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

"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 $\mathbb R$ "

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 $$\mathsf{ZF}$$.

• Mentioning Theorem 3.15 in Herrlich can be nice as well. Jun 6, 2021 at 14:46