A Hamel basis is nothing more than the correct generalization of the notion of a basis for a finite dimensional vector space to vector spaces which are possibly not of finite dimension.
Definition B.1.1 Let X be a vector space. A Hamel basis is a subset of X,Λ such that every vector of X can be written as a finite linear combination of vectors of Λ and the vectors of Λ are linearly independent in the sense that if

then each c_{k} = 0.
The main result is the following theorem.
Proof: Let x_{1} ∈ X and x_{1}≠0. Let ℱ denote the collection of subsets of X, Λ containing x_{1} with the property that the vectors of Λ are linearly independent as described in Definition B.1.1 partially ordered by set inclusion. By the Hausdorff maximal theorem, there exists a maximal chain, C Let Λ = ∪C. Since C is a chain, it follows that if

it follows each c_{k} = 0. Thus the vectors of Λ are linearly independent. Is every vector of X a finite linear combination of vectors of Λ?
Suppose not. Then there exists z which is not equal to a finite linear combination of vectors of Λ. Consider Λ ∪

where the x_{k} are vectors of Λ, then if c≠0 this contradicts the condition that z is not a finite linear combination of vectors of Λ. Therefore, c = 0 and now all the c_{k} must equal zero because it was just shown Λ is linearly independent. It follows C∪
This section is devoted to Stone’s theorem which says that a metric space is paracompact, defined below. See [23] for this which is where I read it. First is the definition of what is meant by a refinement.
Definition C.0.1 Let S be a topological space. We say that a collection of sets D is a refinement of an open cover S, if every set of D is contained in some set of S. An open refinement would be one in which all sets are open, with a similar convention holding for the term “ closed refinement”.
Definition C.0.2 We say that a collection of sets D, is locally finite if for all p ∈ S, there exists V an open set containing p such that V has nonempty intersection with only finitely many sets of D.
Definition C.0.3 We say S is paracompact if it is Hausdorff and for every open cover S, there exists an open refinement D such that D is locally finite and D covers S.
Proof: It is clear the left side is a subset of the right. Let p be a limit point of

and let p ∈ V , an open set intersecting only finitely many sets of D, D_{1}...D_{n}. If p is not in any of D_{i} then p ∈ W where W is some open set which contains no points of ∪_{i=1}^{n}D_{i}. Then V ∩W contains no points of any set of D and this contradicts the assumption that p is a limit point of

Thus p ∈D_{i} for some i. ■
We say S ⊆P

and each S_{n} is locally finite. The following theorem appeared in the 1950’s. It will be used to prove Stone’s theorem.
Theorem C.0.5 Let S be a regular topological space. (If p ∈ U open, then there exists an open set V such that p ∈ ⊆ U. ) The following are equivalent
1.) Every open covering of S has a refinement that is open, covers S and is countably locally finite.
2.) Every open covering of S has a refinement that is locally finite and covers S. (The sets in refinement maybe not open.)
3.) Every open covering of S has a refinement that is closed, locally finite, and covers S. (Sets in refinement are closed.)
4.) Every open covering of S has a refinement that is open, locally finite, and covers S. (Sets in refinement are open.)
Proof:
1.)⇒ 2.)
Let S be an open cover of S and let B be an open countably locally finite refinement

where B_{n} is an open refinement of S and B_{n} is locally finite. For B ∈B_{n}, let

Thus, in words, E_{n}
Claim: {E_{n}
Proof of the claim: Let p ∈ S. Then p ∈ B_{0} ∈B_{n} for some n. Let V be open, p ∈ V, and V intersects only finitely many sets of B_{1} ∪ ... ∪B_{n}. Then consider B_{0} ∩ V . If m > n,

In words, E_{m}
Claim:
Proof: Let p ∈ S and let n = min{k ∈ ℕ : p ∈ B for some B ∈B_{k}}. Let p ∈ B ∈B_{n}. Then p ∈ E_{n}
The two claims show that 1.)⇒ 2.).
2.)⇒ 3.)
Let S be an open cover and let

Then since S is regular, G covers S. (If p ∈ S, then p ∈ U ⊆U ⊆ V ∈S. ) By 2.), G has a locally finite refinement ℭ, covering S. Consider

This collection of closed sets covers S and is locally finite because if p ∈ S, there exists V,p ∈ V, and V has nonempty intersections with only finitely many elements of ℭ, say E_{1},
3.)⇒ 4.) Here is a table of symbols with a short summary of their meaning.

Let S be an open cover and let B be a locally finite refinement which covers S. By 3.) we can take B to be a closed refinement but this is not important here. Let

Then F covers S because B is locally finite. If p ∈ S, then there exists an open set U containing p which intersects only finitely many sets of B. Thus p ∈ U ∈F. By 3., F has a locally finite closed refinement ℭ, which covers S. Define for B ∈B

Thus these closed sets C do not intersect B and so B is in their complement. We use ℭ

In words, E

(by definition B is in some set of S), and let

The intersection with F
Claim: L covers S.
This claim is obvious because if p ∈ S then p ∈ B for some B ∈B. Hence

Claim: L is locally finite and a refinement of S.
Proof: It is clear L is a refinement of S because every set of L is a subset of a set of S, F
But C_{i} is contained in a set U_{i} ∈F which intersects only finitely many sets of B. Thus each C_{i} intersects only finitely many B ∈B and so each C_{i} intersects only finitely many of the sets, E
It is obvious that 4.)⇒ 1.). ■
The following theorem is Stone’s theorem.
Theorem C.0.6 If S is a metric space then S is paracompact (Every open cover has a locally finite open refinement also an open cover.)
Proof: Let S be an open cover. Well order S. For B ∈S,

Thus B_{n} is contained in B but approximates it up to 2^{−n}. Let

where ≺ denotes the well order. If B, D ∈S, then one is first in the well order. Let D ≺ B. Then from the construction, E_{n}

for all B, D ∈S. Fatten up E_{n}

Thus

It follows that the collection of open sets

is locally finite. In fact, B

because if p ∈ S, let B be the first set in S to contain p. Then p ∈ E_{n}