TITLE: Continuity and the Axiom of Choice
QUESTION [68 upvotes]: In my introductory Analysis course, we learned two definitions of continuity.
$(1)$ A function $f:E \to \mathbb{C}$ is continuous at $a$ if every sequence $(z_n) \in E$ such that $z_n \to a$ satisfies $f(z_n) \to f(a)$.
$(2)$ A function $f:E \to \mathbb{C}$ is continuous at $a$ if $\forall \varepsilon>0, \exists \delta >0:\forall z \in E, |z-a|<\delta \implies |f(z)-f(a)|<\varepsilon$.
The implication $(2)\implies(1)$ is trivial (though I will happily post a proof if there is sufficient interest).  The proof of the implication $(1)\implies(2)$ is worth remarking on, though.
Proof that $(1)\implies(2)$:
Suppose on the contrary that $\exists \varepsilon>0:\forall \delta>0, \exists z \in E:\left (|z-a|<\delta \; \mathrm{and} \; |f(z)-f(a)|\ge \varepsilon\right )$.  Let $A_n$ be the set $\{z\in E:|z-a|<\frac{1}{n} \; \mathrm{ and }\; |f(z)-f(a)|\ge\varepsilon\}$.  Now use the Axiom of Choice to construct a sequence $(z_n)$ with $z_n \in A_n \; \forall n \in \mathbb{N}$.  But now $a-\frac{1}{n}<z_n<a+\frac{1}{n}\; \forall n \in \mathbb{N}$ so $z_n \to a$.  So $f(z_n) \to f(a)$.  But $|f(z_n)-f(a)|\ge\varepsilon\; \forall n \in \mathbb{N}$, which is a contradiction.
You will have noticed that the above proof uses the Axiom of Choice (the lecturer didn't explicitly spell out the dependence, but it's definitely there).  My question is: is it possible to prove that $(1) \implies (2)$ without using the Axiom of Choice.  I strongly suspect that it isn't.  In that case, can anyone prove that we have to use the Axiom of Choice?  I can think of three ways to do this:
(A) Show that $\left( (1) \implies (2)\right)\implies \mathrm{AC}$.  I suspect that this statement is untrue. This is definitely untrue, as Arthur points out, because I only used the axiom of countable choice, which is strictly weaker than AC.
(B) Show that $(1)\implies (2)$ is equivalent to some other statement known to require the Axiom of Choice (the obvious example being the well-ordering of the real numbers).
(C) Construct or show the existence of a model of ZF in which there exist sequences which satisfy $(1)$ but not $(2)$.
Of course, if anyone can think of another way, I would be very interested to hear about it.
One final note - I am aware that very many theorems in Analysis use the Axiom of Choice in one way or another, and that this is just one example of such a theorem.  If there exists a model of ZF like the one described in (C), is the study of Analysis in that model interesting?

REPLY [32 votes]: The implication $(1)\to (2)$ can be proved in ZF alone, though this requires major revision of the argument.
Assume that $f:\Bbb R\to\Bbb R$ is sequentially continuous (i.e., satisfies (1)). Let $x\in\Bbb R$ be arbitrary.

Claim: $f\upharpoonright(\Bbb Q\cup\{x\})$ is continuous at $x$.
Proof: Enumerate $\Bbb Q=\{q_n:n\in\omega\}$. If $f\upharpoonright(\Bbb Q\cup\{x\})$ is not continuous at $x$, there is an $\epsilon>0$ such that for each $k\in\omega$, $$A_k\triangleq\{q\in\Bbb Q:|q-x|<2^{-k}\text{ and }|f(q)-f(x)|\ge\epsilon\}\ne\varnothing\;.$$ Let $$n(0)=\min\{k\in\omega:q_k\in A_0\}\;.$$ Given $n(m)$, let $$n(m+1)=\min\{k\in\omega:k>n(m)\text{ and }q_k\in A_{m+1}\}\;.$$ Then $\langle q_{n(k)}:k\in\omega\rangle\to x$, but $|f(q_{n(k)})-f(x)|\ge\epsilon$ for all $k\in\omega$, which is a contradiction. Note that no choice was used in this construction. $\dashv$

Now let $\epsilon>0$. It follows from the Claim that there is a $\delta>0$ such that $|f(x)-f(q)|\le\epsilon$ whenever $q\in\Bbb Q$ and $|x-q|\le\delta$. Now suppose that $y\in\Bbb R$ with $|x-y|\le\delta$. Let $I$ be the closed interval whose endpoints are $x$ and $y$. For each $k\in\omega$ let $$n(k)=\min\{m\in\omega:q_m\in I\text{ and }|q_m-y|<2^{-k}\}\;.$$ Then $\langle q_{n(k)}:k\in\omega\rangle\to y$, so $\langle f(q_{n(k)}):k\in\omega\rangle\to f(y)$. However, each $q_{n(k)}\in I$, so $|q_{n(k)}-x|\le\delta$ for each $k\in\omega$, and therefore $|f(q_{n(k)})-f(x)|\le\epsilon$ for each $k\in\omega$. Clearly this implies that $|f(y)-f(x)|\le\epsilon$ as well.
Thus, we’ve shown that for each $\epsilon>0$ there is a $\delta>0$ such that $|f(y)-f(x)|\le\epsilon$ whenever $|y-x|<\delta$, which is sufficient.
This argument is expanded from the ZF proof of Theorem 3.15 in Horst Herrlich, Axiom of Choice, Lecture Notes in Mathematics 1876. Note that it is not true in ZF that a function $f:\Bbb R\to\Bbb R$ is continuous at a point $x$ iff it is sequentially continuous at $x$: in Theorem 4.54 he proves that this assertion is equivalent to $\mathbf{CC}(\Bbb R)$, the assertion that every countable family of non-empty subsets of $\Bbb R$ has a choice function.
Added: I failed to notice that the functions in the original question are complex-valued, not real-valued. However, the argument can easily be adapted to $\Bbb R^2$, replacing $\Bbb Q$ by $\Bbb Q^2$ and using the $\max$ norm.
Added2: The only slightly tricky bit is figuring out what to use for $I$. Suppose that $x=\langle x_1,x_2\rangle,y=\langle y_1,y_2\rangle\in\Bbb R^2$ with $0<\|x-y\|\le\delta$. If $x_1\ne y_1$ and $x_2\ne y_2$, let $$I=\big[\min\{x_1,y_1\},\max\{x_1,y_1\}\big]\times\big[\min\{x_2,y_2\},\max\{x_2,y_2\}\big]\;.$$ If $x_1\ne y_1$ and $x_2=y_2$, let $$I=\big[\min\{x_1,y_1\},\max\{x_1,y_1\}\big]\times\big[x_2,x_2+\delta\big]\;,$$ and if $x_1=y_1$ and $x_2\ne y_2$ let $$I=\big[x_1,x_1+\delta\big]\times\big[\min\{x_2,y_2\},\max\{x_2,y_2\}\big]\;.$$
Let $\Bbb Q^2=\{q_n:n\in\omega\}$ be an enumeration of $\Bbb Q^2$, and for $k\in\omega$ let $$n(k)=\min\{m\in\omega:q_m\in I\text{ and }\|q_m-y\|<2^{-k}\}\;.$$ Everything else is the same as before, except that $|\cdot|$ must be replaced throughout by $\|\cdot\|$, where $$\|\langle x,y\rangle\|=\max\{|x|,|y|\}\;.$$
With very minor modification this works for $\Bbb R^n$ for any $n\in\Bbb Z^+$.