equivalent condition for being a fundamental system of entourages
Lemma.
Let $X$ be a set and let $\mathrm{B}$ be a nonempty family of subsets of $X\mathrm{\times}X$. Then $\mathrm{B}$ is a fundamental system of entourages of a uniformity on $X$ if and only if it satisfies the following axioms.

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(B1) If $S$, $T\in \mathcal{B}$, then $S\cap T$ contains an element of $\mathcal{B}$.

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(B2) Each element of $\mathcal{B}$ contains the diagonal $\mathrm{\Delta}(X)$.

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(B3) For any $S\in \mathcal{B}$, the inverse relation of $S$ contains an element of $\mathcal{B}$.

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(B4) For any $S\in \mathcal{B}$, there is an element $T\in \mathcal{B}$ such that the relational composition^{} $T\circ T$ is contained in $S$.
Proof.
Suppose $\mathcal{B}$ is a fundamental system of entourages for a uniformity $\mathcal{U}$. Verification of axiom (B2) is immediate, since $\mathcal{B}\subseteq \mathcal{U}$ and each entourage is already required to contain the diagonal of $X$. We will prove that $\mathcal{B}$ satisfies (B1); the proofs that (B3) and (B4) hold are analogous.
Let $S$, $T$ be entourages in $\mathcal{B}\subseteq \mathcal{U}$. Since $\mathcal{U}$ is closed under^{} binary intersections^{}, $S\cap T\in \mathcal{U}$. By the definition of fundamental system of entourages, since $S\cap T\in \mathcal{U}$, there exists an entourage $B\in \mathcal{B}$ such that $B\subseteq S\cap T$. Thus $\mathcal{B}$ satisfies axioms (B1) through (B4).
To prove the converse^{}, define a family of subsets of $X\times X$ by
$$\mathcal{U}=\{S\subseteq X\times X:B\subseteq S\text{for some}B\in \mathcal{B}\}.$$ 
By construction, each element of $\mathcal{U}$ contains an element of $\mathcal{B}$, so all that remains is to show that $\mathcal{U}$ is a uniformity. Suppose $T$ is a subset of $X\times X$ that contains an element $S\in \mathcal{U}$. By the definition of $\mathcal{U}$, there exists some $B\in \mathcal{B}$ such that $B\subseteq S$. Since $S\subseteq T$, it follows that $B\subseteq T$, so $T$ satisfies the requirement for membership in $\mathcal{U}$. Thus $\mathcal{U}$ is closed under taking supersets. The remaining axioms for a uniformity follow directly from the appropriate axioms for the fundamental system of entourages by applying the axiom we have just checked. Hence $\mathcal{B}$ is a fundamental system of entourages for a uniformity on $X$. ∎
Title  equivalent^{} condition for being a fundamental system of entourages 

Canonical name  EquivalentConditionForBeingAFundamentalSystemOfEntourages 
Date of creation  20130322 16:30:02 
Last modified on  20130322 16:30:02 
Owner  mps (409) 
Last modified by  mps (409) 
Numerical id  5 
Author  mps (409) 
Entry type  Derivation 
Classification  msc 54E15 