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This question already has an answer here:

Disclaimer: I never had (yet c:) a rigorous exposure to set theory (independence proofs, and similar stuff...).

I was wondering if, in the following proof of the rank-nullity theorem form Lang's "Linear Algebra" book (page 61, theorem 3.2 in the third edition), the author makes use of the axiom of choice:

Let $V$, $W$ be vector spaces and let $L:V \to W$ be a linear map. Then we have $\dim V = \dim \ker L + \dim \text{im}\, L$.

Proof: Let $n$ be the dimension of $V$, $q$ the dimension of the kernel of $L$, and $s$ the dimension of the image of $L$. Then, assuming $s > 0$, let $\{w_1,\dots, w_s\}$ be a basis of $\text{im}\, L$; let $v_1,\dots,v_s$ be $s$ elements of $V$ such that $L(v_i) = w_i$ for $i = 1,\dots, s$. [...]

Here, the inverse image $L^{-1}(\{w_i\})$ need not to be unique, but the author claims that he can arbitrarily take $s$ elements of $V$ such that $L(v_i)=w_i$; this can be carried out with a choice function $\varphi: \{L^{-1}(\{w_i\}): i = 1,\dots, s\} \to V$: we have our $\{v_1,\dots, v_s\}$ by considering the image of $\varphi$.

Is the axiom of choice (or any weaker equivalent) required in this case?

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marked as duplicate by Asaf Karagila axiom-of-choice Jul 9 '18 at 23:42

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ There's probably a few other reasonable duplicates for this. $\endgroup$ – Asaf Karagila Jul 9 '18 at 23:43
  • $\begingroup$ What if OP can’t see why this question is a duplicate of the linked one? Is it on topic for OP to post a question asking why their question is a duplicate? I don’t see why we need to be so strict on closing duplicate questions that are packaged very differently, especially ones that relate to set theory which is very rarely offered in most undergraduate US university’s $\endgroup$ – Prince M Jul 10 '18 at 0:04
  • $\begingroup$ I thought it might be a stupid question. Should it be deleted, since it is a special case that will rarely be useful to the community? $\endgroup$ – marco21 Jul 10 '18 at 0:06
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Not at all. One can prove by induction on $n$ that for any finite $n,$ we can make $n$ choices. Choice principles come into play when we need to make infinitely-many choices, with no clear way to specify how said choices should/can be made.

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