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The question arises when I am constructing an elementary proof for the following claim:

Given a normed vector space $V$, the following are equivalent:

  1. $V$ is finite dimensional
  2. Every linear map $T:V\to V$ is continuous

Now, the direction $(1)\Rightarrow(2)$ is obvious, but when trying to show $(2)\Rightarrow(1)$, I notice that a part of my proof says "fix a linearly independent set $\{v_i:i\in\mathbb{N}\}$"; some obvious consideration shows that this is a weak form of choice, since the construction would fail, for example, in a universe with an infinite set which is Dedekind-finite. So what would be the exact form of choice equivalent to this statement, assuming ZF?

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  • $\begingroup$ Consistency strength is an extremely coarse invariant. Since ZF and ZFC are equiconsistent, there's no consistency strength here at all. $\endgroup$
    – Noah Schweber
    Commented Jul 7 at 0:21
  • $\begingroup$ @NoahSchweber Sorry for my incorrect terminology, since I am not quite familiar with set theory. What I meant to say is, "assuming ZF, what form of choice would be equivalent to the statement that every NVS without finite bases has a discontinuous linear map" $\endgroup$
    – Akira Satou
    Commented Jul 7 at 0:59
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    $\begingroup$ That's a more reasonable question (and offhand I'm not sure). For future reference, "consistency strength" refers to how difficult it is to prove a theory consistent (over some "reasonable" base theory, usually first-order Peano arithmetic - the point being that this theory should be much weaker than whatever is being looked at at the moment). All the choice principles "collapse together" in terms of consistency strength. Interestingly, anti-choice principles may add consistency strength (e.g. this is known to occur in the case of the axiom of determinacy)! $\endgroup$
    – Noah Schweber
    Commented Jul 7 at 1:01

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