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:(adsbygoogle = window.adsbygoogle || []).push({});

(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!

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# I Hyperreal Convergence

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