jerrylicious/garden
1
0
Fork 0
Code Issues Pull Requests Actions Packages Projects Releases Wiki Activity

Pseudometricised function spaces.

Browse Source
This commit is contained in:
Bokuan Li
2026-01-23 14:01:37 -05:00
parent e847ead004
commit 3fc32813d8
2 changed files with 80 additions and 46 deletions
Download Patch File Download Diff File Expand all files Collapse all files

85
src/topology/uniform/metric.tex
View File

@@ -23,13 +23,13 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
Let $(X, \fU)$ be a uniform space and $d: X \times X \to [0, \infty)$ be a pseudometric, then the following are equivalent:
\begin{enumerate}
\item $d \in UC(X \times X; [0, \infty))$.
\item For each $r > 0$, $U_r = \bracs{(x, y) \in X \times X| d(x, y) < r} \in \fU$.
\item For each $r > 0$, $E(d, r) = \bracs{(x, y) \in X \times X| d(x, y) < r} \in \fU$.
\end{enumerate}
\end{lemma}
\begin{proof}
(1) $\Rightarrow$ (2): Let $r > 0$, then there exists $V \in \fU$ symmetric such that for any $(x, x'), (y, y') \in V$, $\abs{d(x, y) - d(x', y')} < r$. In particular, for any $(x, y) \in V$, $(x, x), (x, y) \in V$. Thus $d(x, y) < d(x, x) + r = r$, $V \subset U_r$, and $U_r \in \fU$.
(1) $\Rightarrow$ (2): Let $r > 0$, then there exists $V \in \fU$ symmetric such that for any $(x, x'), (y, y') \in V$, $\abs{d(x, y) - d(x', y')} < r$. In particular, for any $(x, y) \in V$, $(x, x), (x, y) \in V$. Thus $d(x, y) < d(x, x) + r = r$, $V \subset E(d, r)$, and $E(d, r) \in \fU$.
(2) $\Rightarrow$ (1): Let $r > 0$, then for any $(x, x'), (y, y') \in U_{r/2}$, $\abs{d(x, y) - d(x', y')} < r$ by the triangle inequality.
(2) $\Rightarrow$ (1): Let $r > 0$, then for any $(x, x'), (y, y') \in E(d, r/2)$, $\abs{d(x, y) - d(x', y')} < r$ by the triangle inequality.
\end{proof}
@@ -41,13 +41,13 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
\]
and
\[
U_{i, r} = \bracs{(x, y) \in X \times X| d(x, y) < r}
E(d_i, r) = \bracs{(x, y) \in X \times X| d(x, y) < r}
\]
then there exists a uniformity $\fU$ on $X$ such that:
\begin{enumerate}
\item The family
\[
\fB = \bracs{\bigcap_{j \in J}U_{j, r} \bigg | J \subset I \text{ finite}, r > 0}
\fB = \bracs{\bigcap_{j \in J}E(d_j, r) \bigg | J \subset I \text{ finite}, r > 0}
\]
forms a fundamental system of entourages consisting of symmetric open sets.
\item For any $x \in X$,
@@ -66,12 +66,12 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
\begin{enumerate}
\item[(FB1)] For any $J, J' \subset I$ finite and $r, r' > 0$,
\[
\bigcap_{j \in J \cup J'}U_{j, \min(r, r')} \subset \paren{\bigcap_{j \in J}U_{j, r}} \cap \paren{\bigcap_{j \in J'}U_{j, r'}}
\bigcap_{j \in J \cup J'}E(d_j, \min(r,r')E(d_j, r \wedge r') \subset \paren{\bigcap_{j \in J}E(d_j, r)} \cap \paren{\bigcap_{j \in J'}E(d_j, r')}
\]
\item[(UB1)] For each $i \in I$, $d(x, x) = 0$ for all $x \in X$. Thus for any $i \in I$ and $r > 0$, $U_{i, r}$ contains the diagonal.
\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 $r > 0$,
\[
\paren{\bigcap_{j \in J}U_{j, r/2}} \circ \paren{\bigcap_{j \in J}U_{j, r}} \subset \bigcap_{j \in J}U_{j, r/2} \circ U_{j, r/2} \subset \bigcap_{j \in J}U_{j, r}
\paren{\bigcap_{j \in J}E(d_j, r/2)} \circ \paren{\bigcap_{j \in J}E(d_j, r)} \subset \bigcap_{j \in J}E(d_j, r/2) \circ E(d_j, r/2) \subset \bigcap_{j \in J}E(d_j, r)
\]
by the triangle inequality.
\end{enumerate}
@@ -84,16 +84,16 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
(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
\[
x \in \paren{\bigcap_{j \in J}U_{j, r}}(x) = \bigcap_{j \in J}B_j(x, r) \subset U
x \in \paren{\bigcap_{j \in J}E(d_j, r)}(x) = \bigcap_{j \in J}B_j(x, r) \subset U
\]
(1, symmetric open): Let $i \in I$ and $r > 0$. Since $d_i$ is symmetric, so is $U_{i, r}$. For any $(x, y) \in U_{i, r}$, 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 $B_i(x, s/2) \times B_i(y, s/2) \subset U_{i, r}$.
(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, $U_{i, r}$ 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 \ref{lemma:openneighbourhood}.
(4): By \ref{lemma:pseudometric-continuous}.
(5): By \ref{lemma:pseudometric-continuous}, $U_{i, r} \in \mathfrak{V}$ for all $i \in I$ and $r > 0$, so $\mathfrak{V} \supset \mathfrak{B}$ by (F2), and $\mathfrak{V} \supset \mathfrak{U}$.
(5): By \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 $\mathfrak{V} \supset \mathfrak{U}$.
\end{proof}
\begin{lemma}
@@ -106,7 +106,7 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
\end{enumerate}
then there exists a pseudometric $d: X \times X \to [0, 1]$ such that
\[
U_{n+1} \subset \bracs{(x, y)| d(x, y) < 2^{-n}} \subset U_{n-1}
U_{n+1} \subset E(d, 2^{-n}) \subset U_{n-1}
\]
for each $n \in \natp$.
\end{lemma}
@@ -131,13 +131,13 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
\end{enumerate}
so $d$ is a pseudometric.
For any $(x, y) \in U_{n+1}$, $d(x, y) \le \rho(x, y) < 2^{-n}$, so $U_{n+1} \subset \bracs{(x, y)| d(x, y) \le 2^{-n}}$.
For any $(x, y) \in U_{n+1}$, $d(x, y) \le \rho(x, y) < 2^{-n}$, so $U_{n+1} \subset E(d, 2^{-n})$.
Let $x, y \in X$ with $d(x, y) < 2^{-n}$. If $\rho(x, y) < 2^{-n}$, then the claim holds directly. Assume inductively that for any $x, y \in X$ with $d(x, y) < 2^{-n}$, if there exists $\seqf[m]{x_j} \subset X$ such that $x_0 = x$, $x_m = y$ and $\sum_{j = 1}^m \rho(x_{j - 1}, x_j) < 2^{-n}$, then $(x, y) \in U_{n - 1}$.
Let $x, y \in X$ and $\seqf[m+1]{x_j} \subset X$ such that $x_0 = x$, $x_{m+1} = y$, and $\sum_{j = 1}^{m+1} \rho(x_{j - 1}, x_j) < 2^{-n}$. Let $1 \le k < m+1$ such that $\sum_{j = 1}^{k}\rho(x_{j - 1}, x_j) < 2^{-n-1}$ and $\sum_{j = 1}^{k+1}\rho(x_{j-1}, x_j) \ge 2^{-n-1}$, then $\sum_{j = k + 1}^{m+1}\rho(x_{j-1}, x_j) < 2^{-n-1}$ as well. By the inductive hypothesis, $(x, x_k), (x_{k+1}, y) \in U_{n+1} \subset U_n$. Given that $\rho(x_k, x_{k+1}) < 2^{-n}$, $(x_k, x_{k+1}) \in U_{n+1}$ too. Thus
\[
(x, y) \subset U_{n+1} \circ U_{n + 1} \circ U_{n + 1} \subset U_n \circ U_n \subset U_{n-1}
(x, y) \in U_{n+1} \circ U_{n + 1} \circ U_{n + 1} \subset U_n \circ U_n \subset U_{n-1}
\]
\end{proof}
@@ -159,9 +159,9 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
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 $n \in \natp$,
\[
U_{n + 1} \subset \bracs{(x, y) \in X \times X|d(x, y) < 2^{-n}} \subset U_{n-1}
U_{n + 1} \subset E(d, 2^{-n}) \subset U_{n-1}
\]
By \ref{definition:pseudometric-uniformity}, $d$ is a uniformly continuous pseudometric on $X$. Since $\bracs{(x, y) \in X \times X|d(x, y) < 1/4} \subset U_1$, $U_1 \in \mathfrak{V}$. Therefore $\fU = \mathfrak{V}$.
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 $\fU = \mathfrak{V}$.
\end{proof}
\begin{proposition}
@@ -176,15 +176,11 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
\begin{proof}
(1) $\Rightarrow$ (2): By assumption, there exists $U \in \fU$ such that $(x, y) \not\in U$, so there exists $J \subset I$ finite and $r > 0$ such that
\[
\bracs{(x', y') \in X \times X| d_j(x', y') < r \forall j \in J} \subset U
\bigcap_{j \in J}E(d_j, r) \subset U
\]
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},
\[
\bracs{(x', y') \in X \times X| d(x', y') < r} \in \fU
\]
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 \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}.
\end{proof}
@@ -202,11 +198,7 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
is equivalent to $d$.
\end{lemma}
\begin{proof}
For any $r \in (0, 1]$,
\[
\bracsn{(x, y) \in X \times X| d(x, y) < r} = \bracsn{(x, y) \in X \times X| \td d(x, y) < r}
\]
Since sets of the above form generate the uniformity induced by $d$ and $\td d$, their induced uniformities coincide.
For any $r \in (0, 1]$, $E(d, r) = E(\td d, r)$. Since sets of the above form generate the uniformity induced by $d$ and $\td d$, their induced uniformities coincide.
\end{proof}
\begin{proposition}
@@ -220,14 +212,7 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
\]
then $d$ is a well-defined a pseudometric.
For each $n \in \nat$ and $r > 0$, denote
\begin{align*}
U_{n, r} &= \bracs{(x, y) \in X \times X| d_n(x, y)< r} \\
U_{r} &= \bracs{(x, y) \in X \times X|d(x, y) < r}
\end{align*}
Let $r > 0$, then there exists $n \in \natp$ such that $2^{-n} < r$. Take $s = r - 2^{-n}$, then $\bigcap_{k = 1}^nU_{k, s} \subset U_{r}$. On the other hand, for any $n \in \natp$ and $r > 0$, $U_{r/2^n} \subset \bigcap_{k = 1}^n U_{k, r}$. Therefore $\seq{d_n}$ and $d$ are equivalent.
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.
\end{proof}
\begin{theorem}[Metrisability of Uniform Spaces]
@@ -250,25 +235,35 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
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 $n \in \natp$,
\[
V_{n+1} \subset \bracs{(x, y) \in X \times X| d(x, y) < 2^{-n}} \subset V_{n-1}
V_{n+1} \subset E(d, 2^{-n}) \subset V_{n-1}
\]
For any $U \in \fU$, there exists $n \in \nat$ such that
\[
U \supset U_n \supset V_n \supset \bracs{(x, y) \in X \times X| d(x, y) < 2^{-n-1}}
U \supset U_n \supset V_n \supset E(d, 2^{-n-1})
\]
so $d$ induces the uniformity on $\fU$.
(3) $\Rightarrow$ (1): By (1) of \ref{definition:pseudometric-uniformity}, if
(3) $\Rightarrow$ (1): By (1) of \ref{definition:pseudometric-uniformity},
\[
U_{n, r} = \bracs{(x, y) \in X \times X| d_n(x, y) < r}
\]
then
\[
\fB = \bracs{\bigcap_{j \in J}U_{j, r} \bigg | J \subset \nat \text{ finite}, r > 0}
\fB = \bracs{\bigcap_{j \in J}E(d_j, r) \bigg | J \subset \nat \text{ finite}, r > 0}
\]
is a fundamental system of entourages for $\fU$. Since for any $r > 0$, there exists $q \in \rational \cap (0, r)$,
\[
\bracs{\bigcap_{j \in J}U_{j, r} \bigg | J \subset \nat \text{ finite}, r \in \rational, r > 0}
\bracs{\bigcap_{j \in J}E(d_j, r) \bigg | J \subset \nat \text{ finite}, r \in \rational, r > 0}
\]
is a countable fundamental system of entourages for $\fU$.
\end{proof}
\begin{proposition}
\label{proposition:uniform-continuous-pseudometric}
Let $X, Y$ be uniform spaces and $f: X \to Y$, then the following are equivalent:
\begin{enumerate}
\item $f \in UC(X; Y)$.
\item For every uniformly continuous pseudometric $d_Y$ on $Y$, there exists a uniformly continuous pseudometric $d_X$ on $X$ such that for all $x, y \in X$, $d_Y(f(x), f(y)) \le d_X(x, y)$.
\end{enumerate}
\end{proposition}
\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 $f \in UC(X; Y)$.
\end{proof}