In Baby Rudin, p. 27, it is stated that the $\infty$ in notations like

$\sum\limits_{i=0}^\infty i$


$\bigcup\limits_{i=0}^\infty A_i$

is not the same as the $+\infty$ in the extended real number system (p. 11-12).

This leads me to think that: we could create an extended natural number system, with $\infty$ as the supremum of every unbounded set in the standard natural number system. However, Rudin's statement would mean the $\infty$ in the extended natural numbers is not the same as the $+\infty$ of the real numbers.

Why not?

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    $\begingroup$ I don't know the answer, but you should look into net convergence, and think about nets $\omega+1 \rightarrow X$ where $X$ is a topological space, and $\omega+1$ is the successor of $\omega$, where $\omega = \{0,1,2,3,\ldots\}$ is the least infinite ordinal numbers. $\endgroup$ – goblin GONE Jan 26 '15 at 5:00

If you did in fact extend the natural numbers in that way, your $\infty$ would be the same as that of the extended real numbers. The real point here is that $\infty$ in the expressions $\sum_{i=0}^\infty a_i$ and $\bigcup_{i=0}^\infty A_i$ is actually not serving as a number of any kind: there is no term $a_\infty$ or set $A_\infty$ to be included in the sum or union.

$\sum_{i=0}^\infty a_i$ is actually (in the calculus context) an abbreviation for a limit: it really means


and the $\infty$ in that limit is again not a number: $\displaystyle\lim_{n\to\infty}$ is itself an abbreviation for a moderately complicated concept.

$\bigcup_{i=0}^\infty A_i$ is something different yet: it’s a conventional notation for what could better be written $$\bigcup_{i\in\Bbb N}A_i$$ or $$\bigcup\{A_i:i\in\Bbb N\}\;,$$ notations that are easily extended to unions over other index sets.

Added: Indeed, with unions and intersections there isn’t even an implicit notion of order involved, and we don’t actually need an index set. If $\mathscr{A}$ is any set of sets, we have

$$\bigcup\mathscr{A}=\left\{x:\exists A\in\mathscr{A}(x\in A)\right\}$$


$$\bigcap\mathscr{A}=\left\{x:\forall A\in\mathscr{A}(x\in A)\right\}\;.$$

  • $\begingroup$ Thank you @Brian M.Scott for the clarifications. Firstly, since there are an infinite number of terms in the union, $\bigcup_{i\in\Bbb N}A_i$ is still a limit concept in some sense, right (albeit without a precise $\epsilon-\delta$ definition)? Secondly, I understand the latter two union notations avoids $\infty$, but it seems we lose the ability to specify the starting index. $\endgroup$ – TSJ Jan 26 '15 at 5:33
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    $\begingroup$ @user208884: No, the union does not require any notion of limit. And for unions and intersections there really is no notion of starting index. We don’t even need an index set. If $\mathscr{A}$ is any set of sets, then $\bigcup\mathscr{A}$ is by definition $$\left\{x:\exists A\in\mathscr{A}(x\in A)\right\}\;,$$ and $\bigcap\mathscr{A}$ is by definition $$\left\{x:\forall A\in\mathscr{A}(x\in A)\right\}\;.$$ $\endgroup$ – Brian M. Scott Jan 26 '15 at 5:35
  • $\begingroup$ My level of mathematical maturity is not letting me grasp why the union of a (countably) infinite number of sets does not require a limit, while the addition of infinite terms requires a limit. In both cases, you cannot find the result by systematically stepping through the terms one by one (that would take infinitely long). In the addition case, you can say you found the result when you know the series approach your result arbitrarily closely. But if you don't have this notion of the limit, how will you be able to say you found the answer in the union case?Is this covered in formal settheory? $\endgroup$ – TSJ Jan 26 '15 at 5:54
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    $\begingroup$ @user208884: It is, but you don’t need anything more formal than the definitions that I gave in the previous comment. The infinite series and the union of infinitely many sets are two fundamentally different kinds of animal. The former really is defined as a limit; the latter has a definition that does not rely on the sets being ordered in any way at all. Something is in the union of a bunch of sets if it’s in at least one of them; it’s in their intersection if it’s in all of them. That’s it: no first set, no ordering of the sets — just an unstructured collection of them. $\endgroup$ – Brian M. Scott Jan 26 '15 at 5:58
  • $\begingroup$ That is enlightening. I've taken a few calculus classes but I realized just now order is intrinsic in the limiting process. To prove the set B is the result of a countably infinite intersection, one way is to show every member of B is in fact in $\forall A\in\mathscr{A}$ and every non-member is not in some A. But there is no definite sequence for going through all the A's, so this "notion" is different from a limit. Interesting. Thanks Brian M.Scott. $\endgroup$ – TSJ Jan 26 '15 at 6:21

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