Is it always possible to extend a linearly independent set to a basis in infinite dimensional vector space?
I was proving with the following argument:
If S is a linearly independent set, if it spans the vector space then done else keep on adding the elements such that the resultant set is also linearly independent, till it spans the vector space . But the problem is how can we guarantee that the process will stop?
Best Answer
Let $V$ be a vector space, $S\subseteq V$ a linearly independent subset and $\mathcal{A}=\{T\subseteq V: S\subseteq T \text{and $T$ is linearly independent}\}$. It is easy to see that any chain on $\mathcal{A}$ has an upper bound on $\mathcal{A}$ (we can take the union). Then, it follows from Zorn's lemma that $\mathcal{A}$ has a maximal element $R$. If $\langle R\rangle\neq V$ then we can consider $R\cup\{v\}$ for some $v\notin \langle R\rangle$ and we obtain an element of $\mathcal{A}$ which is greater than a maximal element. The contradiction comes from our assumption that $\langle R\rangle\neq V$. So, we must have $\langle R\rangle = V$.