Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $C$.

Here,

$\phi(q)$ : 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.

share|improve this question
    
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
1  
    
@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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.