# Alternate definition of quasi-uniform and quasi-metric spaces

Reloid is a triple ${( A ; B ; F)}$ where ${A}$, ${B}$ are sets and ${F}$ is a filter on their cartesian product ${A \times B}$.

Endoreloid is reloid with the same ${A}$ and ${B}$.

Uniform space is essentially a special case of an endoreloid.

The reverse reloid is defined by the formula ${( A ; B ; F)^{- 1} = ( B ; A ; F^{- 1})}$.

See here about the definition of composition of reloids.

I define partial order on the set of reloids as ${( A ; B ; F) \sqsubseteq ( A ; B ; G) \Leftrightarrow F \supseteq G}$. The set of reloids with given source and destination is a complete lattice with join denoted ${\sqcup}$ and meet denotes ${\sqcap}$.

Uniform space is essentially the same as symmetric (${f = f^{- 1}}$), reflexive (every entourage contains the diagonal), and transitive ${( f \circ f \sqsubseteq f)}$ endo-reloid.

Traditionally the definition of quasi-uniform space is formed removing symmetry axiom. Other axioms including transitivity ${( f \circ f \sqsubseteq f)}$ remain unchanged.

I propose an alternate (non equivalent) definition of quasi-uniform spaces: Remove symmetry axiom and replace ${f \circ f \sqsubseteq f}$ with ${f \circ f^{- 1} \sqsubseteq f}$.

Why? Because ${f \circ f^{- 1} \sqsubseteq f}$ (not ${f \circ f = f}$) is a condition used in an important theorem (see this short article and also my book):

Theorem 1 Let ${f}$ is such a endoreloid that ${f \circ f^{- 1} \sqsubseteq f}$. Then ${f}$is ${\alpha}$-totally bounded iff it is ${\beta}$-totally bounded.

If we change the definition of quasi-unifrom spaces, this theorem applies to every quasi-uniform space. But if we follow the historical definition, it seems that this theorem does not work for every quasi-uniform space.

Now we should also change the definition of quasi-metric space to match the definition of quasi-uniform space, so that every quasi-metric induces a quasi-uniform space.

Metric spaces are defined by the axioms:

• ${d ( x ; y) \geqslant 0}$
• ${d ( x ; y) = 0 \Leftrightarrow x = y}$
• ${d ( x ; y) = d ( y ; x)}$ (symmetry)
• ${d ( x ; z) \leqslant d ( x ; y) + d ( y ; z)}$

The quasi-metric spaces are metric spaces without symmetry axiom.

I propose modified quasi-metric space with also

$\displaystyle d ( x ; z) \leqslant d ( x ; y) + d ( y ; z)$

replaced with

$\displaystyle d ( x ; z) \leqslant d ( y ; x) + d ( y ; z)$

(this is the same in presence of symmetry axiom).

Proposition 2 For this modified definition of quasi-metric spaces every quasi-metric induces my modified quasi-uniformity.

Proof: We need to prove that for every entourage ${U}$ there is an entourage ${V}$ such that ${V \circ V^{- 1} \subseteq U}$. Really, let ${U = \left\{ ( x ; y) \hspace{1em} | \hspace{1em} d ( x ; y) \leqslant \varepsilon \right\}}$ for some ${\varepsilon > 0}$. Then take ${V = \left\{ ( x ; y) \hspace{1em} | \hspace{1em} d ( x ; y) \leqslant \varepsilon / 2 \right\}}$ and thus we have

$\displaystyle V \circ V^{- 1} = \left\{ ( y ; z) \hspace{1em} | \hspace{1em} d ( y ; z) \leqslant \varepsilon / 2 \right\} \circ \left\{ ( x ; y) \hspace{1em} | \hspace{1em} d ( y ; x) \leqslant \varepsilon / 2 \right\} = \\ \left\{ ( x ; z) \hspace{1em} | \hspace{1em} \exists y : ( d ( y ; z) \leqslant \varepsilon / 2 \wedge d ( y ; x) \leqslant \varepsilon / 2) \right\} \subseteq \left\{ ( x ; z) \hspace{1em} | \hspace{1em} d ( x ; z) \leqslant \varepsilon \right\} = U$

because ${d ( y ; z) \leqslant \varepsilon / 2 \wedge d ( y ; x) \leqslant \varepsilon / 2 \Rightarrow d ( x ; z) \leqslant \varepsilon}$. $\Box$

The exact terms naming for new modified definitions of quasi-uniform an quasi-metric spaces is open for proposals by mathematical community.

Or maybe there are situations when the traditional definitions are better than mine? If so please comment on this blog post with your alternate opinion.