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


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!)

  • $\begingroup$ 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

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.

  • $\begingroup$ What do you mean by 'spanning set'? I don't understand this definition. $\endgroup$ – user142836 Jul 11 '14 at 21:13
  • 2
    $\begingroup$ @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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