(Parts copied from: 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 D of such a sets 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.
Question: Why does one have to explictly state the union of sets D as the upper bound, wouldn't it suffice to know that such an upper bound exits through the total ordering of the chain (for which linear independence would directly follow) ?
