Added the Singer representation theorem.

src/topology/main/compact.tex

@@ -82,3 +82,28 @@
 
     By compactness of $A$, there exists $A_0 \subset A$ finite such that $\bigcup_{x \in A_0}U_x \in \cn(A)$. Let $U = \bigcup_{x \in A_0}U_x \in \cn(A)$ and $V = \bigcap_{x \in A_0}V_x \in \cn(B)$, then $A \cap B = \emptyset$.
 \end{proof}
+
+\begin{lemma}[Tube Lemma]
+\label{lemma:tube-lemma}
+    Let $X$ be a topological space and $Y$ be a compact space, then
+    \begin{enumerate}
+        \item For any $x \in X$ and $U \in \cn_{X \times Y}^o(\bracs{x} \times Y)$, there exists $V \in \cn_X(x)$ such that $V \times Y \subset U$. 
+        \item For any $A \subset X$ and $U \in \cn_{X \times Y}^o(A \times Y)$, there exists $V \in \cn_X(A)$ such that $V \times Y \subset U$. 
+    \end{enumerate}
+    
+\end{lemma}
+\begin{proof}
+    (1): For each $y \in Y$, let $U_y \in \cn_X(x)$ and $V_y \in \cn_Y(y)$ such that $(x, y) \in U_y \times V_y \subset U$, then $\bracs{V_y|y \in Y}$ is an open cover of $Y$. By compactness, there exists $\seqf{y_j} \subset Y$ such that $\bigcup_{j = 1}^nV_{y_j} = Y$. Let $V = \bigcap_{j = 1}^n U_{y_j}$, then $V \in \cn_X(x)$ and $V \times Y \subset U$. 
+
+    (2): For each $x \in A$, there exists $U_x \in \cn_X(x)$ such that $U_x \times Y \subset U$. Let $V = \bigcup_{x \in A}U_x$, then $V \times Y \subset U$. 
+\end{proof}
+
+
+\begin{theorem}[Tychonoff]
+\label{theorem:tychonoff}
+    Let $\seqi{X}$ be compact topological spaces, then $X = \prod_{i \in I}X_i$ is compact. 
+\end{theorem}
+\begin{proof}
+    Let $\fU \subset 2^X$ be an ultrafilter. By \autoref{proposition:imagefilterbase}, for each $i \in I$, $\pi_i(\fU)$ is an ultrafilter base. Since $X_i$ is compact, there exists $x_i \in X_i$ such that $\pi_i(\fU) \to x_i$. By the axiom of choice, there exists $x \in X$ such that $\pi_i(\fU) \to \pi_i(x)$ for all $i \in I$. Therefore $\fU \to x$ by \autoref{proposition:productfilterconvergence}.
+\end{proof}
+

src/topology/main/filters.tex

@@ -48,12 +48,19 @@
 
 \begin{proposition}
 \label{proposition:imagefilterbase}
-    Let $X, Y$ be sets, $f: X \to Y$ be a function, and $\fB \subset 2^X$ be a filter base, then $f(\fB) = \bracs{f(E)| E \in \fB}$ is also a filter base.
+    Let $X, Y$ be sets, $f: X \to Y$ be a function, and $\fB \subset 2^X$ be a filter base, then
+    \begin{enumerate}
+        \item $f(\fB) = \bracs{f(E)| E \in \fB}$ is also a filter base.
+        \item If $\fB$ is an ultrafilter base, then $f(\fB)$ is also an ultrafilter base. 
+    \end{enumerate}
+    
 \end{proposition}
 \begin{proof}
-    (FB1): Let $E, F \in \fB$, then there exists $G \in \fB$ such that $G \subset E \cap F$. In which case, $f(G) \subset f(E \cap F) \subset f(E) \cap f(F)$.
+    (1): Let $E, F \in \fB$, then there exists $G \in \fB$ such that $G \subset E \cap F$. In which case, $f(G) \subset f(E \cap F) \subset f(E) \cap f(F)$.
+    
+    For any $f(E) \in f(\fB)$, $E \ne \emptyset$ implies that $f(E) \ne \emptyset$.
 
-    (FB2): For any $f(E) \in f(\fB)$, $E \ne \emptyset$ implies that $f(E) \ne \emptyset$.
+    (2): Let $E \subset X$, then either $f^{-1}(E)$ or $f^{-1}(E^c)$ contains an element of $\fB$. Therefore either $E$ or $E^c$ contains an element of $f(\fB)$. 
 \end{proof}
 
 \begin{proposition}

src/topology/main/lch.tex

@@ -237,3 +237,13 @@
 
     Let $x \in X$, then there exists $n \in \natp$ such that $x \in U_n \setminus U_{n-1}$. In which case, if $n > 1$, then $x \in U_{n} \setminus \ol{U_{n - 2}} = V_{n-1}$. If $n = 1$, then $x \in U_{2} = V_1$. Thus $\seq{V_n}$ is an open cover of $X$. In addition, for any $m, n \in \natp$ with $m \le n$, $V_m \cap V_n \ne \emptyset$ implies that $n - m < 2$, so $\seq{V_n}$ is locally finite. By (2) of \autoref{proposition:lch-paracompact}, $X$ is paracompact.
 \end{proof}
+
+\begin{proposition}
+\label{proposition:lch-product}
+    Let $\seqi{X}$ be a family of LCH spaces. If $X_i$ is compact for all but finitely many $i \in I$, then $X = \prod_{i \in I}X_i$ is a LCH space. 
+\end{proposition}
+\begin{proof}
+    By \autoref{proposition:product-hausdorff}, $\prod_{i \in I}X_i$ is Hausdorff. Let $x \in \prod_{i \in I}X_i$ and $i \in I$. If $X_i$ is not compact, let $U_i \in \cn_{X_i}(\pi_i(x))$ be compact. Otherwise, let $U_i = X_i$. Let $U = \prod_{i \in I}U_i$, then since $U_i \ne X_i$ for only finitely many $i \in I$, $U \in \cn_X(x)$. By \hyperref[Tychonoff's Theorem]{theorem:tychonoff}, $U$ is compact. Therefore $X$ is locally compact. 
+\end{proof}
+
+

src/topology/main/product.tex

@@ -112,8 +112,8 @@ Let $\bracsn{(X_i, \topo_i)}_{i \in I}$ be a family of topological spaces and $\
     (1): By (U) of \autoref{definition:product-topology}/\autoref{definition:product-uniform}.
 
     (2): Consider the following diagram
-    \[
     % https://darknmt.github.io/res/xypic-editor/#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
+    \[
     \xymatrix{
     \prod_{i \in I}Y_i \ar@{->}[rd]^{\pi_i} &  \\
     \iota_P(\prod_{i \in I}X_i) \ar@{->}[u] & Y_i \\