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

Suppose that;

$\exists \epsilon>0$ such that [ $\forall \delta>0, \exists x\in C$ such that $0<d(p,x)<\delta \bigwedge d(f(x),q)≧\epsilon$. ($q$ is an arbitrary point in $\mathbb{R}$)

Now, define $A_n = \{x\in C|0<d(p,x)<1/n \bigwedge d(f(x),q)≧\epsilon \}$.

Then, it can be shown that $\forall m\in \omega, m\preceq A_n$.

Q1. Does this imply that $A_n$ is dedekind-infinite in ZF? If so, how?

Q2. $\{A_n\}$ defined above is a decreasing sequence and $\mathbb{R}$ is complete. I guess we can form a sequence $\{p_n\}$ in $C$ such that $p_n \rightarrow p \bigwedge p_n≠p \bigwedge \lim_{n\to\infty} f(p_n)≠q$. Can we? (in ZF)

This post is related to ;

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

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Katlus
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1 Answers1

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No, you cannot deduce that $A_n$ is Dedekind-infinite. Let me give an example:

Suppose that $D\subseteq\mathbb R$ is a dense infinite Dedekind-finite set (e.g. in Cohen's first model). Now take $f$ to be the indicator function of $D$, namely $f(x)=1$ if and only if $x\in D$, and $0$ otherwise. Now take $\varepsilon=\frac12$.

If $p\notin D$, then $A_n$ is Dedekind-finite, because any point outside of $D$ have the same value as $p$, so it certainly cannot satisfy the defining formula of $D$. And subsets of a Dedekind-finite sets are Dedekind-finite.

Now it is obvious that the second question also takes a negative answer. In the above example, we cannot construct a sequence which converges to $p$ and all its members are from $D$.

Asaf Karagila
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