Extension linear independent set to Hamel basis If we have a linear independent set, then it is well known that by using Zorn's lemma it can be extended to Hamel basis. My question I have a linear independent set, call it $B_{0}$, I want to extended by transfinite induction to Hamel basis B by the set $ B_{00}=B\setminus B_0.$ such that $B_{00}$ must satisfy a specific condition. I am using perfect as example but in my case is not perfect. I have not seen construction like this before we usually use Zorn's lemma to extend any linear independent set. Any help will be useful.
 A: Let me repeat the proof that you want to do. Assume that we have a linearly independent set $X$, and extend it by adding elements. We may iterate adding elements, and it results in some linearly independent set. (We take a union in limit case.) Formally, we can think of it as follows:

*

*$B_0= X$,

*$B_{\alpha+1} = B_\alpha\cup\{v\}$, where $v$ is not generated by elements of $B_\alpha$, and

*$B_\delta=\bigcup_{\alpha<\delta}B_\alpha$.

The iterating process will stop when we cannot choose any $v$ which is not generated by $B_{\alpha}$: in that case, $B_\alpha$ is a basis.
We may ask there is such an $\alpha$. How can we assure the existence? Mathematicians use proof by contradiction when they have no idea how to start. Let us do it in that way.
Assume that for each $\alpha$, we can find $v$ which is not generated by $B_\alpha$. Then we can repeat the iteration for all ordinals. Let $\langle B_\alpha \mid \alpha\in\mathrm{Ord}\rangle$ be the resulting sets.
We can see that $B_\alpha\subsetneq B_\beta$ if $\alpha<\beta$.
From this, you can see that the vector space $V$ contains $\mathrm{Ord}$-many elements: that is, for each $\alpha$, we have $v_\alpha$ such that $B_{\alpha+1}= B_\alpha\cup\{\alpha\}$. $\{v_\alpha\mid \alpha\in\mathrm{Ord}\}$ is a collection of pairwise different elements of $V$, and it makes the collection a proper class.
Is it possible? No, this is not possible, a set (in this case, $V$) cannot contain a proper subclass. Thus we have a contradiction.
