5.3 Pseudometrics

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 $d(x, y) < d(x, x) + r = r$, $V \subset E(d, r)$, and $E(d, r) \in \fU$.

(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.$\square$

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 Proposition 5.1.8.

1. For any $J, J' \subset I$ finite and $r, r' > 0$,

   \[\bigcap_{j \in J \cup J'}E(d_{j}, \min(r,r')E(d_{j}, r \wedge r') \subset \paren{\bigcap_{j \in J}E(d_j, r)}\cap \paren{\bigcap_{j \in J'}E(d_j, r')}\]

2. 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.

3. 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.

(2): Since $\fB$ is a fundamental system of entourages for $\fU$,

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

(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 Lemma 4.4.3.

(4): By Lemma 5.3.2.

(5): By Lemma 5.3.2, $E(d_{i}, r) \in \mathfrak{V}$ for all $i \in I$ and $r > 0$, so $\mathfrak{V}\supset \mathfrak{B}$ by (F2), and $\mathfrak{V}\supset \mathfrak{U}$.$\square$

Proof. Let

and $d: X \times X \to [0, 1]$ by

then

1. For any $x \in X$, $x \in \bigcap_{n \in \natp}U_{n}$. Thus $\rho(x, x) = 0$ and $d(x, x) \le \rho(x, x) = 0$.

2. Let $x, y \in X$. By assumption (b), $\rho(x, y) = \rho(y, x)$. Thus $d(x, y) = d(y, x)$ as well.

3. Let $x, y, z \in X$, then for any $\seqf{x_j}$ and $\seqf[m]{y_j}$ with $x_{0} = x$, $x_{n} = y = y_{0}$, and $y_{m} = z$,

   \[d(x, z) \le \sum_{j = 1}^{n} \rho(x_{j - 1}, x_{j}) + \sum_{j = 1}^{m} \rho(y_{j - 1}, y_{j})\]

   As this holds for all such $\seqf{x_j}$ and $\seqf[m]{y_j}$, $d(x, z) \le d(x, y) + d(y, z)$.

so $d$ is a pseudometric.

For any $(x, y) \in U_{n+1}$, $d(x, y) \le \rho(x, y) < 2^{-n}$, so $U_{n+1}\subset E(d, 2^{-n})$.

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

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 Definition 5.3.3, $\fU \supset \mathfrak{V}$.

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 Lemma 5.3.4. Thus there exists a pseudometric $d: X \times X \to [0, \infty)$ such that for all $n \in \natp$,

By Definition 5.3.3, $d$ is a uniformly continuous pseudometric on $X$. Since $E(d, 1/4) \subset U_{1}$, $U_{1} \in \mathfrak{V}$. Therefore $\fU = \mathfrak{V}$.$\square$

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

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 Theorem 5.3.7, $E(d, r) \in \fU$. Therefore $\bigcup_{U \in \fU}U = \Delta$, and $X$ is separated by Definition 5.1.17.$\square$

Proof. 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.$\square$

Proof. Using Lemma 5.3.10, assume without loss of generality that for each $n \in \natp$, $d_{n}$ takes values in $[0, 1]$. Let

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 $\bigcap_{k = 1}^{n}E(d_{k}, s) \subset E(d, r)$. 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.$\square$

Proof. (1) $\Rightarrow$ (2): Let $\seq{U_n}\subset \fU$ be a fundamental system of entourages for $X$ and $V_{1} = U_{n}$. Assume inductively that $\bracs{V_k|1 \le k \le n}\subset \fU$ has been constructed such that

1. For each $1 \le k \le n$, $V_{k}$ is symmetric.

2. For each $1 \le k \le n$, $V_{k} \subset U_{k}$.

3. For each $1 \le k < n$, $V_{k+1}\circ V_{k+1}\subset V_{k}$.

Let $W = V_{n} \cap U_{n+1}$, then by Lemma 5.1.9, 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 Lemma 5.3.4, there exists a pseudometric $d: X \times X \to [0, \infty)$ such that for each $n \in \natp$,

For any $U \in \fU$, there exists $n \in \nat$ such that

so $d$ induces the uniformity on $\fU$.

(3) $\Rightarrow$ (1): By (1) of Definition 5.3.3,

is a fundamental system of entourages for $\fU$. Since for any $r > 0$, there exists $q \in \rational \cap (0, r)$,

is a countable fundamental system of entourages for $\fU$.$\square$

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 Theorem 5.3.7, 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 Definition 5.3.3, so $f \in UC(X; Y)$.$\square$