Let $A\subseteq\mathbb{R}$, $p\in\mathbb{R}$. I proved that the following are equivalent:
- $\exists\left(x_{n}\right)_{n\in\mathbb{N}}\subseteq A\cap\left\{ p\right\} ^{c}$ such that $x_{n}\rightarrow p$
- $\text{hal}\left(p\right)\cap{^{*}}A\cap\left\{ p\right\} ^{c}\neq\emptyset$
- $\forall\epsilon\in\mathbb{R}_{>0},\#(B_{\epsilon}\left(p\right)\cap A)=\infty$
- $\forall\epsilon\in\mathbb{R}_{>0},B_{\epsilon}\left(p\right)\cap A\cap\left\{ p\right\} ^{c}\neq\emptyset$
- $p$ is a limit point of $A$
where $\text{hal}\left(p\right)$ is the set of the hyperreals infinitely close to $p$, ${^{*}}A$ is the hyperreal extension of $A$, $B_{\epsilon}\left(p\right):=(p-\epsilon,p+\epsilon)$.
Clearly $\#(\text{hal}\left(p\right)\cap{^{*}}A)=\infty$ implies 2. My problem is: is the converse true?
Attempt 1: Suppose that $\exists x\in\text{hal}\left(p\right)\cap{^{*}}A\cap\left\{ p\right\} ^{c}$, i.e., $\exists\epsilon\in\mathbb{I}\smallsetminus\left\{ 0\right\}$ such that $p+\epsilon\in{^{*}}A$ (where $\mathbb{I}$ is the ring of infinitesimals). In particular, $A$ is infinite. If $p$ is an interior point of $A$, then $\forall n\in\mathbb{N}, p+n\epsilon\in{^{*}}A$. If not, I don't know how to find infinite members of $\text{hal}\left(p\right)\cap{^{*}}A$.
Attempt 2: Suppose that $\exists(x_n)_n\subseteq B_\epsilon(p)\cap A$, with all $x_n$ diferents and $\epsilon\in\mathbb{R}_{>0}$ fixed. I.e., $\forall n\in \mathbb{N},x_n\in B_\epsilon(p)\cap A$ and $\forall n\in \mathbb{N}, \forall m\in \mathbb{N}\smallsetminus{n}, x_n\neq x_m$. By transfer, I obtain infinite (differents) members of $^*A$, etc.
I also tried to find a counterexample, without luck. Examples:
- $A=\{1/n\}_{n\in\mathbb{N}}$, $p=0$. Then $(1/n)_{n\in \mathbb{N}}\rightarrow p$ but $\forall n\in\mathbb{N},1/n\in A$ implies $\forall n\in{^*}\mathbb{N},1/n\in{^*}A$ by transfer and $1/n\simeq p$ for $n\in\mathbb{N}_\infty$.
- Let $A$ be an interval with end points $a,b$ and let $p$ be a limit point of $A$ that it's not an interior point of $A$. Then $p\in\{a,b\}$ and therefore $\forall n\in\mathbb{N},a+n\epsilon\in\text{hal}\left(p\right)\cap{^{*}}A$ or $\forall n\in\mathbb{N},b-n\epsilon\in\text{hal}\left(p\right)\cap{^{*}}A$.
Okay, I solved it, and it's trivial, and it's a simple problem of calculus: if $x_{n}\rightarrow p$ then exists infinite differents $x_n$ (proof by contradicton). Then, by transfer, exists infinite differents $x_N$.
If someone want more details, tell me.