If $X$ is a locally compact $T_2$ space then the Alexandroff compactification is the smallest compactification of $X$

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.

Definition (1)

If $(h_1,K_1)$ and $(h_2,K_2)$ are two compactification of some topological space $X$ we say that $(h_1,K_1)\preccurlyeq(h_2,K_2)$ if there exist a continuous function $p$ such that $p\circ h_2=h_1$.

Lemma (2)

If $X$ a locally compact space and $f:X\rightarrow Y$ a surjective continuous function onto the topological space $Y$ then too $Y$ is locally compact.

Lemma (3)

If X is an Hausdorff space locally compact, then any dense locally compact subspace $Y$ is open.

Lemma (4)

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 $X$.

Lemma (5)

Let be $X$ a not compact Hausdorff space: so the Alexandroff compactification $(i,X^\infty)$ of $X$ is a Hausdorff space iff $X$ is locally compact.

Statement (6)

The Alexandroff compactification $(i,X^\infty)$ is the smallest compactification of any locally compact $T_2$ space $X$.

Proof. Well let be $X$ a locally compact $T_2$ space and $(h,K)$ an its compactification. So we consider the function $p:K\rightarrow X^\infty$ defined as $$ p(k)=\begin{cases}i(x),\text{if } k=h(x)\text{ for some }x\in X\\\infty,\text{ if }x\in K\setminus h[X]\end{cases} $$ and we observe that $p\circ h=i$ so if we will prove that $p$ is continuous we will have proved the statement. So let be $U\in\mathcal{T^\infty}$ and we prove that $p^{-1}(U)$ is open in $K$. Previously we observe that $h[X]$ is open in $K$ since $h[X]$ is locally compact and dense in $K$. So if $\infty\notin U$ and so if $U\subseteq X$ it results that $U$ is open in $X$ and by the definition of $p$ we have $p^{-1}(U)=h[U]$ that is open in $h[X]$ and so in $K$. Then if $\infty\in U$ then $X\setminus U$ is compact in $X$...

Unfortunately I can't prove the continuity of $p$ so I ask for your assistance. Then I used the lemma $4$ on $h[X]$ but I don't know if $K$ is $T_2$. Could someone help me, please?

Answer 1

If $\infty\in U$, then $X\setminus U$ is compact, so $p^{-1}[X\setminus U]=h[X\setminus U]$ is compact. If $K$ is Hausdorff, $p^{-1}[X\setminus U]$ is closed in $K$, and therefore $p^{-1}[U]$ is open in $K$, as desired.

Let $S=\{2^{-n}:n\in\Bbb Z^+\}$, and let $X=\Bbb N\cup S$ with the topology that it inherits from the real line, where $\Bbb N$ is the set of non-negative integers; clearly $X$ is locally compact and Hausdorff. Let $X_0=X\setminus\{0\}$, the set of isolated points of $X$. Let $K=\{q\}\cup X$, where $q$ is any point not in $X$, and topologize $K$ as follows: $X$ is an open subspace of $K$, and the sets of the form $\{q\}\cup(X_0\setminus F)$ for $F$ a finite subset of $X_0$ form a local base at $q$. Clearly $K$ is compact, and $X$ is dense in $K$, so $\langle h,K\rangle$ is a compactification of $X$, where $h:X\to K:x\mapsto x$.

Now $\{0\}\cup S$ is a compact subset of $X$, so $\{\infty\}\cup\Bbb Z^+$ is open in $X^\infty$. However,

$$p^{-1}[\{\infty\}\cup\Bbb Z^+]=\{q\}\cup\Bbb Z^+\;,$$

which is not open in $K$, so $p$ is not continuous.