Y\}$ and $f(x)=x$ is the embedding. (The only open set covering $\infty$ is $Y$ so any cover of $Y$ has a one-element subcover, and denseness of $X=f[X]$ in $Y$ is also trivial for the same reason.)

So for any non-empty $X$ (locally compact or not, Hausdorff or not, compact or not) there is no set of all compactifications of $X$ for trivial set theory reasons.

And if $Y$ has to be Hausdorff then $X\neq \emptyset$ has a compactification iff $X$ is Tychonofff, so for a non-Tychonoff space there is a set of compactifications, namely the empty set. If $X$ is Tychonoff there is always at least one, so not set of them.

There is not the set of all compactification.

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