Equicontinuous at a Point
Let F be a family of functions of the form S → R m where S ⊆ R n , then we say that F is equicontinuous at a point a ∈ S if for every ϵ ∈ R + there exists a δ ∈ R + such that for all x ∈ S and f ∈ F ‖ x − a ‖ < δ ⟹ ‖ f ( x ) − f ( a ) ‖ < ϵ
Equicontinuous Family
We say that F is an equicontinuous family of functions of the form S → R diff for every a ∈ S we have that F is is equicontinuous at a .
Uniformly Equicontinuous Family
Let F be a family of functions of the form S → R m where S ⊆ R n , then we say that F is uniformly equicontinuous if for every ϵ ∈ R + there exists a δ ∈ R + such that for all x , y ∈ S and f ∈ C ‖ x − y ‖ < δ ⟹ ‖ f ( x ) − f ( y ) ‖ < ϵ
Compact Subsets of Continuous Functions with Compact Domain is Equicontinuous
Let K be a compact subset of R n , a compact subset F of C ( K , R m ) is equicontinuous
An Equicontinuous Family of Functions whose Domain is Compact is also Uniformly Equicontinuous
If F is an equicontinuous family of functions of the form K → R m where K is compact, then F is uniformly equicontinuous
Totally Bounded
We say that a subset S ⊆ R m is totally bounded if for any ϵ ∈ R + there exists a 1 , … , a n such that S ⊆ ⋃ i = 1 n B ( a i , ϵ )
A Bounded subset of Rm is Totally Bounded
Suppose that S ⊆ R m is bounded, then S is totally bounded.
Arzela-Ascoli
Let K be a compact subset of R n , a subset F of C ( K , R m ) is compact iff it is closed, bounded and equicontinuous
Functions Bounded by 1 are Not Compact
Show that B = { f ∈ C [ 0 , 1 ] : ‖ f ‖ ∞ ≤ 1 } is not compact.

Recall that B is compact iff it is closed bounded and equicontinuous, if we are able to show that B is not equicontinuous, then it wouldn't be compact, we do so by considering the subset of functions f n ( x ) = x n defined on [ 0 , 1 ] .

We show that these functions are not equicontinuous, we do this by showing that there exists a point at which the function is not equicontinuous, so let a = 1 , ϵ = 1 2 and let δ ∈ R we will prove that there exists an x ∈ [ 0 , 1 ] and f j such that | x − 1 | < δ and | f j ( x ) − f j ( 1 ) | ≥ 1 2 .

What we will do for x is to simply use x = 1 − δ 2 which will satisfy the first inequality, then we will note that for any α ∈ ( 0 , 1 ) that the sequence α n → 0 , since x ∈ [ 0 , 1 ] then it's also true for our value x and so there is some j such that x j < 1 2 note that x j = f j ( x ) and thus we have | f j ( x ) − f j ( 1 ) | = | x j − 1 | = | 1 − x j | ≥ 1 2