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We know that for countable sets , the existence of a choice function is a consequence of the well-ordering principle ; and it is also known that the results like "every vector space has a maximal linearly independent set " , " every proper ideal in a ring with identity is contained in some maximal ideal " , " intersection of all prime ideals is a subset of the set of all nilpotent elements " etc. depends on Zorn's lemma which is equivalent to axiom of choice ; so my question is can we give proofs of these above statements when the corresponding Vector-spaces or Rings are countable , using well ordering principle directly ? Demonstration of at-least one of the proofs for countable case in such a way ( if possible ) would be really helpful , thanks in advance .

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  • $\begingroup$ Have you seen my answer, have you read it, should I clarify some things in it? $\endgroup$ – Asaf Karagila Feb 3 '15 at 16:10
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The answer, is generally yes. The idea is that we enumerate the countable set, then proceed one by one to collect the maximal subset.

For example. If $V$ is a countable vector space, then it has a basis.

Enumerate $V\setminus\{0\}=\{v_n\mid n\in\Bbb N \}$ and define by recursion the sets $B_k$.

$B_0=\{v_0\}$ and if $B_n$ was defined, then $B_{n+1}$ is $B_n$ added $v_k$ where $k$ is the least index of a vector not in the span of $B_n$, if it exists; if not such vector exists then just $B_{n+1}=B_n$.

Now define $B=\bigcup B_n$, and we claim that it is a basis for $V$. To see that, simply prove by induction that any linear dependence would contradict the construction (meaning we chose a vector that was in the span of the previous ones), and similarly if $v$ is any vector then at some point we could create it from some $B_n$.

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  • $\begingroup$ I'd be happy to add details for that final part, but you should try and prove them first with the guidance I've added. $\endgroup$ – Asaf Karagila Feb 1 '15 at 14:47

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