Hyperreal Convergence: Is It 0 or Infinitesimal?

[Someone2841]

I have always thought that non-constant sequences that converge toward 0 in the reals converge toward an infinitesimal in the hyperreals, but recently I have questioned my presumption. If $(a_n)\to0$ in $R$, wouldn't the same seuqnece converge to 0 in $*R$? These two statements should capture convergence of $(a_n)\to 0$ in the reals and hyperreals, respectively:

(i) $\forall \epsilon \in \mathbb{R}^+ \; \exists N \in \mathbb{N} \; \forall n \in \mathbb{N} : n \geq N \implies |a_n| < \epsilon$

(ii) $\forall \epsilon \in *\mathbb{R}^+ \; \exists N \in *\mathbb{N} \; \forall n \in *\mathbb{N} : n \geq N \implies |a_n| < \epsilon$

For example, take $(a_n=1/n)$. Clearly this converges to 0 in the reals. Choose a $\epsilon \in *\mathbb{R}^+$; just for fun, say it is an infinitesimal. If $\epsilon$ is infinitesimal, then $H=1/\epsilon$ is hyperfinite. Any real number x has a natural number $\lceil x \rceil$ such that $x \leq \lceil x \rceil < x+1$, by the transfer principle there much be a hypernatural $\lceil H \rceil$ such that $H \leq \lceil H \rceil < H+1$ and $a_{\lceil H \rceil} \leq \epsilon$. Because the sequence is strictly decreasing, this means that all terms beyond $\lceil H \rceil$ will be strictly less than $\epsilon$, and so the sequence must to 0 and not any infinitesimal. Is this correct? Thanks!

 

[Zafa Pi]

I have always thought that non-constant sequences that converge toward 0 in the reals converge toward an infinitesimal in the hyperreals, but recently I have questioned my presumption. If $(a_n)\to0$ in $R$, wouldn't the same seuqnece converge to 0 in $*R$? These two statements should capture convergence of $(a_n)\to 0$ in the reals and hyperreals, respectively:

(i) $\forall \epsilon \in \mathbb{R}^+ \; \exists N \in \mathbb{N} \; \forall n \in \mathbb{N} : n \geq N \implies |a_n| < \epsilon$

(ii) $\forall \epsilon \in *\mathbb{R}^+ \; \exists N \in *\mathbb{N} \; \forall n \in *\mathbb{N} : n \geq N \implies |a_n| < \epsilon$

For example, take $(a_n=1/n)$. Clearly this converges to 0 in the reals. Choose a $\epsilon \in *\mathbb{R}^+$; just for fun, say it is an infinitesimal. If $\epsilon$ is infinitesimal, then $H=1/\epsilon$ is hyperfinite. Any real number x has a natural number $\lceil x \rceil$ such that $x \leq \lceil x \rceil < x+1$, by the transfer principle there much be a hypernatural $\lceil H \rceil$ such that $H \leq \lceil H \rceil < H+1$ and $a_{\lceil H \rceil} \leq \epsilon$. Because the sequence is strictly decreasing, this means that all terms beyond $\lceil H \rceil$ will be strictly less than $\epsilon$, and so the sequence must to 0 and not any infinitesimal. Is this correct? Thanks!

Though you and I have different notations/sources on non-standard analysis I'll give what I consider the best answer from my perspective since no one else is responding. If you merely say that the sequence {a_n} converges to 0 in R, then it converges to 0 whether you're thinking of the usual or non-standard reals.

If you say, |ε| < 1/n for each standard n, then ε is not necessarily 0, it could be an infinitesimal. But that doesn't contradict the above.

My understanding of non-standard analysis come from the 11 simple pages of Chapters 4, 5, and 6 of Ed Nelson's hyper-beautiful book: https://web.math.princeton.edu/~nelson/books/rept.pdf.

 

[math771]

Yes, your reasoning is correct. The transfer principle states that any property that holds for real numbers also holds for hyperreal numbers, so if a sequence converges to 0 in the reals, it will also converge to 0 in the hyperreals. This is because the hyperreals contain all the real numbers, plus additional infinitesimal and infinite numbers, so any statement about real numbers can be extended to the hyperreals.