Extension of linearly independent set to a basis in an infinite dimensional vector space 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?
 A: Let be $V$ a vector space and $W\leq V$ a subspace of $V$ with basis $Y$. If we consider the quotient space $V/W$ , by Zorn's lemma, we can obtain a basis of $V/W$, denoted $\overline{S}$.
If $v\in V$, $\overline{v}=\beta_1\overline{\alpha_1}+\cdots+\beta_k\overline{\alpha_k}$ 
 where $\alpha_i\in S$, then $v-(\beta_1\alpha_1+\cdots+\beta_k\alpha_k)\in W$ so $v=\beta_1\alpha_1+\cdots+\beta_k\alpha_k+\gamma_1\zeta_1+\cdots+\gamma_r\zeta_r$, where $\zeta_i\in Y$. Thus, $S$ (yes, without bar) and $Y$ generate $V$.
Finally, whit the same notation above, if $\{{\alpha_1,\cdots,\alpha_k}\}\subseteq S$ and $\{{\zeta_1,\cdots,\zeta_r}\}\subseteq Y$ the equation $$\beta_1\alpha_1+\cdots+\beta_k\alpha_k+\gamma_1\zeta_1+\cdots+\gamma_r\zeta_r=0$$ implies that each escalar is zero. Indeed, the equation implies that $$\beta_1\alpha_1+\cdots+\beta_k\alpha_k\in  W,$$ then $\beta_1\overline{\alpha_1}+\cdots+\beta_k\overline{\alpha_k}=\overline{0}$ and it follows that each $\beta_i=0$ and by linear independence of  $\{{\zeta_1,\cdots,\zeta_r}\}$ it follows that $\{{\alpha_1,\cdots,\alpha_k,\zeta_1,\cdots,\zeta_r}\}$ is linearly independent. Thus $Y$ can be extended to a basis of $V$.
PDT: I'm sorry, english is not my mother tongue.
A: 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 $\langle R\rangle = V$.
