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Let $E$ be a vector space and $F$ a subspace of $E$. Show that if $E$ is finite dimensional then $F$ is finite dimensional too.

It's easy to prove by contradiction by taking a family linearly independent with $n+1$ vectors of $F$.

I can also prove it by using axiom of choice, I am curious if we can avoid axiom of choice to prove the result 'directly' (by directly I mean without using the method by contradiction)? Thanks

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The usual proof (which you mentioned) doesn't use the axiom of choice at all.


You can prove this directly using induction. Pick some non-zero $w_1\in F$.

Suppose that we chose $w_1,\ldots,w_n$ and $F_n$ is their span. If $F_n=F$ then $F$ is finite dimensional. Otherwise, $F_n$ is a proper subspace of $F$, then we choose some $w_{n+1}\in F\setminus F_n$.

The induction stops at a stage which is at most $\dim E$, since the $w_i$'s are linearly independent. Since we only made finitely many choices, we didn't need to invoke the axiom of choice.

(Note that the proof doesn't cover the case $F=\{0\}$ but I'm sure you can manage proving this case on your own!)

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  • $\begingroup$ Perhaps my question is poorly worded. Can we prove this directly without axiom of choice. I will add the mention 'directly'. Sorry. $\endgroup$ – user142836 Jul 11 '14 at 21:04
  • $\begingroup$ I hope it's better now. $\endgroup$ – Asaf Karagila Jul 11 '14 at 21:15
  • $\begingroup$ Oh great (+1), thanks a lot. $\endgroup$ – user142836 Jul 11 '14 at 21:17
  • $\begingroup$ Did we not need to arguing by contradiction to prove that the induction stops? $\endgroup$ – user146010 Jul 11 '14 at 22:01
  • $\begingroup$ @Edwin: No, we argued positively. At step $\dim E$, we have a set of $\dim E$-many linearly independent vectors in $E$, so they must form a basis, therefore $E=F$. $\endgroup$ – Asaf Karagila Jul 11 '14 at 22:13
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Linear independence does not depend on which vector space you're in since $F\subseteq E$. By definition a basis is a maximal, linearly-independent, spanning set. Let $x_1,\ldots, x_n$ be any $n$ vectors in $F$. Then because they are also vectors in $E$ the maximal size of a linearly independent subset is $\text{dim}(F)$, hence $E$ has finite dimension by definition.

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  • $\begingroup$ What do you mean by 'spanning set'? I don't understand this definition. $\endgroup$ – user142836 Jul 11 '14 at 21:13
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    $\begingroup$ @Yass: A spanning set is a set of vectors $\{v_\alpha\}\subseteq V$ such that the set of finite linear combinations $\left\{\displaystyle\sum_{n=1}^N c_nv_{\alpha_n}\right\}$ is equal to the whole vector space $\endgroup$ – Eric Stucky Jul 11 '14 at 21:32

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