Lemma 14.2.1 Let A be a closed set in a metric space and let x_{n}
Proof: By assumption,
Note that there was nothing magic about 6 in the above. Another number would work as well.
In the proof of the following theorem, you get a covering of A^{C} with open balls B such that for each of these balls, there exists a ∈ A such that for all x ∈ B,
A Banach space is a normed vector space which is also a complete metric space where the metric comes from the norm.

Thus you can add things in a Banach space. Much more will be considered about Banach spaces a little later.
Definition 14.2.2 A Banach space is a complete normed linear space. If you have a subset B of a Banach space, then conv
In the following theorem, we have in mind both X and Y are Banach spaces, but this is not needed in the proof. All that is needed is that X is a metric space and Y a normed linear space or possibly something more general in which it makes sense to do addition and scalar multiplication.
Theorem 14.2.3 Let A be a closed subset of a metric space X and let F : A → Y, Y a normed linear space. Then there exists an extension of F denoted as
Proof: For each c
So for x ∈ B_{c} what about dist

Now the following is also valid. Letting x ∈ B_{c} be arbitrary, it follows from the assumption on the diameter that there exists a_{0} ∈ A such that d

 (14.2.1) 
It follows from 14.2.1,

Thus for any x ∈ B_{c}, there is an a_{0} ∈ A such that d
By Stone’s theorem, there is a locally finite open refinement ℛ. These are open sets each of which is contained in one of the balls just mentioned such that each of these balls is the union of sets of ℛ. Thus ℛ is a locally finite cover of A^{C}. Since x ∈ A^{C} is in one of those balls, it was just shown that there exists a_{R} ∈ A such that for all x ∈ R ∈ℛ we have d
Now let ϕ_{R}

The sum in the bottom is always finite because the covering is locally finite. Also, this sum is never 0 because ℛ is a covering. Also
Suppose this does not happen. Then there is a sequence converging to some a ∈ ∂A and ε > 0 such that

For x_{n} ∈ R, it was shown above that d

By local finiteness of the cover, each x_{n} involves only finitely many R Thus, in this limit process, there are countably many R involved
It also appears that
Definition 14.2.4 Let S be a subset of X, a Banach space. Then it is a retract if there exists a continuous function R : X → S such that Rs = s for all s ∈ S. This R is a retraction. More generally, S ⊆ T is called a retract of T if there is a continuous R : T → S such that Rs = s for all s ∈ S.
Proof: By Theorem 14.2.3, there is a continuous function Î extending I to all of X. Then also Î has values in conv
Sometimes people call the set a retraction also or the function which does the job a retraction. This seems like strange thing to call it because a retraction is the act of repudiating something you said earlier. Nevertheless, I will call it that. Note that if S is a retract of the whole metric space X, then it must be a retract of every set which contains S.