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I am trying to understand the definition of a limit (for a sequence) regarding hyperreal numbers converging to $L$.

The definition (see link here) states a real sequence of numbers converges to $L$ if every infinite hypernatural $H$, $x_H$ is infinitely close to $L$.

Does $L$ have to be a real number or can it be an element in the set of all hyperreal numbers?

Also, I am confused as to what an infinite hypernatural $H$ (see here) is defined to be.

For instance, what does ${}^*\lfloor 4.4\omega+5.9 \rfloor$ equal? For instance, does ${}^*\lfloor 4.4\omega+5.9 \rfloor=4\omega+5$ hold true?

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W. G.
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1 Answers1

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Does $L$ have to be a real number?

It better be, because otherwise something like the constant zero sequence would converge to a nonzero value, which is something we would normally want to avoid.

confused as to what an infinite hypernatural $H$ is defined to be

For applications, and in "Elementary Calculus: An Infinitesimal Approach", all that matters is that it's the fixed points of the extension of the floor function, as described at Wikipedia. You don't need anything else to prove facts about it.

what does ${}^*\lfloor4.4\omega+5.9\rfloor$ equal?

But if you want something you can sort of "get your hands on", so to speak, then you'll need a particular construction of a collection of hyperreals. The ultrapower construction is a good one, but the details are much thornier/require much more math experience than just using a bunch of hyperreals to do Calculus with.

does ${}^*\lfloor4.4\omega+5.9\rfloor=4\omega+5$ hold true?

The details may depend on what you mean by $\omega$, but the answer is "no". For instance, does $\lfloor4.4*3+5.9\rfloor=4*3+5$ hold true? It's the same sort of situation.

Mark S.
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  • Why does it say for any hyperreal number $x$, that $x$ is a hyperinteger if $x={}^*\lfloor x \rfloor$? What would be an example of an infinite hyperreal number then if this does not work? – W. G. May 16 '20 at 02:08
  • That's a definition of "hyperinteger" that works no matter what construction of hyperreals you're using. If you meant "infinite hyperreal integer", take any infinite hyperreal number $J$, and then put it in the extension of the floor function to get $H=\lfloor J\rfloor$. Since $\lfloor\lfloor x\rfloor\rfloor=\lfloor x\rfloor$ is true for all real $x$, $\lfloor\lfloor x\rfloor\rfloor=\lfloor x\rfloor$ for all hyperreal $x$ by the transfer principle. So $\lfloor H\rfloor=\lfloor \lfloor J\rfloor\rfloor=\lfloor J\rfloor=H$, and $H$ is a hyperinteger. Then $H+1$ and $2H$ are ones too, etc. – Mark S. May 16 '20 at 02:17