"Sequential continuity is equivalent to $\epsilon$-$\delta $ continuity " implies Axiom of countable choice for collection of subsets of $\mathbb R$?

Question

"A function $f: \mathbb R \to \mathbb R$ is continuous at $x \in \mathbb R$ , if and only if it is sequentially continuous " , does this statement imply "the Axiom of Choice for countable collections of non-empty subsets of $\mathbb R$ "

Answer 1

Theorem $\mathbf{4.54}$ in Horst Herrlich, Axiom of Choice:

Equivalent are:

1. $\Bbb R$ is Fréchet.
2. Each subspace of $\Bbb R$ is sequential.
3. $\Bbb R$ is Lindelöf.
4. Each subspace of $\Bbb R$ is Lindelöf.
5. Each second countable topological space is Lindelöf.
6. Each subspace of $\Bbb R$ is separable.
7. Each second countable topological space is separable.
8. A function $f:\Bbb R\to\Bbb R$ is continuous at some point $x$ iff it is sequentially continuous at $x$.
9. A function $f:X\to\Bbb R$, defined on some subspace $X$ of $\Bbb R$, is continuous iff it is sequentially continuous.
10. $\operatorname{CC}(\Bbb R)$.

Here $\operatorname{CC}(\Bbb R)$ is the axiom of choice for countable families of non-empty subsets of $\Bbb R$.

I’ll not prove the full equivalence, but I will show the equivalence of $(8),(9)$, and $(10)$; the argument is adapted from Herrlich.

Assume $(9)$. I’ll show first that every unbounded subset $A$ of $\Bbb R$ contains an unbounded sequence. Let $h:\Bbb R\to(0,1)$ be a homeomorphism; without loss of generality, $0$ is an accumulation point of $h[A]$. Let $X=h[A]\cup\{0\}$, and define

$$f:X\to\Bbb R:x\mapsto\begin{cases} 0,&\text{if }x\in h[A]\\ 1,&\text{if }x=0\;. \end{cases}$$

Then $f$ is not continuous at $0$, so by $(9)$ there is a sequence $\langle y_n:n\in\Bbb N\rangle$ in $A$ such that $\langle h(y_n):n\in\Bbb N\rangle$ converges to $0$ in $X$; clearly $\langle y_n:n\in\Bbb N\rangle$ is unbounded in $\Bbb R$.

Now let $\{X_n:n\in\Bbb N\}$ be a countable family of non-empty subsets of $\Bbb R$, and let $h:\Bbb R\to(0,1)$ be as before. For each $n\in\Bbb N$ let $\varphi_n:\Bbb R^n\to\Bbb R$ be a bijection, and let $X_n'=\varphi_n\left[\prod_{k\le n}X_k\right]$. Define $t_n:\Bbb R\to\Bbb R:x\mapsto n+x$, and let $Y_n=t_n[h[X_n']]\subseteq(n,n+1)$. Finally, let $Y=\bigcup_{n\in\Bbb N}Y_n$; clearly $Y$ is unbounded in $\Bbb R$, so there is an unbounded sequence $\langle y_n:n\in\Bbb N\rangle$ in $Y$.

Let $M=\{m\in\Bbb N:\exists n\in\Bbb N(y_n\in Y_m)\}$. Then $\prod_{m\in M}Y_m\ne\varnothing$, so fix any $$\langle y_m':m\in M\rangle\in\prod_{m\in M}Y_m\;.$$ For $m\in M$ define $x_m'$ to be the unique element of $X_m'$ such that $t_m(h(x_m'))=y_m'$; clearly $\langle x_m':m\in M\rangle\in\prod_{m\in M}X_m$.

For each $n\in\Bbb N$ let $m(n)=\min\{m\in M:n\le m\}$. Then $x_{m(n)}'=\langle x_0^{(n)},\ldots,x_{m(n)}^{(n)}\rangle$ for some $x_k^{(n)}\in X_k$, $k=0,\ldots,m(n)$, and we set $x_n=x_n^{(n)}$. Then $x_n\in X_n$ for each $n\in\Bbb N$, and $\operatorname{CC}(\Bbb R)$ follows.

That $(10)$ implies $(8)$ is straightforward, so it only remains to show that $(8)$ implies $(9)$. Assume $(8)$, and suppose that $f:X\to\Bbb R$ is sequentially continuous, where $X\subseteq\Bbb R$. Let $F\subseteq\Bbb R$ be closed; $f$ is sequentially continuous, so $C=f^{-1}[F]$ is sequentially closed in $X$. Suppose that $C$ is not closed in $X$, and let $x\in(\operatorname{cl}_XC)\setminus C$. Since $C$ is sequentially closed, there is no sequence in $C$ converging to $x$. Define

$$g:\Bbb R\to\Bbb R:x\mapsto\begin{cases} 1,&\text{if }x\in C\\ 0,&\text{otherwise}\;; \end{cases}$$

then $g$ is sequentially continuous at $x$ but not continuous at $x$, contradicting $(8)$.

Added 6 June 2021: As noted below by Watson, Herrlich’s Theorem $\mathbf{3.15}$ is also relevant, as it shows that matters are quite different if we assume that $f$ is sequentially continuous at every point of $\Bbb R$: it says that although the usual proof requires the axiom of choice, the theorem that every sequentially continuous $f:\Bbb R\to\Bbb R$ is continuous can actually be proved in $\mathsf{ZF}$.