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/Part 2: General Topology/Chapter 6: Uniform Spaces

6.2 Uniform Continuity

Definition 6.2.1 (Uniform Continuity).label Let $(X, \fU)$ and $(Y, \mathfrak{V})$ be uniform spaces and $f: X \to Y$, then the following are equivalent:

  1. (1)

    For every $V \in \mathfrak{V}$, there exists $V' \in \fU$ such that $(f(x), f(y)) \in V$ whenever $(x, y) \in V'$.

  2. (2)

    For every $V \in \mathfrak{V}$, $(f \times f)^{-1}(V) \in \fU$.

If the above holds, then $f$ is a uniformly continuous function.

The collection $UC(X; Y)$ denotes the set of all uniformly continuous functions from $X$ to $Y$.

Proof. $(1) \Rightarrow (2)$: Since $(f(x), f(y)) \in V$ whenever $(x, y) \in V'$, $V' \subset f^{-1}(V)$. By (F1) of $\fU$, $f^{-1}(V) \in \fU$.

$(2) \Rightarrow (1)$: Take $V' = (f \times f)^{-1}(V)$, then $(f(x), f(y)) \in V$ whenever $(x, y) \in V'$.$\square$

Proposition 6.2.2.label Let $(X, \fU)$ and $(Y, \mathfrak{V})$ be uniform spaces and $f \in UC(X; Y)$, then $f \in C(X; Y)$.

Proof. Let $x \in X$ and $V(f(x)) \in \cn(f(x))$, then since $f^{-1}(V(f(x))) =[f^{-1}(V)](x)$, by (2) of Definition 6.2.1, $f^{-1}(V(f(x))) \in \cn(x)$. As this holds for all $x \in X$, $f$ is continuous.$\square$

Definition 6.2.3 (Initial Uniformity).label Let $X$ be a set, $\bracsn{(Y_i, \fU_i)}_{i \in I}$ be a family of uniform spaces, and $\seqi{f}$ be a family of maps such that $f_{i}: X \to Y_{i}$ for each $i \in I$, then there exists a uniformity $\fU$ on $X$ such that:

  1. (1)

    For each $i \in I$, $f_{i} \in UC(X; Y_{i})$.

  2. (U)

    If $\mathfrak{V}$ is a uniformity on $X$ satisfying $(1)$, then $\mathfrak{V}\supset \fU$.

Moreover,

  1. (3)

    The family

    \[\fB = \bracs{\bigcap_{j \in J}(f_j \times f_j)^{-1}(U_j) \bigg | J \subset I \text{ finite}, U_j \in \fU_j}\]

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

  2. (4)

    For any uniform space $Y$ and map $f: Y \to X$, $f \in UC(Y; X)$ if and only if $f_{i} \circ f \in UC(Y; Y_{i})$ for all $i \in I$.

known as the initial uniformity on $X$ generated by $\seqi{f}$.

Proof. (3): Since the diagonal is mapped to the diagonal and $\fB$ is closed under intersections, it is sufficient to verify (UB3) for $\fB$. Let $J \subset I$ be finite and $\bigcap_{j \in J}(f_{j} \times f_{j})^{-1}(U_{j}) \in \fB$, then there exists $\bracs{V_j}_{j \in J}$ such that $V_{j} \circ V_{j} \subset U_{j}$ for each $j \in J$. In which case, for any $(x, y), (y, z) \in f_{j}^{-1}(V_{j})$, $(f(x), f(y)), (f(y), f(z)) \in V_{j}$ and $(f(x), f(z)) \in U_{j}$. Thus $(f_{j} \times f_{j})^{-1}(V_{j}) \circ (f_{j} \times f_{j})^{-1}(V_{j}) \subset (f_{j} \times f_{j})^{-1}(U_{j})$, and

\[\paren{\bigcap_{j \in J}(f_j \times f_j)^{-1}(V_j)}\circ \paren{\bigcap_{j \in J}(f_j \times f_j)^{-1}(V_j)}\subset \bigcap_{j \in J}(f_{j} \times f_{j})^{-1}(U_{j})\]

By Proposition 6.1.8, there exists a uniformity $\fU$ such that $\fB$ is a fundamental system of entourages for $\fU$.

(1): $\fU \supset (f_{i} \times f_{i})^{-1}(\fU_{i})$ for all $i \in I$.

(U): For any $i \in I$, $\mathfrak{V}\supset (f_{i} \times f_{i})^{-1}(\fU_{i})$. By (F2), $\mathfrak{V}\supset \fB$, so $\mathfrak{V}\supset \fU$.

(4): Let $J \subset I$ finite and $\seqj{U_j}$ such that $U_{j} \in \fU_{j}$ for each $j \in J$, then

\[(f \times f)^{-1}\paren{\bigcap_{j \in J}(f_j \times f_j)^{-1}(U_j)}= \bigcap_{j \in J}[(f_{j} \circ f) \times (f_{j} \circ f)]^{-1}(U_{j})\]

is an entourage of $Y$.$\square$

Proposition 6.2.4.label Let $X$ be a set, $\bracsn{(Y_i, \fU_i)}_{i \in I}$ be a family of uniform spaces, and $\seqi{f}$ where $f_{i}: X \to Y_{i}$ for each $i \in I$, then the initial topology $\topo$ on $X$ coincides with the topology $\topo_{U}$ induced by the initial uniformity.

Proof. By Proposition 6.2.2 and (1) of Definition 6.2.3, $f_{i}: X \to Y_{i}$ is continuous with respect to $\topo_{U}$ for all $i \in I$. By (U) of Definition 5.1.7, $\topo_{U} \supset \topo$.

On the other hand, let $x \in X$ and $U \in \cn_{\topo_U}(x)$, then there exists an entourage $V$ such that $U = V(x)$. By (3) of Definition 6.2.3, there exists $J \subset I$ finite and $\seqj{V}$ such that $\bigcap_{j \in J}(f_{j} \times f_{j})^{-1}(V_{j}) \subset V$ and each $V_{j} \in \fU_{j}$. In which case, $f_{j}^{-1}(V_{j}(f_{j}(x))) \in \cn(x)$ for all $j \in J$. Using (F2),

\[W = \bigcap_{j \in J}f_{j}^{-1}(V_{j}(f_{j}(x))) \in \cn(x)\]

where for any $y \in W$, $(f_{j}(x), f_{j}(y)) \in V_{j}$ for all $j \in J$. Thus $f(y) \in V(x) = U$, and $\topo_{U} \subset \topo$.$\square$

Definition 6.2.5 (Product Uniformity).label Let $\bracs{(X_i, \fU_i)}_{i \in I}$ be a family of uniform spaces, then the product uniformity $\fU$ on $X = \prod_{i \in I}X_{i}$ is the initial uniformity generated by the projections $\seqi{\pi}$, and:

  1. (1)

    The family

    \[\fB = \bracs{\bigcap_{j \in J}(\pi_j \times \pi_j)^{-1}(U_j) \bigg | J \subset I \text{ finite}, U_j \in \fU_j}\]

    forms a fundamental system of entourages for $\fU$.

  2. (U)

    For any uniform space $(Y, \mathfrak{V})$ and $\seqi{f}$ where $f_{i} \in UC(Y; X_{i})$ for all $i \in I$, there exists a unique $f \in UC(Y; X)$ such that the following diagram commutes

    \[\xymatrix{ Y \ar@{->}[d]_{f} \ar@{->}[rd]^{f_i} & \\ X \ar@{->}[r]_{\pi_i} & X_i }\]

    for all $i \in I$.

Proof. (1): By (3) of Definition 6.2.3.

(U): Let $f = \prod_{i \in I}f_{i}$, then $f: Y \to X$ is the unique function such that the diagram commutes for all $i \in I$.

For each $J \subset I$ finite and $\bigcap_{j \in J}(\pi_{j} \times \pi_{j})^{-1}(U_{j}) \in \fU$,

\[(f \times f)^{-1}\paren{\bigcap_{j \in J}(\pi_j \times \pi_j)^{-1}(U_i)}= \bigcap_{j \in J}(f_{j} \times f_{j})^{-1}(U_{j}) \in \mathfrak{V}\]

by (F2) of $\mathfrak{V}$.$\square$

Proposition 6.2.6.label Let $\bracs{(X_i, \fU_i)}_{i \in I}$ be a family of uniform spaces, then the topology $\topo_{U}$ induced by the product uniformity coincides the product topology $\topo_{P}$.

Proof. By Proposition 6.2.4.$\square$

Proposition 6.2.7.label Let $X$, $\seqi{Y}$ be uniform spaces, $\seqi{f}$ where $f_{i} \in UC(X; Y_{i})$ for each $i \in I$, then

  1. (1)

    There exists a unique $f \in UC(X; \prod_{i \in I}Y_{i})$ such that the following diagram commutes

    \[\xymatrix{ & \prod_{i \in I} Y_i \ar@{->}[rd]^{\pi_i} & \\ X \ar@{->}[rr]_{f_i} \ar@{->}[ru]^{f} & & Y_i }\]

    for all $i \in I$.

  2. (2)

    If $X$ is T0 and equipped with the initial uniformity induced by $\seqi{f}$, then $f$ is an isomorphism onto its image.

Proof. (1): By (U) of Definition 6.2.5.

(2): By (2) of Proposition 5.7.3, $f$ is injective.

Let $U$ be an entourage in $X$. Assume without loss of generality that there exists $J \subset I$ finite and $\bracs{U_j}_{j \in J}$ such that $U = \bigcap_{j \in J}(f_{j} \times f_{j})^{-1}(U_{j})$. In which case, $\bigcap_{j \in J}(\pi_{j} \times \pi_{j})^{-1}(U_{j})$ is open in $\prod_{i \in I}Y_{i}$ and

\[(f \times f)(U) = (f \times f)(X) \cap \bigcap_{j \in J}(\pi_{j} \times \pi_{j})^{-1}(U_{j})\]

$\square$

Proposition 6.2.8.label Let $X$ be a compact uniform space, $Y$ be a uniform space, and $f \in C(X; Y)$, then $f \in UC(X; Y)$.

Proof. Let $V$ be an entourage of $Y$. For each $x \in X$, let $U_{x}$ be an entourage of $X$ such that $(f(y), f(z)) \in V$ for all $y, z \in (U_{x} \circ U_{x})(x)$. Since $X$ is compact, there exists $\seqf{x_j}\subset X$ such that $X = \bigcup_{j = 1}^{n}U_{x_j}(x_{j})$.

Let $U = \bigcap_{j = 1}^{n} U_{x_j}$, then for any $(x, y) \in U$, there exists $1 \le j \le n$ such that $x \in U_{x_j}(x_{j})$. In which case, $x, y \in (U_{x_j}\circ U_{x_j})(x_{j})$, so $(f(x), f(y)) \in V$.$\square$

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