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?
Best Answer
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.