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Heine's theorem stated that $\lim_{x\to a}f(x)=b$ if and only if:
for any sequence ${a_n},\lim_{n\to \infty}a_n=a,a_n \neq a$ ,we have$$\lim_{n \to \infty}f(a_n)=b$$ I have just learned that this theorem can be used when we accepted axiom of choice.However,can we prove that without axiom of choice,Heine's theorem is not valid or we just don't have a proof of this theorem in ZF since now.

J.Guo
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  • If ZF is logically consistent, you won't be able to disprove Henie's theorem, since it is true in ZFC, and the consistency of ZF is equivalent to the consistency of ZFC. But, as I think you're asking, it may be true that Henie's theorem cannot be proven in ZF. – Theo Bendit Oct 07 '18 at 04:26
  • I think the relationship between ZFC and ZF is quite like the eucilidean geometry and noneucilidean geometry if we can find another axiom instead of axiom of choice.But if we still expect the new axiom may help us prove Henie's theorem,I think we need to now why it can not be proved in ZF. – J.Guo Oct 07 '18 at 04:34
  • Is there really a mathematician named Henie or did you typo Heine? – bof Oct 07 '18 at 05:23
  • Sorry,I made a mistake there, it should be Heine. – J.Guo Oct 07 '18 at 05:43
  • This has been discussed in detail on this site. Difficulties arise when trying to prove without AC that if $\neg (f(x)\to b)$ then there exists a sequence $a_n\to a$ with $\neg (f(a_n)\to b).$ If I recall correctly, this can't be done in ZF, although Countable Choice (a corollary of AC) will suffice. It can be done in ZF if $f$ is assumed to be continuous on $\Bbb R \setminus {a} $. – DanielWainfleet Oct 07 '18 at 09:52

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