One-point compactification and embedding in a cube

Question

I am currently reading Folland's Real Analysis. On page 145 of his book, he claims the following:

Claim: If $X$ is a noncompact, locally compact Hausdorff space, then the closure of the image of the embedding $e:X\hookrightarrow I^\mathcal F$ associated to $\mathcal F=C_c(X)\cap C(X,I)$ is the one-point compactification of $X$.

A few remarks:

• The "associated embedding" refers to the map $X\to I^\mathcal F$ obtained by mapping $X\ni x\mapsto \hat x\in I^{\mathcal F}$, where $\hat x$ is given by $\hat x(f)=f(x)$ for $f\in \mathcal F$.
• $C_c(X)$ is the set of all continuous, compactly supported, complex valued functions on $X$.
• $C(X,I)$ is the set of all continuous functions from $X$ to $I=[0,1]$.

I understand that this is indeed an embedding, because Urysohn's lemma guarantees that $\mathcal F$ separates points and closed sets and we know Theorem 1 below. I also understand that $\overline{e(X)}\setminus X$ contains at least one point because $\overline {e(X)}$ is a compact space containing the noncompact subspace $X$. What I do not understand is that why $\overline {e(x)}\setminus X$ consists of precisely one point.

I managed to prove the above claim. (See below. Any comment on the proof is welcome.) But it's a bit lengthy. Folland states the above claim as if it's very trivial, so I must be missing something. Can someone help me? Thanks in advance.

---

Theorem 1: Let $X$ be a completely regular space, and suppose that $\mathcal F\subset C(X,I)$ is a collection that separates points and closed sets (that is, if $E\subset X$ is a closed set and $x\in X\setminus E$, then there is some $f\in \mathcal F$ such that $f(x)\not\in \overline{f(E)}$). Then the map $X\hookrightarrow I^{\mathcal F}$ defined as above is an embedding.

---

My attempt: I first proved the following:

Lemma 1: Let $X$ be a completely regular space, let $Y$ be a compact Hausdorff space, and let $\phi\in C(X,Y)$. Suppose that $\mathcal{F}\subset C(X,I)$ separates points and closed sets, and the collection $\mathcal{G}=(\phi^{*})^{-1}(\mathcal{F})\subset C(Y,I)$ also separates points and closed sets. Then $\phi$ has a unique continuous extension $\tilde{\phi}:\hat{X}\to Y$, where $\hat{X}$ is the compactification of $X$ associated to $\mathcal F$. If $\phi$ is a compactification of $X$, then $\tilde{\phi}$ is a quotient map.

Here $\phi^*$ is the pullback $\phi^*(g)=g\circ \phi$.

(Proof of Lemma 1.)The map $\phi:X\to Y$ defines the pullback $\phi^{*}:\mathcal{G}\to\mathcal{F}$ given by $\phi^{*}(g)=g\circ\phi$, and then $\phi^{*}$ defines the pullback $\Phi=\phi^{**}:I^{\mathcal{F}}\to I^{\mathcal{G}}$ given by $\Phi(F)=F\circ\phi^{*}$. It is easy to check that the diagram $$\require{AMScd} \begin{CD} X@>>> I^{\mathcal F} \\ @VV\phi V & @V\Phi VV\\ Y@>>> I^{\mathcal G} \end{CD} $$commutes. (The horizontal maps are the embeddings.) For each $g\in\mathcal{G}$, the corresponding coordinate function of $\Phi$ is given by $\pi_{g}\circ\Phi=\pi_{g\circ\phi}:I^{\mathcal{F}}\to I$. Because $\pi_{g\circ\phi}$ just a projection, it follows that $\pi_{g}\circ\Phi$ is continuous. Therefore $\Phi$ is continuous.

Now by the closed map lemma, $Y$ is closed in $I^{\mathcal{G}}$, so $Y=\overline{Y}$. Thus we have $\Phi(\hat{X})=\overline{\Phi(X)}\subset Y$, where the equality follows again from the closed map lemma. So $\Phi$ restricts to a map $\tilde{\phi}:\overline{X}\to Y$ which extends $\phi$. Because $X$ is dense in $\hat{X}$ and $Y$ is Hausdorff, this is a unique extension. If $\phi$ is a compactification of $X$, then $\overline{\Phi(X)}=Y$ , so $\tilde{\phi}$ is surjective and hence is a quotient map by the closed map lemma. $\blacksquare $

Lemma 2:In the situation of lemma 1, if every function $f\in\mathcal{F}$ has a continuous extension to $Y$, then $\tilde{\phi}:\hat{X}\to Y$ is a homeomorphism.

(Proof.) In view of the preceding proposition, it suffices to show that $\tilde{\phi}$ is injective. So suppose that $F,F'\in\hat{X}$ are distinct elements of $\hat{X}$. Then there is some $f\in\mathcal{F}$ such that $F(f)\neq F'(f)$. If $g:Y\to I$ denotes the continuous extension of $f$, then $\tilde{\phi}(F)(g)=F(g\circ\phi)=F(f)$, and similarly, $\tilde{\phi}(F')(g)=F'(f)$. Therefore $\tilde{\phi}(F)\neq\tilde{\phi}(F')$, proving that $\tilde{\phi}$ is injective. $\blacksquare$

Now we can prove our claim. Suppose $X$ is a locally compact Hausdorff space, and let $\phi:X\to X^{*}=X\cup {\infty}$ be the one-point compactification of $X$. By the Urysohn's lemma, the collection $\mathcal{F}=C_{c}(X,I)\cap C(X)$ separates points and closed sets. The collection $\mathcal{G}=(\phi^{*})^{-1}(\mathcal{F})$ consists of continuous functions $g:X^{*}\to I$ with $\operatorname{supp}_{g}\subset X$. Because $X^{*}$ is normal, given a point $x\in X^{*}$ and a closed set $A\subset X^{*}$ disjoint from $X$, there exists a closed set $B\subset X^{*}$ containing a neighborhood of $x$, and then the Urysohn's lemma shows that there is some $g\in\mathcal{G}$ that is equal to 0 on $A$ and $1$ on $B$. So $\mathcal{G}$ separates points and closed sets. Moreover, any function in $\mathcal{F}$ extends continuously to $X^{*}$ (by declaring ). Thus by above lemmas, $\hat{X}$ is homeomorphic to $X^{*}$.

Answer 1

Your proof is correct, although you should explicitly define $\hat X$ is and restate lemma 1 as

Lemma 1: Let $X$ be a completely regular space, and let $Y$ be a compact Hausdorff space. Suppose that $\mathcal{F}\subset C(X,I)$ separates points and closed sets. Then any $\phi\in C(X,Y)$ such that the collection $\mathcal{G}=(\phi^{*})^{-1}(\mathcal{F})\subset C(Y,I)$ also separates points and closed sets, has a unique continuous extension $\tilde{\phi}:\hat{X}\to Y$. If $\phi$ is a compactification of $X$, then $\tilde{\phi}$ is a quotient map.

However, your approach is too complicated. Let $X^+$ denote the one-point compactification of $X$ and write $X^+ \setminus X = \{\infty\}$. Then each $f \in C_c(X)$ has a (trivally unique) extension $f^+ : X^+ \to \mathbb C$ (take $f^+(\infty) = 0$). Clearly if $f \in C(X,I)$, then $f^+ \in C(X^+,I)$. Let $j : \mathcal F \to C(X^+,I), j(f) = f^+$. If $i : X \to X^+$ denotes inclusion, then the diagram $$\require{AMScd} \begin{CD} X@> \epsilon >> I^{\mathcal F} \\ @VVi V @AA j^*A \\ X^+@>>\epsilon^+ > I^{\mathcal C(X^+,I)} \end{CD} $$ commutes because for $x \in X$ and $f \in \mathcal F$ we have $$\big(j^*(\epsilon^+(i(x)))\big)(f) = \big(\epsilon^+(i(x))j\big)(f) = \epsilon^+(i(x))(j(f)) = \epsilon^+(i(x))(f^+) = f^+(i(x)) = f(x) =\epsilon(x)(f) .$$ $X' = (j^*\epsilon^+)(X^+)$ is compact, thus $\overline{\epsilon (X)} \subset X'$. But $X' = (j^*\epsilon^+)(X^+) = (j^*\epsilon^+)(X \cup \{\infty\}) = \epsilon(X) \cup \{ (j^*\epsilon^+)(\infty) \}$.

Answer 2

I can offer a "direct" proof that $\overline{e(X)} \backslash e(X)$ consists of just the point all of whose coordinates are zero, though I would not say the proof is short.

Now I need to prove some lemmas:

Lemma 1: If $w \neq x$ are points in $X$, then $e(w)$ and $e(x)$ differ in at least two coordinates.

Proof: By Urysohn's lemma for LCH spaces (and since a single point is compact), there is a function $f\in \mathscr{F}$ such that $f(w)=1$ and $f(z)=0$. Similarly, there is a function $g\in \mathcal{F}$ such that $g(z)=1$ and $g(w)=0$. Then $e(w)$ and $e(x)$ differ in the coordinates associated with $f$ and $g$. $\square$

Now let $p\in \mathcal{F}\backslash e(X)$ be such that it has a non-zero coordinate. We show that $p$ cannot be in $\overline{e(X)}$. For the purpose of notation, suppose $g\in \mathcal{F}$ is a coordinate such that $\pi_g (p) \neq 0$.

Lemma 2: There can only be at most one $x\in X$ such that $\pi_f (e(x)) = \pi_f(p)$ for all $f \in \mathcal{F}\backslash\{g\}$.

Proof: If there are two distinct $w,x\in X$ satisfying the hypothesis of the lemma, then $e(w)$ and $e(x)$ differ only in at most 1 coordinate, which contradicts lemma 1. $\square$

If there is such an $x^*$ that satisfies the hypothesis of lemma 2, let $c = \pi_g(e(x^*))$. Note that $c \neq \pi_g(p)$ because $p$ is not in $e(X)$. Also since $\pi_g(p) \neq 0$, we can find a closed interval $I\subseteq [0,1]$ containing $\pi_g(p)$ but neither 0 nor $c$ (if $x^*$ exists). Also make sure $I$ is not just a single point. Now, $g^{-1}(I)$ is a closed subset of the support of $g$ which is compact, and hence compact. Call $g^{-1}(I) =: K$.

For $f\in \mathcal{F}\backslash\{g\}$ and $\epsilon > 0$, consider the open set

$$ f^{-1} ( [\pi_f(p) - \epsilon, \pi_f(p) + \epsilon]^c) $$

I claim as $f$ and $\epsilon$ varies, we have an open cover of $K$. Indeed, any element $x\in K$ (which excludes $x^*$ if that exist) must have $e(x)$ differ from $p$ in at least some coordinate $f\neq g$. Then as $\epsilon$ shrinks, $x$ would fall into the corresponding open set.

Since $K$ is compact, there is a finite subcover of the open cover above. After "combining common coordinates", this means I can find distinct $f_1,\dots,f_n$ and $\epsilon_1, \dots, \epsilon_n$ such that

$$ K \subseteq \bigcup_{i=1}^n f_i^{-1} ( [\pi_{f_i}(p) - \epsilon_i, \pi_{f_i}(p) + \epsilon_i]^c)$$

Consider the open neighbourhood of $p$ defined by

$$ \pi_g^{-1} (I_g) \cap \bigcap_{i=1}^n \pi_{f_i}^{-1} (I_{f_i})$$

where $I_g \subseteq I$ is an open interval containing $\pi_g(p)$ and $I_{f_i} \subseteq [\pi_{f_i}(p) - \epsilon_i, \pi_{f_i}(p) + \epsilon_i]$ is an open interval containing $\pi_{f_i}(p)$. Then we can easily check that this neighbourhood does not intersect $e(X)$. Indeed, if $x\notin K$, then $\pi_g(e(x))$ lies outside $I$ and if $x\in K$, then for some $1\le i \le n$, we have $\pi_{f_i}(e(x))$ lies in $[\pi_{f_i}(p) - \epsilon_i, \pi_{f_i}(p) + \epsilon_i]^c$ and hence not in $I_{f_i}$. This shows that $p$ is not in $\overline{e(X)}$ as desired.

Finally, we show that the point q consisting of all coordinates 0 is in $\overline{e(X)} \backslash e(X)$. $q$ is not in $e(X)$ because each $x\in X$ must have a non-zero coordinate in $e(x)$ (e.g. use Urysohn's lemma for LCH space). Next, let $\bigcap_{i=1}^n \pi_{f_i}^{-1} ([0,c_i))$ be a basis neighbourhood of $q$. Note that $K_i := $ supp$(f_i)$ are all compact. Since finite union of compact sets is compact, $\cup K_i$ is compact. Since $X$ is not compact, there exists a point $x\in X\backslash \cup K_i$ and $f_i(x)=0$ for all $1\le i \le n$. Then $e(x)$ lies in our neighbourhood basis.

Answer 3

Simply use the fact mentioned in the preceding sentence (on p.145): Y consists of e(X) together with the single point of $I^{\mathcal{F}}$ all of whose coordinates are zero.

To prove the aforementioned fact, just observe that if $y \in Y \setminus e(X)$, then $y \notin K$ for all compact subsets $K \subseteq e(X)$.