Exercise 4.47, Folland's Real Analysis, Alexandroff compactification

Question

Let $X$ be a locally compact Hausdorff space (LCH). Let $\infty$ denote an element that is not in $X$ and $X^\ast=X\cup\{\infty\}$. Let $\mathcal{T}$ be the collection of all subsets of $X^\ast$ such that either (i) $U$ is an open subset of $X$, or (ii) $\infty\in U$ and $U^c=X\setminus U$ is a compact subset of $X$.

Proposition 4.36. If $X, X^\ast$ and $\mathcal T$ are LCH, the one-point compactification and the topology on $X^\ast$ respectively, then $(X^\ast,\mathcal{T})$ is compact Hausdorff and the inclusion map $i: X \to X^\ast$ is an embedding. Moreover, if $f\in C(X)$, then $f$ extends continuously to $X^\ast$ iff $f=g+c$ where $g\in C_0(X)$ and $c$ is a constant, in which case the continuous extension is given by $f(\infty)=c$.

Exercise. Prove Proposition 4.36. Also, show that if $X$ is Hausdorff but not locally compact, Proposition 4.36 remains valid except that $X^\ast$ is not Hausdorff.

My solution so far consists of showing $(X^\ast,\mathcal{T})$ is actually a topological space, that it is compact, that it is Hausdorff and then showing the equivalence in the last sentence of the proposition. Since I only require local compactness to show $X^\ast$ is Hausdorff, I reason that the proposition remains valid even if $X$ is not locally compact, though $X^\ast$ is then not implied to be Hausdorff. Or do I need to show something more here regarding the last sentence of the exercise?

The equivalence in the last sentence in the proposition, however, leaves me doubtful.

Moreover, if $f\in C(X)$, then $f$ extends continuously to $X^\ast$ iff $f=g+c$ where $g\in C_0(X)$ and $c$ is a constant, in which case the continuous extension is given by $f(\infty)=c$.

A function $f\in C_0(X)$ if for every $\epsilon>0$, the set $\{x:|f(x)|\geq\epsilon\}$ is compact. Inspired by this answer (I really struggle following the second equivalence in that answer), here's my attempt.

Attempt

Given that $f\in C(X)$ extends continuously to $X^\ast$, let $F:X^\ast\to\mathbb{R}$ be such an extension, with $F|X=f$ and $c:=F(\infty)$, where $c$ is some constant. We show $F-c\in C_0(X)$. Let $\epsilon>0$ be given and define $$U=\{y\in X^\ast:|F(y)-c|<\epsilon\}.$$We aim to show $X^\ast\setminus U$ is a compact subset of $X$. Notice that $U=h^{-1}((-\infty,\epsilon))$ for the function $h:X^\ast\to \mathbb{R}$ defined via $h(y)=|F(y)-c|$. Since $F$ is continuous, so is $h$ and thus $U$ is open in $X^\ast$. Therefore $K=X^\ast\setminus U$ is closed. But as $K$ is a closed subset of a compact set $X^\ast$, it is compact. And $\infty\notin K$ since $\infty\in U$ (indeed, $|F(\infty)-c|=0<\epsilon$). So given $\epsilon$, we found a $K\subset X$ compact such that $|F-c|<\epsilon$ on $X^\ast\setminus K$ (in particular on $X\setminus K$).

Conversely, if $f=g+c$ for $g\in C_0(X)$ and $c$ a constant such that $f(\infty)=c$, then we need to show $f$ is continuous. It suffices to check the preimages of open intervals of the form $(a,b)\subset\mathbb{R}$, since these form a base for the topology on $\mathbb{R}$. We have $f^{-1}((a,b))=g^{-1}((a,b))$ if $c\notin (a,b)$, which is open. If $c\in(a,b)$, then, $(a,b)=(a,c-\delta)\cup(c-\epsilon,c+\epsilon)\cup(c+\delta,b)$ for some $\epsilon,\delta>0$ small such that $a<c-\epsilon,a<c-\delta$ and $c+\epsilon<b,c+\delta<b$ and $\delta<\epsilon$. Since $g\in C_0(X)$, take a compact $K\subset X$ such that $|g|=|f-c|<\epsilon$ on $X\setminus K$, i.e. $f(x)\in (c-\epsilon,c+\epsilon)$ for $x\in X\setminus K$. Then \begin{align*}f^{-1}((a,b))&=f^{-1}((a,c-\delta))\cup f^{-1}((c-\epsilon,c+\epsilon))\cup f^{-1}((c+\delta,b))\\ &=g^{-1}((a,c-\delta))\cup (X\setminus K)\cup g^{-1}((c+\delta,b)) .\end{align*}This is a union of open sets in $X^\ast$ since $K$ is closed (compact subsets of Hausdorff spaces are closed) and $g$ continuous. So $f^{-1}((a,b))$ is open also when $c\in (a,b)$, proving continuity of $f$.

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Any feedback to this attempt is appreciated very much. It became a bit lengthy, hence my doubt. Also the thing with introducing both $\epsilon$ and $\delta$ feels complicated, but I don't know how to do it differently.

Answer 1

Suppose that $X$ is a locally compact (and non-compact) Hausdorff topological space.

$\textbf{Definition :}$ A continuous function $f:X\to\mathbb{C}$ is said to vanish at $\infty$ if for any $\epsilon>0$ the set, $K_\epsilon=\{x\in X\,:\,|f(x)|\geq\epsilon\}$ is compact.

$\textbf{Lemma :}$ Suppose that $X$ is a locally compact (and non-compact) Hausdorff topological space and $f:X\to\mathbb{C}$ is continuous. Then, $f$ vanish at $\infty$ if and only if for any $\epsilon>0$ there exists a compact subset $C_\epsilon$ of $X$ such that, $|f(x)|<\epsilon$ for all $x\in X\setminus C_\epsilon$.

$\textbf{Proof :}$ We first assume that $f$ vanish at $\infty$. Let, $\epsilon>0$. Choose, $C_\epsilon=\{x\in X\,:\,|f(x)|\geq\epsilon\}$. Then, by definition, $C_\epsilon$ is compact and $|f(x)|<\epsilon$ for all $x\in X\setminus C_\epsilon$.

Conversely, suppose that the condition holds. We claim that, $f$ vanish at $\infty$. Take any $\epsilon>0$ and set, $K_\epsilon=\{x\in X\,:\,|f(x)|\geq\epsilon\}$. For this $\epsilon$ let, $C_\epsilon$ be a compact subset of $X$ such that, $|f(x)|<\epsilon$ for all $x\in X\setminus C_\epsilon$. Since, $f$ is continuous $K_\epsilon$ is then a closed subset of $C_\epsilon$. Since, $C_\epsilon$ is compact, we get, $K_\epsilon$ is also compact. So, $f$ vanish at $\infty$.

This completes the proof.

$\textbf{Lemma :}$ Suppose that $X$ is a locally compact (and non-compact) Hausdorff topological space. If $f,g:X\to\mathbb{C}$ are continuous function which vanish at $\infty$ and $\alpha\in\mathbb{C}$ then, all of the functions $\bar{f}$, $\alpha f$, $f+g$, $fg$ vanish at $\infty$. Furthermore, if $f:X\to\mathbb{C}$ is a continuous function which vanish at $\infty$ then, $f$ is bounded.

$\textbf{Proof :}$ Since, $f$, $g$ both vanish at $\infty$ for any $\epsilon>0$ there exist compact sets $A_\epsilon$ and $B_\epsilon$ in $X$ such that, $|f(x)|<\min\{\sqrt{\epsilon},\epsilon/2\}$ for all $x\in X\setminus A_\epsilon$ and $|g(x)|<\min\{\sqrt{\epsilon},\epsilon/2\}$ for all $x\in X\setminus B_\epsilon$. Let, $C_\epsilon=A_\epsilon\cup B_\epsilon$. Then those inequalities hold for all $x\in X\setminus C_\epsilon$ and $C_\epsilon$ is also compact. So, we get, $|(fg)(x)|=|f(x)|.|g(x)|<\epsilon$ and $|(f+g)(x)|\leq |f(x)|+|g(x)|<\epsilon$ for all $x\in X\setminus C_\epsilon$. It shows that $fg$ and $f+g$ are continuous functions which vanish at $\infty$. Now, let $\alpha\in\mathbb{C}$. If $\alpha=0$ then, $\alpha f=0$. So, for any $\epsilon>0$ the set, $\{x\in X\,:\,|(\alpha f)(x)|\geq\epsilon\}=\phi$ is compact. So, $\alpha f$ vanish at $\infty$. Now suppose that $\alpha\neq 0$. Then, the set, $\{x\in X\,:\,|f(x)|\geq \epsilon/|\alpha|\}=\{x\in X\,:\,|(\alpha f)(x)|\geq\epsilon\}$ is compact. So, $\alpha f$ is again a continuous function which vanish at $\infty$. The result for $\bar{f}$ follows from the fact that $|z|=|\bar{z}|$ for all complex number $z$.

Finally, suppose that $f:X\to\mathbb{C}$ is continuous and vanish at $\infty$. Then, for $\epsilon=1>0$ the set, $K_1=\{x\in X\,:\,|f(x)|\geq 1\}$ is compact. Then, $|f(x)|<1$ for all $x\in X\setminus K_1$. Since, $f$ is continuous and $K_1$ is compact, it is bounded on $K_1$. Therefore, $f$ is bounded on $X$. This completes the proof.

$\textbf{Remark :}$ Suppose that $X$ is a locally compact (and non-compact) Hausdorff topological space. We denote the set of all continuous functions $f:X\to\mathbb{C}$ which vanish at $\infty$ by, $C_0(X)$. The previous lemma shows that (?) $C_0(X)$ is a non-unital (?) abelian $C^*$-algebra when it is endowed with the sup-norm. Also, Urysohn's lemma for locally compact Hausdorff spaces gurantees existence of lots of elements in $C_0(X)$.

Now we discuss the main theorems.

$\textbf{Theorem :}$ Suppose that $X$ is a locally compact (and non-compact) Hausdorff topological space. Then, $X$ has a unique (?) one-point (ie. Alexandroff) compactification $X^\#=X\cup\{\infty\}$ ie. $X^\#$ is a compact Hausdorff topological space containing $X$ as a dense (?) subspace.

You have already proven it.

So we discuss the proof of the next theorem.

$\textbf{Theorem :}$ Suppose that $X$ is a locally compact (and non-compact) Hausdorff topological space and $f:X\to\mathbb{C}$ is a continuous function. Then, $f$ extends to a continuous (necessarily unique (?)) $F:X^\#\to\mathbb{C}$ if and only if there exists a complex number $c$ and a continuous function $g:X\to\mathbb{C}$ which vanish at $\infty$ such that, $f=g+c$ on $X$. In which case $F$ is given by, $F(\infty)=c$.

$\textbf{Proof :}$ We first assume that $f$ has such an extension $F$. Define, $g:X\to\mathbb{C}$ by, $g=f-c$ where, $c=F(\infty)\in\mathbb{C}$. Then, $g$ is continuous on $X$ and $f=g+c$. We only need to show that $g$ vanish at $\infty$. Note that $g$ has obvious continuous extension $G:X^\#\to\mathbb{C}$ given by, $G=F-c$ and $G(\infty)=0$. Now take any $\epsilon>0$. Then, $|G(\infty)|=0<\epsilon$. So, there exists an open set $U$ in $X^\#$ such that $|G(x)|<\epsilon$ for all $x\in U=X^\#\setminus K_\epsilon$ for some compact set $K_\epsilon$ in $X$. So, $\{x\in X\,:\,|g(x)\geq\epsilon\}$ is compact in $X$. So, $g$ vanish at $\infty$. Conversely, suppose that there exists such a $g$. Then checking continuity of $F$ is routine.

$\textbf{Further Clarifications :}$ Suppose that $X$ is a locally compact (and non-compact) Hausdorff topological space and $f\in C_0(X)$ then, there exists a unique $F\in C(X^\#)$ extending $f$.

$\textbf{Proof :}$ Define, $F:X^\#\to\mathbb{C}$ by, $F(x)=f(x)$ for each $x\in X$ and $F(\infty)=0$. We will now check local continuity of $F$ at each point in $X^\#$. Let, $p\in X^\#$ and $p\neq\infty$. Let, $V$ be any open set in $\mathbb{C}$ with, $F(p)=f(p)\in V$. Since, $f$ is continuous at $p\in X$ there is an open set $U$ in $X$ containing $p$ (this $U$ is also open in $X^\#$) such that, $F(U)=f(U)\subseteq V$. So, $F$ is continuous at $p$. Now it is enough to check the continuity of $F$ at $\infty$. Let, $V$ be open set in $\mathbb{C}$ with $0=F(\infty)\in V$. Then, there exists a $\epsilon>0$ such that, $D_\epsilon=\{z\in\mathbb{C}\,:\,|z|<\epsilon\}\subseteq V$. Since, $f$ vanish at $\infty$ the set, $K_\epsilon=\{x\in X\,:\,|f(x)|\geq\epsilon\}$ is a compact set in $X$. So, $U=X^\#\setminus K_\epsilon$ is open in $X^\#$ containing $\infty$. Now see that, $F(U)\subseteq D_\epsilon\subseteq V$. So, $F$ is continuous at $\infty$. So, $F$ is continuous on $X^\#$ and it extends $f$. Uniqueness follows from the density of $X$ in $X^\#$. This completes the proof.

Now can you use this result ?

Answer 2

Your proof seems to be correct, but it is too complicated.

First note that it is sufficient to prove that $f$ has a continuous extension $F$ with $F(\infty) = 0$ if and only if $f \in C_0(X)$.

Observe that each $f : X \to \mathbb R$ can be extended to $f^* : X^* \to \mathbb R$ by setting $f^* \mid_X = f$ and $f^*(\infty) = 0$. However, in general $f^*$ will not be continuos. The claim is therefore:

• $f^*$ is continuous if and only if $f \in C_0(X)$.

Note that we always have $$\{x \in X :|f(x)|\geq\epsilon\} = f^{-1}((\mathbb R \setminus (-\epsilon,\epsilon)) = X \setminus f^{-1}((-\epsilon,\epsilon)) = X^* \setminus (f^*)^{-1}((-\epsilon,\epsilon)) . $$

Proof of the "if" part: We have to show that $f^*$ is continuous. Since $X$ is open in $X^*$, $f^*$ is continuous at all points of $X$. It remains to show that it is continuous at $\infty$. Let $\epsilon > 0$. We know that $C = \{x \in X :|f(x)|\geq\epsilon\}$ is a compact subset of $X$. Hence $U = X^* \setminus C = (f^*)^{-1}((-\epsilon, \epsilon))$ is an open neigborhood of $\infty$ in $X^*$. By construction $f^*(U) \subset (-\epsilon, \epsilon)$.

Proof of the "only if" part: Let $\epsilon >0$. Since $f^*$ is continuous, there exists an open neighborhood $U$ of $\infty$ in $X^*$ with $f^*(U) \subset (-\epsilon, \epsilon)$, which is equivalent to $U \subset (f^*)^{-1}((-\epsilon,\epsilon))$. The set $X^* \setminus U$ is then a compact subset of $X$. But $\{x \in X :|f(x)|\geq\epsilon\} = X^* \setminus (f^*)^{-1}((-\epsilon,\epsilon)) \subset X^* \setminus U$. Since $\{x \in X :|f(x)|\geq\epsilon\}$ is closed in $X$, it is a closed subset of the compact set $X^* \setminus U$ and therefore itself compact.

What happens if $X$ is Hausdorff, but not locally compact?

In that case $X$ is still open in $X^*$ and the proof that $X^*$ is compact still works. However, $X^*$ is not Hausdorff. It it were, then $X$ would be locally compact as an open subset of a compact Hausdorff space.

Moreover, the above proof concerning $f^*$ still works.