# Definition of uniform equicontinuity of a family of mappings between uniform spaces

From Wiki

A set $A$ of continuous functions between two uniform spaces $X$ and $Y$ is uniformly equicontinuous if for every element $W$ of the uniformity on $Y$, the set $\{ (u,v) ∈ X × X: \forall f ∈ A. (f(u),f(v)) ∈ W \}$ is a member of the uniformity on $X$.

1. I was wondering if it is equivalent to

A set $A$ of continuous functions between two uniform spaces $X$ and $Y$ is uniformly equicontinuous if for every element $W$ of the uniformity on $Y$, there is a member $V$ of the uniformity on $X$ such that $\forall (u,v) ∈ V, \forall f ∈ A. (f(u),f(v)) ∈ W$.

This is because in the special case when the uniform spaces are metric spaces, for a family $F$ of continuous functions between two metric spaces:

The family $F$ is uniformly equicontinuous if for every $ε > 0$, there exists a $δ > 0$ such that $d(f(x_1), f(x_2)) < ε$ for all $f ∈ F$ and all $x_1, x_2 ∈ X$ such that $d(x_1, x_2) < δ$.

Also this is because I think that the relation between equicontinuity and uniform equicontinuity should be similar to the relation between continuity and uniform continuity.

2. Does the definition require the set of functions to be continuous? Doesn't the definition itself, when without requiring continuity, imply uniform continuity and therefore continuity?

Thanks and regards!

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1. Yes, the two are equivalent. This follows from clause (2) of this definition of a uniformity. Let $U=\{\langle u,v\rangle\in X\times X:\forall f\in A(\langle f(u),f(v)\rangle\in W\}$. If $U$ belongs to the uniformity on $X$, then $U$ itself is a member of the uniformity on $X$ such that for each $\langle u,v\rangle\in U$, $\langle f(u),f(v)\rangle\in W$. Conversely, if there is a member $V$ of the uniformity on $X$ such that for each $\langle u,v\rangle\in V$, $\langle f(u),f(v)\rangle\in W$, then $U\supseteq V$, so $U$ is in the uniformity on $X$.
2. Yes, uniform equicontinuity of a family of functions implies that the functions are uniformly continuous. It is not necessary in the definition to specify that the members of $A$ are continuous, but it does no real harm.