# General Topology/Uniform spaces

Definition (uniform structure):

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

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

Definition (uniform space):

A uniform space is a set ${\displaystyle X}$ together with a uniform structure on it.

In this definition, if ${\displaystyle (x,y)}$ 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 ${\displaystyle x,y\in X}$ 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 ${\displaystyle X}$ be a set with a uniform structure ${\displaystyle {\mathcal {U}}}$. An entourage is simply an element of ${\displaystyle {\mathcal {U}}}$.

Definition (entourage-induced neighbourhood):

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

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

A uniform structure induces a topology on its space.

Proposition (uniform structure induces topology):

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

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

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

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

Definition (V-small):

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

Proposition (every uniform space is regular):

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

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

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

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

Definition (Cauchy filter):

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

Definition (completeness):

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

Definition (total boundedness):

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

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

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

Theorem (compact iff totally bounded and complete):

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

(On the condition of the ultrafilter lemma.)

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

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

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

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

Definition (uniform continuity):

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

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

Proposition (uniform continuity implies continuity):

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

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

Definition (fundamental system of entourages):

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

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

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

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

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

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

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

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

Proposition (least upper bound of uniform structures):

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

$\displaystyle \mathcal B := \left\{U_{\alpha_1} \cap \cdots \cap U_{\alpha_n} \middle| \alpha_1, \ldots, \alpha_n \in A, \forall j \in [n]: U_{\alpha_j} \in \mathcal U_{\alpha_j} \right\}$ ,

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

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

Definition (initial uniform structure):

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

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

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

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

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

Proposition (the topology induced by a final uniform structure coincides with the respective final topology):

Let ${\displaystyle X}$ be a set, and let ${\displaystyle (X_{\alpha })_{\alpha \in A}}$ be a collection of uniform spaces. Suppose further that we are given a function ${\displaystyle f_{\alpha }:X_{\alpha }\to A}$ for each ${\displaystyle \alpha \in A}$. Then the final topology on ${\displaystyle X}$ that is induced by the functions ${\displaystyle (f_{\alpha })_{\alpha \in A}}$ coincides with the topology that is induced by the final uniform structure induced by the functions ${\displaystyle (f_{\alpha })_{\alpha \in A}}$.

Proof: It suffices to show that the neighbourhood filters of the two topologies coincide. Hence, we select an arbitrary point ${\displaystyle x\in X}$. Suppose first that ${\displaystyle V}$ is an enclosure of ${\displaystyle X}$ with respect to the final uniform structure induced by the ${\displaystyle f_{\alpha }}$s. ${\displaystyle \Box }$

## Exercises

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