The family of functions {f} is called uniformly equicontinuous iff .....
This is the question i have to complete the blanks and then prove the statement.
What i can think of is The family of functions {f} is called uniformly equicontinuous iff each member is uniformly continuous. I don't know if i am right or wrong because i know if the family was finite then this is true. I am confused about the fact that if the family is infinite. I need to understand what would be the blank.
Start with an equicontinuous family of functions: this means that we have continuity (at a point) for each point and we can choose the same $\delta$ for a given $\varepsilon$ for all functions at the same time. So
$$\forall x \in X: \forall \varepsilon>0: \exists \delta >0: \forall f \in \mathcal{F} \forall y \in X: ( d(x,y) < \delta ) \to (d(f(x),f(y) < \varepsilon$$
Uniformly equicontinuous is the same but then for uniform continuity (so the same $\delta$ should work for all $x$ at the same time. So:
$$\forall \varepsilon>0: \exists \delta >0: \forall f \in \mathcal{F}: \forall x,y \in X: ( d(x,y) < \delta ) \to (d(f(x),f(y) < \varepsilon$$
Note the subtle order difference in quantifiers. If the domain is compact the first definition implies the second, just as continuity implies uniform continuity for single functions.
