Definitions of uniform space

Question

Let $X$ be a topological space. The collection $\mathcal{A}=\{U_\lambda(x)\}_{\lambda\in \Lambda, x\in X}$, with $U_\lambda(x)\subseteq X$, is called a uniform structure of $X$ if

1. If $x\in X$ and $\lambda\in \Lambda$, then $x\in U_\lambda(x)$.

2. If $\alpha, \beta\in\Lambda$, then there is some $\gamma\in \Lambda$ such that whenever $U_\gamma(x)\subseteq U_\alpha(x)\bigcap U_\beta(x)$ for all $x\in X$.

3)For every $\epsilon\in \Lambda$ there is $\lambda\in\Lambda$ such that whenever $x,y, z\in X$ with $x,y \in U_\lambda(z)$ there follows $x\in U_\epsilon(y)$

Also Let $(X, \mathcal{V})$ be as uniform space $\mathcal{V}=\{V_\alpha\}$ with entourage $\Delta_X\subseteq V_\alpha$.

What is difference between $(X, \mathcal{V})$ and $(X, \mathcal{A})$ in above?

Let uniform space $(X, \mathcal{V})$ with $\mathcal{V}=\{V_\alpha\}_{\alpha\in \Lambda}$ be given. Define $U_\alpha(x)= V_\alpha[x]=\{y: (x, y)\in V_\alpha\}$. I think that $\mathcal{A}= \{U_\alpha(x)\}_{\alpha\in \Lambda, x\in X}$ is satisfies items 1, 2, 3 in above. What can say about reverse inclusion? Is it true that if $\mathcal{A}=\{U_\lambda(x)\}_{\lambda\in \Lambda, x\in X}$ satisfies in items 1, 2, 3 and take $U_\alpha=\bigcup_{x\in X}U_\alpha(x)\times U_\alpha(x)$, then $\mathcal{U}=\{U_\alpha\}_{\alpha\in \Lambda}$ is an an uniformity for $X$?

Would you please help me to know it or introduce a reference to study difference between them.

Answer 1

Having a "uniformity" $(X,\mathcal{A})$ as described, gives rise to a base for a uniformity $(X,\mathcal{U})$ in "entourage style" (see the Wikipedia page). Just let $\mathcal{B}(A) = \bigcup\{U_\lambda(x) \times U_\lambda(x): x \in X\}$ where $A = A_\lambda= \{U_\lambda(x): x \in X\}$ is in $\mathcal{A}$ (so $\lambda \in \Lambda$), and check that this forms a base for an entourage uniformity.

And this $\mathcal{A}$ is itself a base for a uniformity in "cover style" : just take the collection of all covers (any kind) that are refined by a member of $\mathcal{A}$. And conversely too if $\mathcal{U}$ is a (base for an) entourage uniformity then all covers $\{U[x]\mid x \in X\}$, where $U$ ranges over $\mathcal{U}$, form a uniformity as described in your first paragraph.