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In the definition of limit of a function ($\epsilon-\delta$ definition) we say certain statements such as for every $\epsilon>0$ there exist $\delta>0$ .... Now my question is, is a choice function required to ensure that for every $\epsilon$ there exist a $\delta>0$? Moreover, how we can we test whether a mathematical statement depends on axiom of choice or not? I really don't know whether this question has any meaning or not, so please help me.

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  • $\begingroup$ I don't understand how choice could be involved. It's a definition, not a theorem. I.e., you're not making a claim of the form "the limit of a function is $x$, therefore there exists a $\delta$...". Rather you're making a statement of the form "the limit of a function is x, which is to say there exists a $\delta$..." $\endgroup$ – hasnohat Aug 21 '13 at 14:19
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The question doesn't make much sense, and it seems that you are misunderstanding the definition.

We define continuity by $\varepsilon$-$\delta$ at a point $x$ in such way. If you want to show that a function is continuous at $x$ it suffices to show that every $\varepsilon$ has such $\delta$. We are not required to assign in a single stroke all the $\varepsilon$'s with fitting $\delta$'s. We just have to show that given $\varepsilon>0$ there is some $\delta$ which satisfies the statement.

On the other hand, the axiom of choice is required in order to show that continuity by $\varepsilon$-$\delta$ at a point is equivalent to continuity using sequences (i.e. $f(x)=\lim_{x_n\to x_0}f(x_n)$ for every $x_n\to x$). The axiom of countable choice suffices here.

For more details on that last part, see the entire discussion on: Continuity and the Axiom of Choice

Finally, to prove that the axiom of choice is required to prove $\varphi$ we can either show that $\varphi$ implies something we already know to require the axiom of choice (or a fragment thereof), or that we can construct a model of $\sf ZF$ in which $\varphi$ fails (of course it makes sense only if $\sf ZFC$ proves $\varphi$ in the first place).

For more on that, you can read:

  1. How do we know we need the axiom of choice for some theorem?
  2. We cannot write this function
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    $\begingroup$ I think the confusinon is: There is a $\delta$ (depending on $\varepsilon$). Which one? If we pick one, did we use choice? Of course, we do not need to assign a single $\delta$ to each $\varepsilon$ (knowing that there are possible assignments suffices). But, regardless, if we were to assign a $\delta$ (in a situation where the limit indeed exists), this is still possible without any use of choice, because if $\delta>0$ works for $\varepsilon$, then so does any $\delta'$ with $0<\delta'<\delta$, so we may restrict ourselves to rational (thus explicitly enumerated) values of $\delta$. $\endgroup$ – Andrés E. Caicedo Aug 21 '13 at 14:52
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The answer to your first question is: usually no. In $\epsilon-\delta$ questions, you usually have to construct a specific $\delta$ given $\epsilon$; i.e. your $\delta$ will be some function or equation in terms of $\epsilon$.

Having such a construction is exactly when you don't need the Axiom of Choice. Very basically, you need the Axiom of Choice when you need to make an infinite number of choices that you cannot explicitly construct.

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I don't get your first question, could you elaborate? As for the second one, a mathematical statement can't "depend" on the axiom of choice, a proof of this statement can.

Now, if your question was "does a proof of the statement require the axiom of choice", then this is very difficult to answer. More formally, for a formula $\phi$, you are asking whether $\mathrm{ZF} \not\vdash \phi$ but $\mathrm{ZFC} \vdash \phi$ is true. To illustrate how this is difficult, take $\phi$ to be the countable axiom of choice. Then it has not been known for a long time whether $\mathrm{ZF} \not\vdash \phi$ was true, altough $\mathrm{ZFC}\vdash \phi$ is trivial.

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