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Bokuan Li
2026-03-06 14:06:15 -05:00
parent 173727665b
commit 5034bc4220
109 changed files with 1184 additions and 410 deletions
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47
src/topology/uniform/metric.tex
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@@ -39,21 +39,25 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
\[
B_i(x, r) = \bracs{y \in X| d(x, y) < r}
\]
and
\[
E(d_i, r) = \bracs{(x, y) \in X \times X| d(x, y) < r}
\]
then there exists a uniformity $\fU$ on $X$ such that:
\begin{enumerate}
\item The family
\[
\fB = \bracs{\bigcap_{j \in J}E(d_j, r) \bigg | J \subset I \text{ finite}, r > 0}
\]
forms a fundamental system of entourages consisting of symmetric open sets.
\item For any $x \in X$,
\[
\cb(x) = \bracs{\bigcap_{j \in J}B_j(x, r) \bigg | J \subset I \text{ finite}, r > 0}
\]
is a fundamental system of neighbourhoods at $x$.
\item For each $U \subset X$, $U$ is open if and only if for every $x \in U$, there exists $J \subset I$ finite and $r > 0$ such that $\bigcap_{j \in J}B_j(x, r) \subset U$.
\item For each $i \in I$, $d_i \in UC(X \times X; [0, \infty))$.
@@ -62,17 +66,19 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
The uniformity $\fU$ is the \textbf{pseudometric uniformity} induced by $\seqi{d}$, and the topology induced by $\fU$ is the \textbf{pseudometric topology} on $X$ induced by $\seqi{d}$.
\end{definition}
\begin{proof}
(1, fundamental system): To see that $\fB$ is a fundamental system of entourages for a uniformity on $X$, it is sufficient to verify the conditions of \ref{proposition:fundamental-entourage-criterion}.
(1, fundamental system): To see that $\fB$ is a fundamental system of entourages for a uniformity on $X$, it is sufficient to verify the conditions of \autoref{proposition:fundamental-entourage-criterion}.
\begin{enumerate}
\item[(FB1)] For any $J, J' \subset I$ finite and $r, r' > 0$,
\[
\bigcap_{j \in J \cup J'}E(d_j, \min(r,r')E(d_j, r \wedge r') \subset \paren{\bigcap_{j \in J}E(d_j, r)} \cap \paren{\bigcap_{j \in J'}E(d_j, r')}
\]
\item[(UB1)] For each $i \in I$, $d(x, x) = 0$ for all $x \in X$. Thus for any $i \in I$ and $r > 0$, $E(d_i, r)$ contains the diagonal.
\item[(UB2)] For each $J \subset I$ finite and $r > 0$,
\[
\paren{\bigcap_{j \in J}E(d_j, r/2)} \circ \paren{\bigcap_{j \in J}E(d_j, r)} \subset \bigcap_{j \in J}E(d_j, r/2) \circ E(d_j, r/2) \subset \bigcap_{j \in J}E(d_j, r)
\]
by the triangle inequality.
\end{enumerate}
@@ -80,6 +86,7 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
\[
\cb(x) = \bracs{U(x)| U \in \fB}
\]
is a fundamental system of neighbourhoods at $x$.
(3): By definition of the uniform topology, for any $U \subset X$, $U$ is open if and only if for any $x \in U$, there exists $V \in \fU$ such that $x \in V(x) \subset U$. As $\fB$ is a fundamental system of entourages for $\fU$, this is equivalent to the existence of $J \subset I$ finite and $r > 0$ such that
@@ -87,13 +94,14 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
x \in \paren{\bigcap_{j \in J}E(d_j, r)}(x) = \bigcap_{j \in J}B_j(x, r) \subset U
\]
(1, symmetric open): Let $i \in I$ and $r > 0$. Since $d_i$ is symmetric, so is $E(d_i, r)$. For any $(x, y) \in E(d_i, r)$, let $s = r - d_i(x, y)$, then for any $(x', y') \in X \times X$ with $d_i(x, x') < s/2$ and $d_i(y, y') < s/2$, $d(x', y') < s + d_i(x, y) = r$. Thus $B_i(x, s/2) \times B_i(y, s/2) \subset E(d_i, r)$.
By (3), $B_i(x, s/2) \in \cn(x)$ and $B_i(y, s/2) \in \cn(y)$, so $B_i(x, s/2) \times B_i(y, s/2) \in \cn((x, y))$. As such, $E(d_i, r)$ is open by \ref{lemma:openneighbourhood}.
By (3), $B_i(x, s/2) \in \cn(x)$ and $B_i(y, s/2) \in \cn(y)$, so $B_i(x, s/2) \times B_i(y, s/2) \in \cn((x, y))$. As such, $E(d_i, r)$ is open by \autoref{lemma:openneighbourhood}.
(4): By \ref{lemma:pseudometric-continuous}.
(4): By \autoref{lemma:pseudometric-continuous}.
(5): By \ref{lemma:pseudometric-continuous}, $E(d_i, r) \in \mathfrak{V}$ for all $i \in I$ and $r > 0$, so $\mathfrak{V} \supset \mathfrak{B}$ by (F2), and $\mathfrak{V} \supset \mathfrak{U}$.
(5): By \autoref{lemma:pseudometric-continuous}, $E(d_i, r) \in \mathfrak{V}$ for all $i \in I$ and $r > 0$, so $\mathfrak{V} \supset \mathfrak{B}$ by (F2), and $\mathfrak{V} \supset \mathfrak{U}$.
\end{proof}
\begin{lemma}
@@ -108,6 +116,7 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
\[
U_{n+1} \subset E(d, 2^{-n}) \subset U_{n-1}
\]
for each $n \in \natp$.
\end{lemma}
\begin{proof}
@@ -115,10 +124,12 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
\[
\rho: X \times X \to [0, 1] \quad (x, y) \mapsto \inf\bracs{2^{-n}| n \in \natz, (x, y) \in U_n}
\]
and $d: X \times X \to [0, 1]$ by
\[
d(x, y) = \inf\bracs{\sum_{j = 1}^n\rho(x_{j-1}, x_j) \bigg | \seqf{x_j} \subset X, x_0 = x, x_1 = y, n \in \natp}
\]
then
\begin{enumerate}
\item[(PM1)] For any $x \in X$, $x \in \bigcap_{n \in \natp}U_n$. Thus $\rho(x, x) = 0$ and $d(x, x) \le \rho(x, x) = 0$.
@@ -127,6 +138,7 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
\[
d(x, z) \le \sum_{j = 1}^n \rho(x_{j - 1}, x_j) + \sum_{j = 1}^m \rho(y_{j - 1}, y_j)
\]
As this holds for all such $\seqf{x_j}$ and $\seqf[m]{y_j}$, $d(x, z) \le d(x, y) + d(y, z)$.
\end{enumerate}
so $d$ is a pseudometric.
@@ -139,6 +151,7 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
\[
(x, y) \in U_{n+1} \circ U_{n + 1} \circ U_{n + 1} \subset U_n \circ U_n \subset U_{n-1}
\]
\end{proof}
\begin{remark}
@@ -156,13 +169,14 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
Let $(X, \fU)$ be a uniform space, then $\fU$ is the pseudometric uniformity induced by the family of all uniformly continuous pseudometrics on $X$.
\end{theorem}
\begin{proof}
Let $\seqi{d}$ be the family of all uniformly continuous pseudometrics on $X$, and $\mathfrak{V}$ be the pseudometric uniformity induced by $\seqi{d}$. By (U) of \ref{definition:pseudometric-uniformity}, $\fU \supset \mathfrak{V}$.
Let $\seqi{d}$ be the family of all uniformly continuous pseudometrics on $X$, and $\mathfrak{V}$ be the pseudometric uniformity induced by $\seqi{d}$. By (U) of \autoref{definition:pseudometric-uniformity}, $\fU \supset \mathfrak{V}$.
On the other hand, let $U_1 \in \mathfrak{U}$. By (U3), there exists $\seq{U_n} \subset \mathfrak{U}$ such that $U_{n + 1} \circ U_{n + 1} \subset U_n$ for all $n \in \natp$. Let $U_0 = X \times X$, then $\bracsn{U_n}_0^\infty \subset \fU$ satisfies the hypothesis of \ref{lemma:uniform-sequence-pseudometric}. Thus there exists a pseudometric $d: X \times X \to [0, \infty)$ such that for all $n \in \natp$,
On the other hand, let $U_1 \in \mathfrak{U}$. By (U3), there exists $\seq{U_n} \subset \mathfrak{U}$ such that $U_{n + 1} \circ U_{n + 1} \subset U_n$ for all $n \in \natp$. Let $U_0 = X \times X$, then $\bracsn{U_n}_0^\infty \subset \fU$ satisfies the hypothesis of \autoref{lemma:uniform-sequence-pseudometric}. Thus there exists a pseudometric $d: X \times X \to [0, \infty)$ such that for all $n \in \natp$,
\[
U_{n + 1} \subset E(d, 2^{-n}) \subset U_{n-1}
\]
By \ref{definition:pseudometric-uniformity}, $d$ is a uniformly continuous pseudometric on $X$. Since $E(d, 1/4) \subset U_1$, $U_1 \in \mathfrak{V}$. Therefore $\fU = \mathfrak{V}$.
By \autoref{definition:pseudometric-uniformity}, $d$ is a uniformly continuous pseudometric on $X$. Since $E(d, 1/4) \subset U_1$, $U_1 \in \mathfrak{V}$. Therefore $\fU = \mathfrak{V}$.
\end{proof}
\begin{proposition}
@@ -179,9 +193,10 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
\[
\bigcap_{j \in J}E(d_j, r) \subset U
\]
In which case, there must exist $j \in J$ such that $d_j(x, y) \ge r > 0$.
(3) $\Rightarrow$ (1): Let $x, y \in X$ with $x \ne y$ and $d$ be a continuous pseudometric on $X$ such that $r = d(x, y) > 0$. By \ref{theorem:uniform-pseudometric}, $E(d, r) \in \fU$. Therefore $\bigcup_{U \in \fU}U = \Delta$, and $X$ is separated by \ref{definition:uniform-separated}.
(3) $\Rightarrow$ (1): Let $x, y \in X$ with $x \ne y$ and $d$ be a continuous pseudometric on $X$ such that $r = d(x, y) > 0$. By \autoref{theorem:uniform-pseudometric}, $E(d, r) \in \fU$. Therefore $\bigcup_{U \in \fU}U = \Delta$, and $X$ is separated by \autoref{definition:uniform-separated}.
\end{proof}
@@ -196,6 +211,7 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
\[
\td d: X \times X \to [0, \infty) \quad (x, y) \mapsto d(x, y) \wedge 1
\]
is equivalent to $d$.
\end{lemma}
\begin{proof}
@@ -207,10 +223,11 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
Let $X$ be a set and $\seq{d_n}$ be pseudometrics on $X$, then there exists a pseudometric $d: X \times X \to [0, \infty)$ equivalent to $\seq{d_n}$.
\end{proposition}
\begin{proof}
Using \ref{lemma:pseudometric-clamp}, assume without loss of generality that for each $n \in \natp$, $d_n$ takes values in $[0, 1]$. Let
Using \autoref{lemma:pseudometric-clamp}, assume without loss of generality that for each $n \in \natp$, $d_n$ takes values in $[0, 1]$. Let
\[
d(x, y) = \sum_{n \in \natp}\frac{d_n(x, y)}{2^n}
\]
then $d$ is a well-defined a pseudometric.
Let $r > 0$, then there exists $n \in \natp$ such that $2^{-n} < r$. Take $s = r - 2^{-n}$, then $\bigcap_{k = 1}^nE(d_k, s) \subset E(d, r)$. On the other hand, for any $n \in \natp$ and $r > 0$, $E(d, r/2^n) \subset \bigcap_{k = 1}^n E(d_k, r)$. Therefore $\seq{d_n}$ and $d$ are equivalent.
@@ -232,26 +249,30 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
\item[(b)] For each $1 \le k \le n$, $V_k \subset U_k$.
\item[(c)] For each $1 \le k < n$, $V_{k+1} \circ V_{k+1} \subset V_{k}$.
\end{enumerate}
Let $W = V_n \cap U_{n+1}$, then by \ref{lemma:symmetricfundamentalentourage}, there exists $V_{n+1} \in \fU$ symmetric such that $V_{n+1} \circ V_{n+1} \subset W$. Thus $\bracs{V_k|1 \le k \le n + 1} \subset \fU$ satisfies (a), (b), and (c) for $n + 1$.
Let $W = V_n \cap U_{n+1}$, then by \autoref{lemma:symmetricfundamentalentourage}, there exists $V_{n+1} \in \fU$ symmetric such that $V_{n+1} \circ V_{n+1} \subset W$. Thus $\bracs{V_k|1 \le k \le n + 1} \subset \fU$ satisfies (a), (b), and (c) for $n + 1$.
Let $V_0 = X \times X$, then by \ref{lemma:uniform-sequence-pseudometric}, there exists a pseudometric $d: X \times X \to [0, \infty)$ such that for each $n \in \natp$,
Let $V_0 = X \times X$, then by \autoref{lemma:uniform-sequence-pseudometric}, there exists a pseudometric $d: X \times X \to [0, \infty)$ such that for each $n \in \natp$,
\[
V_{n+1} \subset E(d, 2^{-n}) \subset V_{n-1}
\]
For any $U \in \fU$, there exists $n \in \nat$ such that
\[
U \supset U_n \supset V_n \supset E(d, 2^{-n-1})
\]
so $d$ induces the uniformity on $\fU$.
(3) $\Rightarrow$ (1): By (1) of \ref{definition:pseudometric-uniformity},
(3) $\Rightarrow$ (1): By (1) of \autoref{definition:pseudometric-uniformity},
\[
\fB = \bracs{\bigcap_{j \in J}E(d_j, r) \bigg | J \subset \nat \text{ finite}, r > 0}
\]
is a fundamental system of entourages for $\fU$. Since for any $r > 0$, there exists $q \in \rational \cap (0, r)$,
\[
\bracs{\bigcap_{j \in J}E(d_j, r) \bigg | J \subset \nat \text{ finite}, r \in \rational, r > 0}
\]
is a countable fundamental system of entourages for $\fU$.
\end{proof}
@@ -266,5 +287,5 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
\begin{proof}
(1) $\Rightarrow$ (2): $d_X(x, y) = d_Y(f(x), f(y))$ is a uniformly continuous pseudometric.
(2) $\Rightarrow$ (1): Let $U$ be an entourage of $Y$. By \ref{theorem:uniform-pseudometric}, there exists a uniformly continuous pseudometric $d_Y$ on $Y$ and $r > 0$ such that $E(d_Y, r) \subset U$. By assumption, there exists a uniformly continuous pseudometric $d_X$ on $X$ such that $d_Y(f(x), f(y)) \le d_X(x, y)$. In which case, $(f \times f)(E(d_X, r)) \subset U$, and the pre-image of $U$ by $(f \times f)$ is an entourage in $X$ by \ref{definition:pseudometric-uniformity}, so $f \in UC(X; Y)$.
(2) $\Rightarrow$ (1): Let $U$ be an entourage of $Y$. By \autoref{theorem:uniform-pseudometric}, there exists a uniformly continuous pseudometric $d_Y$ on $Y$ and $r > 0$ such that $E(d_Y, r) \subset U$. By assumption, there exists a uniformly continuous pseudometric $d_X$ on $X$ such that $d_Y(f(x), f(y)) \le d_X(x, y)$. In which case, $(f \times f)(E(d_X, r)) \subset U$, and the pre-image of $U$ by $(f \times f)$ is an entourage in $X$ by \autoref{definition:pseudometric-uniformity}, so $f \in UC(X; Y)$.
\end{proof}