Limit of function on connected set in $\mathbb{R}$

Question

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.

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
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
http://math.stackexchange.com/questions/126010/continuity-and-the-axiom-of-choice/ — Asaf Karagila, Oct 16 '12 at 10:46
@Asaf I have trouble with understanding Brian's argument. Would you please tell me what are names of the notions ↑and 'delta on equality' (literally) so I can search for it? — Katlus, Oct 16 '12 at 10:59
Wait, something in the question is unclear to me. How do you define $\lim_{x\to p} f(x)$ to begin with? Do you also require that $f(p)=q$? — Asaf Karagila, Oct 16 '12 at 14:42
@Asaf No. $q$ may not be in $f(C)$. And the definition of $\lim_{x\to\p} f(x)$ is the same as usual. That is, $\forall \epsilon >0, \exists \delta >0$ such that $\forall x\in C, 0<d(x,p)<\delta \Rightarrow d(f(x),q)<\epsilon$. — Katlus, Oct 16 '12 at 14:51
@Katlus: I have learned [as a freshman] that $\lim_{x\to p} f(x)=q$ is to say that every sequence $p_n\to p$ has the property $f(p_n)\to q$. Do you require some sort of continuity, by the way? — Asaf Karagila, Oct 16 '12 at 14:57
@Asaf No. Not at all :) I just wanted to check 'the definition of limit I know' and the 'the definition you know' are equivalent when $C$ is connected. — Katlus, Oct 16 '12 at 15:05
In that sense, my post may not be related to the link. Still, it's very nice understanding Brian's argument. — Katlus, Oct 16 '12 at 15:15
@Katlus: Do you assume that $f$ is continuous in any sense? Namely, $f(p)=q$, if $p\in C$? (I think we can assume that, but if you already have this working assumption it will be easier for me to write an answer.) — Asaf Karagila, Oct 16 '12 at 15:23
@Asaf $f$ need not to be continuous. I think I may not get what you are asking since it seems to me that you keep asking same questions. $f$ is just any function and $q$ may not be in $f(C)$. If continuity of $f$ is essential in the proof, please let me know.. — Katlus, Oct 16 '12 at 16:17
@Katlus: I didn’t use $\uparrow$; did you mean $\upharpoonright$, as in $f\upharpoonright A$? That’s a standard notation for the restriction of the function $f$ to the set $A$. — Brian M. Scott, Oct 16 '12 at 18:53
To complement on @Brian's last comment: http://math.stackexchange.com/questions/208448/ — Asaf Karagila, Oct 16 '12 at 19:15
@Brian Yes! Since i didn't know that notation, i couldn't use LaTeX for it. — Katlus, Oct 16 '12 at 22:15
You forgot in $\Phi(q)$ to end the proposition with $d(f(x),q)<\epsilon$. — Asaf Karagila, Oct 18 '12 at 13:48

score 0 · Accepted Answer · edited Apr 13 '17 at 12:20

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.

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 06:17

Limit of function on connected set in $\mathbb{R}$

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