View on GitHub

memo

Alexandroff Extension

Alexandroff Extension

Theorem

\[\begin{eqnarray} \mathcal{T} & := & \{ (X \setminus C) \cup \{\infty\} \mid C \subseteq X: \text{ compact and closed} \}, \nonumber \\ \mathcal{O}^{*} & := & \mathcal{T} \cup \mathcal{O}_{X} . \nonumber \end{eqnarray}\]

\((X^{*}, \mathcal{O}^{*})\) is a topological space. Moreover,

(1) The embedding \(c: X \rightarrow X^{*}\) is continuous and $c(X)$ is open in $X^{*}$.

(2) $X^{*}$ is compact

(3) If $X$ is not compact, $c(X)$ is dense in $X^{*}$

(4) $X^{*}$ is Hausdorff if and only if $X$ is Hausdorff and locally compact

(5) $X^{*}$ is $T_{1}$ if and only if $X$ is $T_{1}$

(6) $(X, \mathcal{O}_{X})$ is a subspace of $X^{*}$ with the subspace topology.

proof

First of all,

\[\begin{eqnarray} \mathcal{A} & := & \{ A \subseteq X^{*} \mid X^{*} \setminus A: \text{ closed and compact in } X \} \nonumber \\ \mathcal{O}^{*} & = & \mathcal{O}_{X} \cup \mathcal{A} . \nonumber \end{eqnarray}\]

Indeed, if $A \in \mathcal{T}$,

\[\begin{eqnarray} X^{*} \setminus A & = & X^{*} \setminus ((X \setminus C) \cup \{\infty\}) \nonumber \\ & = & X \setminus (X \setminus C) \nonumber \\ & = & C . \nonumber \end{eqnarray}\]

If $A \in \mathcal{A}$,

\[\begin{eqnarray} A & = & X^{*} \setminus (X^{*} \setminus A) \nonumber \\ & = & (X \cup \{\infty\}) \setminus (X^{*} \setminus A) \nonumber \\ & = & (X \setminus (X^{*} \setminus A)) \cup (\{\infty\} \setminus (X^{*} \setminus A)) \nonumber \\ & = & (X \setminus (X^{*} \setminus A)) \cup \{\infty\} \nonumber \end{eqnarray}\]

where the third euqality holds $A$ always contians \(\{\infty\}\).

(6)

If $A \in \mathcal{O}^{*}$,

\[\begin{eqnarray} A \cap X & = & ((X \setminus A) \cap \{\infty\}) \cap X \nonumber \\ & = & (X \setminus A) \cap X \nonumber \\ & = & (X \setminus A) \in \mathcal{O}_{X} . \nonumber \end{eqnarray}\]

$\mathcal{O}_{X}$ is a subspace topology of $X^{*}$ if \(X^{*}\) is a topological space.

($X^{*}$ is Topology)

Obviously, \(X^{*}, \emptyset \in \mathcal{O}^{*}\)

Let $A_{i} \in \mathcal{O}^{*}$ $(i = 1, \ldots, N)$. And show \(\cap_{i} A_{i} \in \mathcal{O}^{*}\). If \(A_{i} \in \mathcal{O}_{X}\) for all $i$, it’s obvious. If $A_{i} \in \mathcal{T}$ for all $i$,

\[\begin{eqnarray} \cap_{i} (X \setminus C_{i} \cup \{\infty\}) & = & \cap_{i}(X \setminus C_{i}) \cup \cap_{i}(\{\infty\}) \nonumber \\ & = & (X \setminus \cap_{i} C_{i}) \cup \{\infty\} \nonumber \\ & = & (X \setminus \cap_{i} C_{i}) \cup \{\infty\} \nonumber \end{eqnarray}\]

For the finite intersection of the comapct set is compact, so $\cap_{i} C_{i}$ is compact. $\cap_{i} C_{i}$ is closed so the finite intersection of $A_{i}$ is $\mathcal{T}$.

\[\begin{eqnarray} A_{1} \cap A_{2} & = & ((X \setminus C_{1}) \cup \{\infty\}) \cap A_{2} \nonumber \\ & = & (X \setminus C_{1}) \cap A_{2} \in \mathcal{O}_{X} \nonumber \end{eqnarray}\]

Thus, the finite intersection of $A_{i} \in \mathcal{O}^{*}$ is in $\mathcal{O}$.

Let \(A_{i} \in \mathcal{O}^{*}\) for all $i$. We will show that \(\cup_{i} A_{i} \in \mathcal{O}^{*}\).

If $A_{i} \in \mathcal{A}$, $\cup_{i} A_{i} \in \mathcal{A}$. Inded,

\[X^{*} \setminus \cup_{i} A_{i} = \cap_{i} (X^{*} \setminus A_{i}) .\]

The RHS is the intersection of closed sets so the RHS is closed. For some $j$,

\[X^{*} \setminus \cup_{i} A_{i} \subseteq X^{*} \setminus A_{j} .\]

The closed subset of the compact set is compact so the LHS is compact. If $A_{1} \in \mathcal{A}$ and $A_{2} \in \mathcal{O}_{X}$,

\[\begin{eqnarray} X^{*} \setminus (A_{1} \cup A_{2}) & = & (X^{*} \setminus A_{1}) \cap (X^{*} \setminus A_{2}) \nonumber \\ & = & (X \setminus A_{1}) \cap (X^{*} \setminus A_{2}) \nonumber \\ & = & (X \setminus A_{1}) \cap (X \setminus A_{2}) \nonumber \end{eqnarray}\]

then it is closed. Thus, it’s compact as well. Finally, $A_{i} \in \mathcal{O}^{*}$ is either in $\mathcal{A}$ or $\mathcal{O}_{X}$. So

\[\bigcup_{i} A_{i} = \bigcup_{i} A_{i} \cup \bigcup_{i} A_{i} .\]

The first term in $\mathcal{A}$ and the second term in $\mathcal{O}_{X}$. This is the case of the two union.

(2)

($\Rightarrow$)

Let \(U_{i} \in \mathcal{O}^{*}\) be a open cover of $X^{*}$. There is a $U_{k}$ such that \(\{\infty\} \in U_{k}\). That is, $U_{k} \in \mathcal{A}$. By the definition, \(X^{*} \setminus U_{k}\) is compact and closed in $X$.

Let

\[U_{i}^{\prime} := U_{i} \setminus \{\infty\} = S \cap U_{i} \in \mathcal{O}_{X} .\]

$U_{i}^{\prime}$ is an open cover of $X$. There is a finite open cover \(U_{j}^{\prime}\) of $X^{*} \setminus U_{k}$. Thus,

\[\bigcup_{j} U_{j} \cup U_{k}\]

is an finite open cover of $X^{*}$.

(4)

We show if $X$ is locally compact and Hasudorff, \(X^{*}\) is Hausdorff. Let \(x, y \in X^{*}\) and $x \neq y$.

If $x, y \in X$, there are neighbors which separate $x, y$.

Let $x \in X$ and \(y = \infty\). Since $X$ is locally compact, there is a compact neighbor $U$ of $x$. Since $X$ is Hausodorff, $U$ is closed. That is, $X^{*} \setminus U \in \mathcal{A}$. It’s separated by $U$ and \(X^{*} \setminus U\).

($\Leftarrow$)

We show if \(X^{*}\) is Hausdorff, $X$ is locally compact and Hausdorff. Since $X$ is supace of \(X^{*}\) which is Hausdorff. Hence $X$ is Hausdorff.

Let $x \in X$. Since $X^{*}$ is Hausdorff, there are neighbors $U$ and $V$ of $x$ and $\infty$ in \(X^{*}\) such that $U \cap V = \emptyset$.

\[\mathrm{cl}_{X^{*}}(U) \cap V = \emptyset .\]

This implies \(\infty \notin \mathrm{cl}_{X^{*}}(U)\). Hence \(\mathrm{cl}_{X^{*}}(U) \subseteq X\).

Since \(X^{*}\) is Hausdorff and \(\mathrm{cl}_{X^{*}}(U)\) is closed, \(\mathrm{cl}_{X^{*}}(U)\) is compact in \(X^{*}\). $X$ is subspace of $X^{*}$ so \(\mathrm{cl}_{X^{*}}(U)\) is compact in $X$.

Thus, \(\mathrm{cl}_{X^{*}}(U)\) is a compact neighbor of $x$ in $X$.

$\Box$

Example

\[\begin{eqnarray} \mathbb{Q}^{*} & := & \mathbb{Q} \cup \{\infty\} \nonumber \\ \mathcal{O}^{*} & := & \{ O \subseteq \mathbb{Q}^{*} \mid \mathbb{Q}^{*} \setminus O: \text{ closed and compact in } \mathbb{Q} \} \cup \mathcal{O} . \nonumber \end{eqnarray}\]

(1) non-Hausdorff and compact

(2) KC

(3) weak Hausdorff

■

Example

\[X := \mathbb{Q}^{*} \times \mathbb{Q}^{*}\]

with product topology.

(1) non-Hausdorff and compact

(2) not KC

\(\Delta: \mathbb{Q}^{*} \rightarrow X\) $\Detal(x) := (x, x)$. \(D := \Delta(\mathbb{Q}^{*})\) is compact but not closed in $X$. $\Delta$ and $\Delta^{-1}$ is continuous. $\mathbb{Q}^{*}$ is compact in $X$ so $D$ is compact.

$D$ is not closed in $X$.

(3) weak Hausdorff

■

Reference