# While proving that every vector space has a basis, why are only finite linear combinations used in the proof?

Statement: Every vector space has a basis

Standard Proof:It is observed that a maximal linearly independent set is a basis. Let $\mathscr{Y}$ be a chain of linearly independent subsets of a vector space $\mathscr{V}$. The union of such a set can serve as an upper bound for it.To apply Zorn's lemma,we have to check whether the union is linearly independent? Well, if $t_1,\dots,t_n$ belong to the union, then each $t_i$ belongs to some linearly independent set $L_i\in \mathscr{Y}$. Because $\mathscr{Y}$ is a chain, one of these sets $L_i$ contains all the others. If that is $L_j$, then the linear independence of $L_j$ implies that no non-trivial linear combination of $t_1,\dots,t_n$ can be zero, which proves that the union of the sets in $\mathscr{Y}$ is linearly independent. Therefore, by Zorn’s lemma, there is a maximal linearly independent set and hence a basis.

My question: Why are we using only finite linear combinations to show that the union is linearly independent. Surely, if the union is infinite,then there do exist many infinite linear combinations of elements of the union, which cannot be proven to be linearly independent by the same reasoning. I suspect, that perhaps we are not concerned with infinite linear combinations due to issues of convergence, but I'm not sure.

• Linear dependence, basis, are precisely about finite linear combinations. In a restricted number of cases, some infinite sums make sense, but that is another subject. Jul 2 '12 at 12:59
• There needs to be a way of ascribing a meaning to infinite sums. Jul 2 '12 at 13:00
• The problem with infinite bases is that you have to worry about convergence of infinite sums. There are infinite dimensional vector spaces. See Hilbert Spaces. Feb 2 '16 at 1:12