5.25 Compactifications

Proof. Let $e: X \to [0, 1]^{C(X; [0, 1])}$ be the embedding of $X$ into $[0, 1]^{C(X; [0, 1])}$ associated with $C(X; [0, 1])$ in Definition 5.24.4, and $\beta X = \ol{e(X)}$.

(1): By Theorem 5.16.8 and Proposition 5.8.3, $[0, 1]^{C(X; [0, 1])}$ is a compact Hausdorff space. By definition, $e(X)$ is dense in $\beta X$.

(U1): For each $f \in C(X; [0, 1])$, $\pi_{f} \in C([0, 1]^{C(X; [0, 1])}; [0, 1])$ is an extension of $f$ to $e(x)$.

(U2): Let $(Y, \varphi)$ be a compactification of $X$. For each $f \in C(Y; [0, 1])$, by (U1), there exists a unique $\beta(f \circ \varphi) \in C(X; [0, 1])$ such that the following diagram commutes:

Let $e': Y \to [0, 1]^{C(Y; [0, 1])}$ be the embedding of $Y$ into $[0, 1]^{C(Y; [0, 1])}$ associated with $C(Y; [0, 1])$, then by (U) of the product topology, there exists $\beta(e' \circ \varphi) \in C(\beta X; [0, 1])$ such that the following diagram commutes:

Since $Y$ is a compact Hausdorff space, $e'(Y)$ is closed by Proposition 5.16.3 and Proposition 5.16.4. As $e'$ is an embedding, identify $Y$ as a subspace of $[0, 1]^{C(Y; [0, 1])}$. Given that $e(X)$ is dense in $\beta X$, the the image of $\beta (e' \circ \varphi)$ lies in $Y$ by Proposition 5.5.3. Therefore under the identification, the following diagram commutes:

$\square$

Proof. (1): Firstly, $\beta X = U_{X}$. For any $A, B \subset X$, $U_{A \cap B}= U_{A} \cap U_{B}$. Thus $\fB$ is a base for the topology that it generates.

Let $\mathfrak{U}, \mathfrak{V}\in \beta X$ be distinct, then there exists $A \subset X$ with $A \in \mathfrak{U}$ and $A^{c} \in \mathfrak{V}$, so $\mathfrak{U}\in U_{A}$, $\mathfrak{V}\in U_{A^c}$, and $U_{A} \cap U_{A^c}= \emptyset$. Thus $\beta X$ is Hausdorff.

Let $\mathcal{U}\subset 2^{\beta X}$ be an open cover of $X$. Since $\fB$ is a base for $\beta X$, assume without loss of generality that $\mathcal{U}$ is of the form $\bracsn{U_A|A \in \mathcal{S}}$, where $\mathcal{S}\subset 2^{X} \setminus \bracsn{\emptyset}$. Suppose for contradiction that $\mathcal{U}$ admits no finite subcover, then for any $F \subset \mathcal{S}$ finite,

In particular, $\bigcap_{A \in F}A^{c} \ne \emptyset$. By the ultrafilter lemma, $\bracsn{A^c|A \in \mathcal{S}}$ generates an ultrafilter $\fU \in \beta X$. Since $\fU \not\in U_{A}$ for all $A \in \mathcal{S}$, $\mathcal{U}$ does not cover $X$, which is a contradiction. Thus $\beta X$ is compact.

Finally, since $X$ is discrete, $e$ is automatically continuous. On the other hand, for each $x \in X$, $U_{\bracsn{x}}= e(\bracsn{x})$ is open in $\beta X$. In addition, for any $\emptyset \ne A \subset X$, $e(x) \in U_{A}$ for all $x \in A$, so $U_{A} \cap e(X) \ne \emptyset$, and $e(X)$ is dense in $\beta X$. Therefore $e(X)$ is embedded as a dense subspace of $\beta X$.

(U1): Let $f \in C(X; [0, 1])$, and let $\beta f: \beta X \to [0, 1]$ be defined by $\beta f(\fU) = \lim_{\fU}f$. Fix $\fU \in \beta X$ and $\eps > 0$, then there exists $A \in \fU$ such that $f(A) \subset B_{[0, 1]}(\lim_{\fU}f, \eps)$. Thus for any $\fV \in U_{A}$,

and $|\beta f(\fU) - \beta f(\fV)| \le \eps$. Since $A \in \fU$, $U_{A} \in \cn_{\beta X}(\fU)$, so $\beta f$ is continuous at $\fU$. As this holds for all $\fU \in \beta X$, $\beta f$ is continuous.$\square$

Proof. For each $x \in X$, let $U \in \cn_{X}(x)$ be a compact neighbourhood. Since $Y$ is a compact Hausdorff space, $\varphi(U)$ is closed by Proposition 5.16.4. As $\varphi \in C(X; Y)$ is an embedding, there exists $V \in \cn_{Y}(\varphi(x))$ such that $\varphi(U) = \varphi(X) \cap V$. Given that $\varphi(X)$ is dense in $Y$, $\varphi(U) = \ol{\varphi(X) \cap V}\supset V$. Therefore $\varphi(U) \in \cn_{Y}(\varphi(x))$, and $\varphi(X)$ is open in $Y$.$\square$

Proof. Let $\fB \subset 2^{X}$ be a base for the topology on $\beta X$, then for each $A \subset \beta X$ clopen, there exists $\mathcal{F}\subset \fB$ finite such that $A = \bigcup_{U \in \mathcal{F}}U$. Should $\fB$ be countable, then there are at most countably many clopen subsets of $\beta X$.

However, for each $A \subset X$, let $U_{A} = \bracsn{\fU \in \beta X|A \in \fU}$, then $\bracsn{U_A|A \subset X}$ represents uncountably many clopen sets. Therefore $\fB$ cannot be countable, $\beta X$ is not second countable.

If $\beta X$ was metrisable, then it is second countable by compactness and Proposition 8.1.2. Therefore $\beta X$ is not metrisable.$\square$

Proof. (1): Let $\infty$ be a point not in $X$, $X^{*} = X \sqcup \bracs{\infty}$, and $\mathcal{T}^{*} \subset 2^{X^*}$ such that for each $U \in \mathcal{T}^{*}$, one of the following holds:

1. (a)

   $U \in \mathcal{T}$.

2. (b)

   $\infty \in U$ and $U^{c} \subset X$ is compact.

Let $\seqi{U}\subset \mathcal{T}^{*}$ be an open cover of $X$, then there exists $i \in I$ such that $\infty \in U$. In which case, $U_{i}$ must satisfy (b), so there exists $J \subset I$ finite such that $\bigcup_{j \in J}U_{j} \supset U_{i}^{c}$, and $\bracsn{U_j|j \in J \cup \bracs{i}}$ is a finite subcover. Now, let $x \in X$, then since $X$ is locally compact, there exists a relatively compact neighbourhood $U \in \cn_{X}^{o}(x)$. In which case, $\ol{U}^{c} \in \cn_{X^*}(\infty)$ with $U \cap \ol{U}^{c} = \emptyset$. Therefore $X^{*}$ is a compact Hausdorff space.

Let $\iota: X \to X^{*}$ be the inclusion map. For each $U \in \mathcal{T}^{*}$ satisfying (b), $\iota^{-1}(U) = U \cap X$. Since $U^{c} \subset X$ is compact, $U \cap X$ is open by Proposition 5.16.4, so $\iota \in C(X; X^{*})$. By (a), $\iota$ is an embedding.

(U): Let

Let $U \subset X^{*}$ with $\infty \not\in U$, then $(\varphi^{*})^{-1}(U) = \varphi(U)$ is open in $\varphi(X)$ because $\varphi$ is an embedding, and open in $Y$ by Lemma 5.25.4. On the other hand, for each $V \in \cn_{X^*}^{o}(\infty)$,

Since $\varphi \in C(X; Y)$ is an embedding, $V$ is relatively open in $\varphi(X)$, so $V \cup (Y \setminus \varphi(X))$ is open in $Y$.$\square$