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Let $C$ be a connected set in $\mathbb{R}$. Let $f:C\rightarrow \mathbb{R}$ be a function. Let $p$ be a limit point of $C$.


$\phi(q)$ : For every sequence $\{p_n\}$ in $C$ where $p_n \rightarrow p$ and $p_n ≠ p$, $\lim_{n\to\infty} f(p_n) = q$

$\Phi(q)$ : $\forall \epsilon >0, \exists \delta>0$ such that $\forall x\in C, 0<d(x,p)<\delta \Rightarrow d(f(x),q)$

Then, is $\phi(q) \Rightarrow \Phi(q)$ provable, $\forall q\in \mathbb{R}$?

Till now, I have proved that there exists a sequence $\{p_n\}$ in $C$ such that $p_n ≠ p$ and $p_n \rightarrow p$.

Edit; To clarify definition of limit and $q$, I edited my original post.

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I have trouble understanding your $p$ and $q$; it appears as if you wanted $p$ to be the limit point of $C$, not $q$. –  Lord_Farin Oct 16 '12 at 10:42
@Lord You're right. Edited –  Katlus Oct 16 '12 at 10:43
Both me and Brian discussed this in a question about continuity and choice. Note that a connected set is an interval so it is the same as saying the function is defined on everything. –  Asaf Karagila Oct 16 '12 at 10:44
@Asaf So, it is unprovable in ZF. Can i get the link? –  Katlus Oct 16 '12 at 10:48

1 Answer 1

up vote 0 down vote accepted

If we require that whenever $p\in C$, $f(p)=q$, then we require that the function is sequentially continuous everywhere. In which case we can prove the continuity of $f$, as shown by Brian in Continuity and the Axiom of Choice.

If we do not require this, then the function I constructed in Connected set in $\mathbb{R}$ and constructing a sequence to a limit point is a counterexample. Namely, if $D$ is a dense [infinite] Dedekind-finite set of reals, its indicator function is a counterexample.

Note that this function is not sequentially continuous, despite the fact that there are no sequences can meet $D$ more than finitely many times. If $a\in D$ then there is a sequence of rational numbers $q_n$ approaching $a$, and almost all of those are not in $D$, therefore $$\lim_{n\to\infty}f(q_n)=0\neq f(a)=f\left(\lim_{n\to\infty} q_n\right)$$

So there is no contradiction with the first paragraph.

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It's clear now, thank you! It's interesting that "If $C$ is a connected set in $\mathbb{R}$ and $X$ is a metric space, $f:C→X$ is sequentially continuous on $C$ iff $f$ is continuous on $C$." is true. Thus, "If $f:(a,c)→X$ is sequentially continuous on $(a,b)$ and $(b,c)$, then $f(b+)=f(b-)=q$(defined by sequence) iff $\lim_{x\to b} f(x)=q$ (defined by $\epsilon - \delta$)" is true. –  Katlus Oct 19 '12 at 6:17

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