Uniform Spaces: the Three Definitions

Choose $R\in\mathcal U$ with $R\circ R\subseteq E$, and put $S:=R\cap R^{-1}$.

Put $E_0:=E$. By finitely iterating the square-root axiom, choose entourages $E_1,\ldots,E_n$ such that $E_{k+1}\circ E_{k+1}\subseteq E_k$ for $0\le k<n$, and put $D:=E_n\cap E_n^{-1}$.

The set $S$ is an entourage, since $R,R^{-1}\in\mathcal U$ and a filter is closed under intersections; also $S=S^{-1}$ and $S\subseteq R\circ R\subseteq E$, because every entourage contains the diagonal.

The entourage $D$ is symmetric and $D\subseteq E_n$. Induction on $k$ gives $D^{\circ 2^k}\subseteq E_{n-k}$ for $0\le k\le n$, hence $D^{\circ 2^n}\subseteq E$. Since every entourage contains the diagonal and $n\le 2^n$, one may insert diagonal factors to obtain $D^{\circ n}\subseteq D^{\circ 2^n}\subseteq E$.

Thus symmetric entourages refine every entourage; their intersections are symmetric entourages and none is empty because each contains the diagonal, so they form a filter base by [L1].

Therefore symmetric entourages form a base and admit the asserted finite-composite control.

The sets $\varnothing$ and $X$ are open: the first has no points to test, and for $x\in X$ every entourage ball is contained in $X$.

An arbitrary union of open sets is open, because a point in the union lies in one member and retains that member's entourage ball.

If $x\in O\cap P$, choose entourage balls $E[x]\subseteq O$ and $F[x]\subseteq P$; then $(E\cap F)[x]\subseteq O\cap P$, so binary intersections are open.

By steps 1.1 to 1.3, the open sets form a topology by [L1].

Let $E$ be an entourage and define $O_E=\{y\in E[x]:F[y]\subseteq E[x]\text{ for some }F\in\mathcal U\}.$ This set is open. Indeed, given $y\in O_E$, choose $F$ as displayed and then a symmetric $G$ with $G\circ G\subseteq F$. If $z\in G[y]$, symmetry gives $G[z]\subseteq(G\circ G)[y]\subseteq F[y]\subseteq E[x]$, so $z\in O_E$; hence $G[y]\subseteq O_E$. Now choose a symmetric $D$ with $D\circ D\subseteq E$. If $y\in D[x]$, then $D[y]\subseteq E[x]$, so $y\in O_E$. Thus $x\in D[x]\subseteq O_E\subseteq E[x]$, proving that $E[x]$ is a neighbourhood of $x$.

Conversely, if $N$ is a neighbourhood of $x$, it contains an open set $O$ with $x\in O$; the definition of the topology supplies an entourage $E$ with $E[x]\subseteq O\subseteq N$. Thus the entourage balls refine every neighbourhood, and by step 3.1 they are themselves neighbourhoods. They form a neighbourhood base by [L2].

Suppose $\mathcal U$ is separated and $x\ne y$. Choose $E$ with $(x,y)\notin E$, then a symmetric $D$ with $D\circ D\subseteq E$.

Conversely, if the induced topology is Hausdorff and $x\ne y$, choose disjoint neighbourhoods of $x,y$ and refine the first by an entourage ball $E[x]$; then $y\notin E[x]$, so $(x,y)\notin E$.

The neighbourhoods $D[x]$ and $D[y]$ are disjoint: if $z$ belonged to both, symmetry would give $(x,z),(z,y)\in D$ and hence $(x,y)\in D\circ D\subseteq E$.

Thus the induced topology is Hausdorff by [L2].

Every distinct pair is excluded by an entourage, so $\mathcal U$ is separated by [A1].

Let $N$ be a neighbourhood of $f(x)$ and choose a target entourage $V$ with $V[f(x)]\subseteq N$.

Uniform continuity supplies a source entourage $U$ whose pairs map into $V$, so $f[U[x]]\subseteq V[f(x)]\subseteq N$.

Since $U[x]$ is a neighbourhood of $x$, [L2] gives continuity at $x$; as $x$ was arbitrary, $f$ is continuous.

The diagonal lies in every $E_\varepsilon$, inverses agree with $E_\varepsilon$ by symmetry, intersections contain $E_{\min(\varepsilon,\delta)}$, and $E_{\varepsilon/2}\circ E_{\varepsilon/2}\subseteq E_\varepsilon$ by the triangle inequality.

The family $(E_\varepsilon)_{\varepsilon>0}$ is nonempty, none of its members is empty because $X\ne\varnothing$, and it is downward directed by step 1.1, so [L4] makes its upward closure a filter. The diagonal, inverse, and square-root properties in step 1.1 then make it a uniformity. Its $E_\varepsilon[x]$ are precisely metric balls, so its induced topology is the metric topology by [L2] and [L5].

The intersection of all $E_\varepsilon$ is the diagonal, since $d(x,y)>0$ for $x\ne y$ and $E_{d(x,y)/2}$ excludes $(x,y)$; hence the uniformity is separated.

The defining entourage implication for $E_\delta$ and $E_\varepsilon$ is exactly the quantified condition of [L3], which proves the final equivalence.

From an entourage $E$, form $\mathcal C_E=\{E[x]:x\in X\}$. Choose a symmetric entourage $D$ with $D^{\circ3}\subseteq E$. If $D[y]\cap D[x]\ne\varnothing$ and $z\in D[y]$, symmetry and a point in the intersection give $(x,z)\in D^{\circ3}\subseteq E$. Hence the star of $D[x]$ in $\mathcal C_D$ lies in $E[x]$, so $\mathcal C_D$ star-refines $\mathcal C_E$.

From a uniform cover $\mathcal V$, form $E_{\mathcal V}=\bigcup_{V\in\mathcal V}V\times V$. It contains the nonempty diagonal, so it is nonempty. A star-refinement $\mathcal W$ has $E_{\mathcal W}\circ E_{\mathcal W}\subseteq E_{\mathcal V}$, while common refinements and coarsenings give the remaining filter axioms.

Declare a cover uniform when it is coarser than some $\mathcal C_E$. Intersections of entourages give common refinements, enlargement of an entourage gives coarsening, and step 1.1 gives star-refinements. Hence these covers satisfy the uniform-cover axioms.

Start with an entourage uniformity. For symmetric $D$, $D\subseteq E_{\mathcal C_D}\subseteq D^{-1}\circ D=D^{\circ2}.$ The first inclusion uses the diagonal, and the second follows because two points in one $D$-ball are $D^{-1}\circ D$-related. Taking a symmetric square root inside any prescribed entourage shows that the recovered entourage filter is exactly the original one.

Start instead with a uniform-cover structure. The $E_{\mathcal V}$-ball at $x$ is $E_{\mathcal V}[x]=\operatorname{St}(x,\mathcal V),$ the union of the members of $\mathcal V$ containing $x$. Thus $\mathcal V$ refines $\mathcal C_{E_{\mathcal V}}$, so the latter is uniform by coarsening. Conversely, if $\mathcal W$ star-refines $\mathcal V$, then for any $x$ and any $W_0\in\mathcal W$ containing $x$, $\operatorname{St}(x,\mathcal W)\subseteq\operatorname{St}(W_0,\mathcal W)$, which lies in some member of $\mathcal V$. Therefore $\mathcal C_{E_{\mathcal W}}$ refines $\mathcal V$. The recovered cover structure is exactly the original one.

Steps 2.2 and 2.3 prove that the two constructions are mutually inverse at the level of generated structures.

Given a symmetric entourage $E$, choose a symmetric $R$ with $R^{\circ2}\subseteq E$, and then a symmetric $D$ with $D^{\circ2}\subseteq R$. Since every entourage contains the diagonal, $D\subseteq R$, and hence $D^{\circ3}\subseteq R^{\circ2}\subseteq E.$ Thus there exists a symmetric $D\subseteq E$ with $D^{\circ3}\subseteq E$.

The relation $E\mathrel R D$ meaning that $D$ is symmetric, $D\subseteq E$, and $D\circ D\circ D\subseteq E$ is serial by step 1.1.

Choose a symmetric entourage $E_1\subseteq U$ using [L1]. Dependent choice applied to the serial relation of step 2.1 starting at $E_1$ gives $E_1,E_2,\ldots$. Adjoin $E_0=X\times X$; then $E_1^{\circ3}\subseteq X\times X=E_0$, and all the required properties hold.

For $x,y\in X$, let $W(x,y)$ be the set of sums $\sum_{i<k}2^{-n_i}$ over all finite chains $x=x_0,\ldots,x_k=y$ with $(x_{i-1},x_i)\in E_{n_i}$. This set is nonempty because the one-edge $E_0=X\times X$ chain has weight $1$. Every dyadic term is positive by [L4], so every such finite sum is nonnegative by [L3]; hence $0$ is a lower bound. By [L2], the infimum exists; define $p(x,y):=\inf W(x,y)$.

We prove by strong induction on the number $k$ of edges, simultaneously for every $n$, that a $k$-edge chain of total weight less than $2^{-n}$ has $E_n$-related endpoints. For $k=0$ the endpoints coincide and hence are $E_n$-related by [A2]. Now let $k\ge1$ and assume the claim for every shorter chain. Its total weight $w$ is positive by [L3] and [L4]. Take the first edge for which the cumulative weight through that edge exceeds $w/2$. The subchain before it has weight at most $w/2$, and the subchain after it has weight less than $w/2$; because $w<2^{-n}$, both are less than $2^{-(n+1)}$. They have fewer than $k$ edges, so the induction hypothesis makes both endpoint pairs $E_{n+1}$-related. The middle edge has weight $2^{-m}\le w<2^{-n}$; strict decrease of the dyadic weights gives $m\ge n+1$, and decreasingness of $(E_j)$ puts that edge in $E_{n+1}$. Thus the endpoints lie in $E_{n+1}^{\circ3}\subseteq E_n$. Strong induction proves the claim for every finite chain.

The empty chain has weight $0$, while all weights are nonnegative, so $p(x,x)=0$. Reversing a chain preserves its weight because each $E_n$ is symmetric, so $p(x,y)=p(y,x)$. For the triangle inequality, suppose instead that $p(x,z)>p(x,y)+p(y,z)$ and put $\delta=(p(x,z)-p(x,y)-p(y,z))/3>0$. By [L2], choose an $x$-to-$y$ chain of weight $a<p(x,y)+\delta$ and a $y$-to-$z$ chain of weight $b<p(y,z)+\delta$. Their concatenation has weight $a+b<p(x,y)+p(y,z)+2\delta<p(x,z)$ by [L3], contradicting that $p(x,z)$ is a lower bound of $W(x,z)$. Thus the triangle inequality holds, and $p$ is a pseudometric by [L1].

A one-edge $E_n$-chain has weight $2^{-n}$, so $E_n\subseteq\{p\le2^{-n}\}$.

If $p(x,y)\le2^{-n}$ with $n\ge1$, then [L4] gives $p(x,y)\le2^{-n}<2^{-(n-1)}$. Apply the epsilon property in [L2] with $\varepsilon=2^{-(n-1)}-p(x,y)>0$ to obtain a chain of weight less than $2^{-(n-1)}$. Step 1.2 gives $(x,y)\in E_{n-1}$. Thus $\{p\le2^{-n}\}\subseteq E_{n-1}$.

Given $\varepsilon>0$, choose $n$ with $2^{-n}<\varepsilon$ by [L4]. Then $E_n\subseteq\{p\le2^{-n}\}\subseteq\{p<\varepsilon\}$, so the latter set is an entourage by upward closure. By A gauge of pseudometrics and, on a nonempty set, the uniformity it generates, $p$ is uniformly continuous for the original uniformity.

Let $\mathcal P$ be the set of all pseudometrics obtained by applying [L2] to normal sequences of [L1] subordinate to some entourage. This definition makes no simultaneous choice. Every $p\in\mathcal P$ is uniformly continuous for $\mathcal U$, and for each entourage $E$, [L1] and [L2] ensure that at least one $p\in\mathcal P$ has a positive-radius $p$-ball contained in $E$.

For each $p\in\mathcal P$ and each positive radius, its ball is an original entourage by step 1.1. Hence every finite intersection defining a basic entourage of the gauge belongs to $\mathcal U$.

Conversely every original entourage $E$ contains a positive-radius ball for at least one $p\in\mathcal P$ by step 1.1, so it belongs to the gauge uniformity.

The two uniformities contain one another and are equal.

The equivalence between entourages and covers is exactly [L1] and uses no choice principle.

Assuming dependent choice, [L2] sends an entourage uniformity to a gauge and [L3] sends every gauge back to an entourage uniformity.

Thus the first two formulations are equivalent in ZF and all three are equivalent under the displayed assumption.

Let $E$ be an entourage and choose a symmetric entourage $D$ with $D\circ D\subseteq E$.

The neighbourhood $D[x]$ belongs to $\mathcal F$ by convergence.

Since $D$ is symmetric, $D[x]\times D[x]\subseteq D^{-1}\circ D=D\circ D\subseteq E$, so $D[x]$ is $E$-small.

As $E$ was arbitrary, $\mathcal F$ is Cauchy by [L3].

Let $E$ be an entourage and choose symmetric $D$ with $D\circ D\subseteq E$; choose $A\in\mathcal F$ with $A\times A\subseteq D$.

The ball $D[x]$ is a neighbourhood of $x$, so it meets $A$ because $x$ is a cluster point; fix $a\in A\cap D[x]$.

For every $b\in A$, symmetry gives $(x,a)\in D$ and smallness gives $(a,b)\in D$, hence $(x,b)\in E$; so $A\subseteq E[x]$.

Every entourage ball about $x$ belongs to $\mathcal F$ by upward closure, and such balls form a neighbourhood base at $x$, so $\mathcal F$ converges to $x$.

Let $\mathcal B$ consist of all $E[A]$ with $A\in\mathcal F$ and symmetric entourage $E$. Every such set contains the nonempty set $A$. Given $E[A],D[B]\in\mathcal B$, the symmetric entourage $E\cap D$ and the member $A\cap B\in\mathcal F$ give $(E\cap D)[A\cap B]\subseteq E[A]\cap D[B].$ Thus $\mathcal B$ is a proper downward-directed filter base. Let $m(\mathcal F)$ be the filter it generates.

Since $A\subseteq E[A]$, every member of $\mathcal B$ belongs to $\mathcal F$, so $m(\mathcal F)\subseteq\mathcal F$. To prove it Cauchy, let $U$ be an entourage and choose a symmetric $E$ with $E^{\circ3}\subseteq U$. Choose $A\in\mathcal F$ with $A\times A\subseteq E$. If $y,z\in E[A]$, take $a,b\in A$ with $aEy$ and $bEz$; symmetry gives $yEaEbEz$, so $(y,z)\in E^{\circ3}\subseteq U$. Hence $E[A]\in m(\mathcal F)$ is $U$-small.

Let $\mathcal G\subseteq\mathcal F$ be Cauchy, and fix $E[A]\in\mathcal B$. Choose a symmetric $D$ with $D\subseteq E$, and a $D$-small $B\in\mathcal G$. Since $A,B\in\mathcal F$, choose $c\in A\cap B$. Then $B\subseteq D[c]\subseteq E[A]$, so $E[A]\in\mathcal G$. Thus every Cauchy filter coarser than $\mathcal F$ contains $m(\mathcal F)$.

If a Cauchy filter is coarser than $m(\mathcal F)$, step 2.2 places $m(\mathcal F)$ inside it, so equality holds; hence $m(\mathcal F)$ is minimal. Any minimal Cauchy filter coarser than $\mathcal F$ contains $m(\mathcal F)$ by step 2.2 and must equal it by minimality. This proves uniqueness.

For $x\in X$, the set $\{x\}$ belongs to $\mathcal P_x$, and $\{x\}\times\{x\}\subseteq\Delta_X\subseteq E$ for every entourage $E$. Thus $\mathcal P_x$ is Cauchy by [L1], and step 3.1 supplies its associated minimal Cauchy filter.

Because a uniformity is a proper filter on $X\times X$, its carrier $X$ is nonempty. Choose $x\in X$; then [L1] gives a minimal Cauchy filter associated to the principal filter at $x$, so $\widehat X$ is nonempty. For symmetric $E$, put $(\mathcal F,\mathcal G)\in\widehat E$ when some $A\in\mathcal F$ and $B\in\mathcal G$ satisfy $A\times B\subseteq E$.

Every $\mathcal F$ is $\widehat E$-close to itself: choose an $E$-small member of the Cauchy filter and use it on both sides. Thus each $\widehat E$ contains the nonempty diagonal of $\widehat X$. The relation $\widehat E$ is symmetric. If $\widehat E$ and $\widehat D$ have respective witnesses $A\times B$ and $C\times K$, then $(A\cap C)\times(B\cap K)\subseteq E\cap D$, so finite intersections are refined by the corresponding hatted intersection.

Choose a symmetric entourage $D$ with $D^{\circ2}\subseteq E$. If $\mathcal F\,\widehat D\,\mathcal G$ via $A\times B\subseteq D$ and $\mathcal G\,\widehat D\,\mathcal H$ via $C\times K\subseteq D$, choose $b\in B\cap C$. Then $aDbDk$ for every $a\in A,k\in K$, so $A\times K\subseteq D^{\circ2}\subseteq E$. Hence $\widehat D\circ\widehat D\subseteq\widehat E$.

Steps 2.1 and 2.2 show that the upward closure of the relations $\widehat E$ is a uniformity. To prove separation, suppose $\mathcal F\,\widehat E\,\mathcal G$ for every entourage $E$. Given $A\in\mathcal F$, minimality gives $\mathcal F=m(\mathcal F)$ by [L1], so some $D[C]\subseteq A$ with $C\in\mathcal F$ and symmetric $D$. Choose an entourage $E\subseteq D$ and witnesses $P\in\mathcal F,Q\in\mathcal G$ with $P\times Q\subseteq E$. Pick $c\in C\cap P$. Then $Q\subseteq E[c]\subseteq D[C]\subseteq A$, so $A\in\mathcal G$. Thus $\mathcal F\subseteq\mathcal G$; symmetry gives equality.

Therefore the standard relations form the asserted separated uniformity.

For $A\subseteq X$, put $A^\#:=\{\,\mathcal M\in\widehat X:A\in\mathcal M\,\},$ and define $\mathcal F:=\{A\subseteq X:A^\#\in\Phi\}$. Since $X^\#=\widehat X$, $\varnothing^\#=\varnothing$, $(A\cap B)^\#=A^\#\cap B^\#$, and $A^\#\subseteq B^\#$ whenever $A\subseteq B$, [L4] shows that $\mathcal F$ is a filter on $X$.

The filter $\mathcal F$ is Cauchy. Given an entourage $U$, choose a symmetric $D$ with $D^{\circ3}\subseteq U$. Choose a $\widehat D$-small $S\in\Phi$, a filter $\mathcal M_0\in S$, and a $D$-small $C\in\mathcal M_0$. For every $\mathcal N\in S$, the relation $\mathcal M_0\,\widehat D\,\mathcal N$ has witnesses $P\in\mathcal M_0$ and $Q\in\mathcal N$ with $P\times Q\subseteq D$. A point of $C\cap P$ shows $Q\subseteq D[C]$, hence $D[C]\in\mathcal N$. Thus $S\subseteq(D[C])^\#$, so $D[C]\in\mathcal F$. Moreover $D[C]\times D[C]\subseteq D^{\circ3}\subseteq U$, making this a $U$-small member of $\mathcal F$.