General Topology/Uniform spaces

Definition (uniform structure):

Let $X$ be a set. A uniform structure on $X$ is a filter ${\mathcal {U}}$ of $X\times X$ such that

1. $\forall U\in {\mathcal {U}}:\Delta (X)\subseteq U$, $\Delta (X):=\{(x,x)\mid x\in X\}$ being the diagonal of $X$
2. $\forall U\in {\mathcal {U}}:U^{-1}\in {\mathcal {U}}$, where $U^{-1}:=\{(x,y)\mid (y,x)\in U\}$
3. $\forall U\in {\mathcal {U}}:\exists V\subseteq U:V\circ V\subseteq U$, where $V\circ W:=\{(x,z)\mid (x,y)\in V\wedge (y,z)\in W\}$ for general $V,W\subseteq X\times X$

Definition (uniform space):

A uniform space is a set $X$ together with a uniform structure on it.

In this definition, if $(x,y)$ are contained in a sufficiently small entourage, they are considered "close" to each other. That is, a uniform structure provides a means of determining when two arbitrary points $x,y\in X$ are close. This is the intuition behind this definition. A very important special case of a uniform space are metric spaces, which we'll learn about in the next chapter. Uniform spaces are a generalisation of metric spaces, and many of the notions and theorems carry over from metric spaces to uniform spaces, and we'll immediately treat them in full generality.

Definition (entourage):

Let $X$ be a set with a uniform structure ${\mathcal {U}}$. An entourage is simply an element of ${\mathcal {U}}$.

Definition (entourage-induced neighbourhood):

Let $X$ be a uniform space and $V$ an entourage of $X$. For $x\in X$, define

$V(x):=\{y|(x,y)\in V\}$.

A uniform structure induces a topology on its space.

Proposition (uniform structure induces topology):

Let $X$ be a space with a uniform structure ${\mathcal {U}}$. Then there exists a unique topology on $X$ so that a basis for each neighbourhood filter of points $x$ is given by $V(x)$ as $V$ ranges over all entourages.

Proof: Define $N(x)$ to be the filter generated by the sets $V(x)$ as $V$ ranges over all entourages, and observe that indeed these sets are a filter subbasis, since they all contain $x$, so that the characterisation of filter subbases is applicable. Claim that $N(x)$ satisfies 1.-4. of the characterisation of a topology by its neighbourhoods.

1. Since $\Delta \subseteq V$ for all entourages $V$ ($\Delta =\{(x,x)|x\in X\}$), $x\in N(x)$
2. Due to $V_{1}(x)\cap \cdots \cap V_{n}(x)=(V_{1}\cap \cdots \cap V_{n})(x)$, $N(x)$ is closed under finite intersections
3. $N(x)$ is closed under supersets by definition
4. Let $M\in N(x)$. Pick $V\in N(x)$ so that $V\subseteq M$ and then $W\subseteq V$ so that $W\circ W\subseteq V$. For all $y\in W(x)$, we have $W(y)\subseteq V(x)$, so that $V(x)$ is in $N(y)$ $\Box$

Henceforth, we shall consider a uniform space as a topological space with this topology.

Definition (V-small):

Let $X$ be a uniform space and let $V$ be an entourage of $X$. A subset $A\subseteq X$ is called $V$-small iff for all $x,y\in A$, we have $(x,y)\in V$.

Proposition (every uniform space is regular):

Let $X$ be a uniform space (with uniformity ${\mathcal {U}}$). Then $X$ is regular.

Proof: Let $A\subset X$ be closed and $x\in X\setminus A$. Since $A$ is closed, $X\setminus A$ is an open neighbourhood of $x$. Hence, pick a symmetric entourage $V\in {\mathcal {U}}$ s.t. $V(x)\subset X\setminus A$, and then another symmetric entourage $W$ s.t. $W\circ W\subseteq V$. Then $W(x)$ and

$W(A):=\bigcup _{y\in A}W(y)$

are disjoint, since otherwise, if $z\in W(x)\cap W(A)$, there is $\alpha \in A$ s.t. $(z,\alpha )\in W$ and also $(z,x)\in W$, so that $(x,\alpha )\in W\circ W\subseteq V$, a contradiction to $V(x)\cap A=\emptyset$. $\Box$

Definition (Cauchy filter):

Let $X$ be a uniform space. A Cauchy filter is a filter $\phi$ on $X$ such that for any entourage $V$ of $X$, there exists $A\in \phi$ such that $x,y\in A\Rightarrow (x,y)\in V$, that is, $A$ is $V$-small.

Definition (completeness):

Let $X$ be a uniform space. $X$ is called complete iff each Cauchy filter $\phi$ on $X$ converges to some point in $X$.

Definition (total boundedness):

Let $X$ be a uniform space, and let ${\mathcal {U}}$ be its entourage filter. $X$ is totally bounded if and only if for each $V\in {\mathcal {U}}$, there exist finitely many points $x_{1},\ldots ,x_{n}\in X$ so that

$X=\bigcup _{j=1}^{n}V(x_{j})$.

The following is a generalisation of the Heine–Borel theorem.

Theorem (compact iff totally bounded and complete):

Let $X$ be a uniform space, and $S\subseteq X$ a subset. $S$ is compact if and only if it is totally bounded and complete.

(On the condition of the ultrafilter lemma.)

Proof: Suppose first that $S$ is compact, and let $V$ be an arbitrary entourage. Note that $(V(x)\cap S)_{x\in S}$ is an open cover of $S$, so that we may choose a finite subcover in order to achieve total boundedness. Let then ${\mathcal {F}}$ be a Cauchy filter of subsets of $S$, and suppose that ${\mathcal {F}}$ does not converge to any point of $S$. For each $x\in S$, select a nonempty set of entourages $\{V_{x,\lambda }\}_{\lambda \in \Lambda _{x}}$ sufficiently small so that $V_{x,\lambda }(x)\cap S\notin {\mathcal {F}}$ for all $\lambda \in \Lambda _{x}$, and then $W_{x,\lambda }$ sufficiently small so that $W_{x,\lambda }\circ W_{x,\lambda }\subseteq V_{x,\lambda }$. Then choose by compactness a finite subcover $S=(S\cap W_{x_{1},\lambda _{1}}(x_{1}))\cup \cdots \cup (S\cap W_{x_{n},\lambda _{n}}(x_{n}))$ (where $\lambda _{j}\in \Lambda _{x_{j}}$ for $j\in [n]$, and define

$V:=V_{x_{1},\lambda _{1}}\cap \cdots \cap V_{x_{n},\lambda _{n}}$, $W:=W_{x_{1},\lambda _{1}}\cap \cdots \cap W_{x_{n},\lambda _{n}}$.

Since ${\mathcal {F}}$ is a Cauchy filter, it will contain a $W$-small set $A$. Then pick $x_{0}\in A$ arbitrary, and $j\in [n]$ so that $x_{0}\in W_{x_{j},\lambda _{j}}(x_{j})$. Then we will have for $z\in A$, that $(z,x_{0})\in W$ and $(x_{0},x_{j})\in W_{x_{j},\lambda _{j}}$, so that $(z,x_{j})\in V_{x_{j},\lambda _{j}}$, that is, $z\in V_{x_{j},\lambda _{j}}(x_{j})$, and we conclude that $A\subseteq V_{x_{j},\lambda _{j}}(x_{j})$ and $V_{x_{j},\lambda _{j}}(x_{j})\in {\mathcal {F}}$, a contradiction.

Suppose now that $S$ is not compact. By the characterisation of compactness by filter convergence, pick a filter $\phi$ on $S$ which does not admit a refinement that converges to a point of $S$ (note that this does not use the axiom of choice). By the ultrafilter lemma, pick a maximal filter $\phi _{\infty }$ that contains $\phi$. Upon proving that $\phi _{\infty }$ is Cauchy, we obtain a contradiction, since Cauchy filters converge in $S$ as $S$ is complete. Let hence $V$ be any entourage of $X$, and pick $W\subseteq V$ so that $W\circ W\subseteq V$. Then $W(x)$ is $V$-small for all $x\in X$. By definition of the subspace topology and since $S$ is totally bounded, pick $x_{1},\ldots ,x_{n}\in S$ so that $S\subseteq W(x_{1})\cup \cdots \cup W(x_{n})$. Suppose that for all $j\in [n]$, there existed $F_{j}\in \phi _{\infty }$ so that $F_{j}\cap W(x_{j})=\emptyset$. Then set $F:=F_{1}\cap \cdots \cap F_{n}$ and observe that $F\cap W(x_{j})=\emptyset$ for all $j\in [n]$, and taking the union over all $j$ we get that $F\cap (W(x_{1})\cup \cdots \cup W(x_{n}))=F=\emptyset$, a contradiction to $\phi _{\infty }$ being a filter. Hence, pick $j\in [n]$ so that $F\cap W(x_{j})\neq \emptyset$ for all $F\in \phi _{\infty }$ and observe that $W(x_{j})\in \phi _{\infty }$, for otherwise we could properly extend $\phi _{\infty }$ by extending $\phi _{\infty }$ by $W(x_{j})$. But $W(x_{j})$ is $V$-small, so that $\phi _{\infty }$ is Cauchy. $\Box$

Definition (uniform continuity):

Let $X,Y$ be uniform spaces with uniform structures ${\mathcal {U}}$ resp. ${\mathcal {V}}$. A function $f:X\to Y$ is said to be uniformly continuous if and only if

$\forall V\in {\mathcal {V}}:\exists U\in {\mathcal {U}}:((x,y)\in U\Rightarrow (f(x),f(y))\in V)$.

Proposition (uniform continuity implies continuity):

Let $f:X\to Y$ be a uniformly continuous function, where ${\mathcal {U}}$ is the uniform structure of $X$ and ${\mathcal {V}}$ is the uniform structure of $Y$. Then $f$ is continuous.

Proof: Let $\phi$ be a filter of $X$ that converges to a point $x\in X$, so that $N(x)\subseteq \phi$. Let ${\mathcal {G}}$ be the filter on $Y$ that is generated by $f(\phi )$, and let $M\in N(y)$ be a neighbourhood of $y\in Y$. By definition of the topology on $Y$ induced by the uniform structure, pick $V\in {\mathcal {V}}$ so that $V(y)\subseteq M$. By uniform continuity, pick $U\in {\mathcal {U}}$ so that $(z,w)\in U\Rightarrow (f(z),f(w))\in V$. Then $U(x)\in \phi$, so that $f(U(x))\in f(\phi )$, but for $z\in X$ we have $(f(x),f(z))=(y,f(z))\in V$ so that $f(z)\in V(y)$, and we get $f(U(x))\subseteq V(y)$. $\Box$

Definition (fundamental system of entourages):

Let $X$ be a uniform space with uniform structure ${\mathcal {U}}$. A fundamental system of entourages is a filter basis for ${\mathcal {U}}$.

Proposition (inverse image of uniform structure generates a uniform structure):

Let $X$ be a set, let $Y$ be a uniform space (with uniform structure ${\mathcal {V}}$) and let $f:X\to Y$ be a function. Then

${\mathcal {B}}:=(f\times f)^{-1}({\mathcal {V}})$

is a filter basis for a uniform structure ${\mathcal {U}}$ on $X$

Proof: First note that whenever $U\in {\mathcal {B}}$, then $U$ contains the diagonal, since we have $U=(f\times f)^{-1}(V)$ for a certain $V\in {\mathcal {V}}$, that is,

$U=\{(x,x')|(f(x),f(x'))\in V\}$,

and clearly, for $x=x'$, we have $(f(x),f(x'))\in V$. Therefore, every $U\in {\mathcal {U}}$ also contains the diagonal (as it contains a set of ${\mathcal {B}}$). Further, ${\mathcal {B}}$ is a filter base, since taking preimages commutes with intersections, and ${\mathcal {V}}$ is closed under finite intersections, being a filter itself. Then let $U\in {\mathcal {U}}$, and pick $U'\in {\mathcal {B}}$ so that $U'\subseteq U$. Pick $V\in {\mathcal {V}}$ so that $U'=(f\times f)^{-1}(V)$. Then pick $W\in {\mathcal {V}}$ so that $W\circ W\subseteq V$. We claim that if we set $W':=(f\times f)^{-1}(W)$, then $W'\circ W'\subseteq U'$ ($\subseteq U$). Indeed, if $(x,x')\in W'$ and $(x',x'')\in W'$, then $(f(x),f(x''))\in V$, so that $(x,x'')\in U'$. Finally, $(f\times f)^{-1}(V^{-1})=(f\times f)^{-1}(V)^{-1}\subseteq U^{-1}$. $\Box$

Proposition (least upper bound of uniform structures):

Let $X$ be a set and let $({\mathcal {U}}_{\alpha })_{\alpha \in A}$ be a family of uniform structures on $X$. Then there exists a (unique) least upper bound uniform structure ${\mathcal {U}}$ on $X$ of the $({\mathcal {U}}_{\alpha })_{\alpha \in A}$'s, namely the filter generated by

${\mathcal {B}}:=\left\{U_{\alpha _{1}}\cap \cdots \cap U_{\alpha _{n}}|\alpha _{1},\ldots ,\alpha _{n}\in A,\forall j\in [n]:U_{\alpha _{j}}\in {\mathcal {U}}_{\alpha _{j}}\right\}$,

and the topology induced by it coincides with the least upper bound topology of the topologies $\tau _{\alpha }$ on $X$ that are induced by the ${\mathcal {U}}_{\alpha }$'s.

Proof: First we claim that ${\mathcal {U}}$ as given above is a uniform structure. Indeed, ${\mathcal {B}}$ is closed under finite intersections and hence forms a filter base. Further, suppose that $U\in {\mathcal {U}}$, and pick $\alpha _{1},\ldots ,\alpha _{n}\in A$ and $U_{\alpha _{1}},\ldots ,U_{\alpha _{n}}$ in ${\mathcal {U}}_{\alpha _{1}}$ resp. ${\mathcal {U}}_{\alpha _{2}}$ resp. ... resp. ${\mathcal {U}}_{\alpha _{n}}$. Then for each $j\in [n]$, observe for one that $U_{\alpha _{j}}^{-1}\in {\mathcal {U}}_{\alpha _{j}}$ (so that $U^{-1}\in {\mathcal {U}}$), and then pick $V_{\alpha _{j}}\in {\mathcal {U}}_{\alpha _{j}}$ so that $V_{\alpha _{j}}\circ V_{\alpha _{j}}\subseteq U_{\alpha _{j}}$. Then $(V_{\alpha _{1}}\cap \cdots \cap V_{\alpha _{n}})\circ (V_{\alpha _{1}}\cap \cdots \cap V_{\alpha _{n}})\subseteq U_{\alpha _{1}}\cap \cdots \cap U_{\alpha _{n}}\subseteq U$. $\Box$

Definition (initial uniform structure):

Let $X$ be a set, let $(Y_{\alpha })_{\alpha \in A}$ be a family of uniform spaces with uniform structures ${\mathcal {V}}_{\alpha }$, and let $f_{\alpha }:X\to Y_{\alpha }$ be functions. The initial uniform structure on $X$ is defined to be the least upper bound uniform structure induced by the uniform structures $f_{\alpha }$.

Proposition (separation of compact subsets of open sets by an entourage):

Let $X$ be a uniform space with uniformity ${\mathcal {U}}$. Let $U\subseteq X$ be open, and let $K\subseteq U$ be compact. Then there exists a symmetric entourage $V\in {\mathcal {U}}$ such that $V(K)\subseteq U$, where

$V(K):=\{x\in X|\exists y\in K:(x,y)\in V\}$.

Proof: For each $x\in K$, pick a symmetric entourage $V_{x}$ such that $V_{x}(x)\subseteq U$ (passing to the union so as to avoid the axiom of choice) and then a symmetric entourage $W_{x}$ such that $W_{x}\circ W_{x}\subseteq V_{x}$. $\Box$

Exercises

1. Let $X$ be a set, and let ${\mathcal {U}}$ and ${\mathcal {V}}$ be uniform structures on $X$ so that they generate the same topology $\tau$ and $X$ is compact with respect to $\tau$. Prove that in fact ${\mathcal {U}}={\mathcal {V}}$.
2. Let $X$ be a topological space whose topology is induced by both of the two uniform structures ${\mathcal {U}}$ and ${\mathcal {V}}$. Suppose that $X$ is complete with respect to the uniform structure induced by ${\mathcal {U}}$. Show that $X$ is complete with respect to the uniform structure induced by ${\mathcal {V}}$.