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I've been studying the construction of Free Vector Spaces and I want to confirm if my conclusions are correct. Given a set $A$ we wish to construct a vector space $F(A)$ which intuitively is the vector space whose elements are linear combination of $A$ (in other words, we think of $F(A)$ as the vector space that has $A$ as it's base).

Well, the problem here is that $A$ is just a set, without additional operations defined on it's elements. So what we do is to consider indicator functions $\delta_a : A \to \left\{0,1\right\}$ such that $\delta_a(x) = 1$ if $x = a$ and $\delta_a(x) = 0$ if not.

Then we consider the set $F(A)$ of all functions with finite support on $A$. Well, this seems very natural: the indicator functions themselves are in $F(A)$ since they have finite support. In this case, $\delta_a\in F(A)$ is the "representative" of $a \in A$ at $F(A)$.

Now, for those functions we can define their sum and multiplication by scalar in the natural pointwise fashion. Now, each $\delta_a$ represents some element of $A$ and the linear combinations of those functions represents the linear combinations of the elements of $A$ that we wanted to define.

Since all functions with finite support can be expressed as linear combination of those indicator functions, and since they are linearly independent, we can say that they indeed form a base for the space. And then we can check that the set $F(A)$ with the operations defined really is a vector space.

With all of this, given a set $A$ we've succesfully constructed a vector space $F(A)$ that intuitively has all linear combinations from the elements of $A$. And since $\delta_a$ represents the element $a \in A$ we usually forget the delta, and just write $a \in F(A)$.

Is all of this correct ? My understanding of the construction is indeed correct ? Thanks very much for your help.

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up vote 2 down vote accepted

Everything you have written above is correct.

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