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For example, can we say: $\infty=\lim\limits_{n\rightarrow\infty} n < \aleph_0$?

These are two different types of structures. The limit being like the length, extension, or just generic magnitude and the other being cardinality of a set. Can we compare magnitude to cardinality?

Intuitively, we can reach $\aleph_0$ by counting the natural numbers on the number line and in the process will be approaching $\infty$. Which leads me to believe $\infty\leq\aleph_0$. But I can't see why it should be a strict inequality. I feel like they should be of equal magnitude.

I saw on a recent comment that $2^\infty=\infty$, but are those infinities really the same? It seems not to me. Of course we (usually) have that $2^{\aleph_0}=\aleph_1$ where ${\aleph_0}$ and $\aleph_1$ are clearly two very distinct infinities, countable vs uncountable at least. Maybe one might argue that as far as the concept of magnitude is concerned, all infinities are "equal".

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    $\begingroup$ Trying to prove (or disprove) $2^{\aleph_0}=\aleph_1$ was allegedly one of the things that drive Cantor to madness. It cannot be done (at least not within standard Zermelo-Fraenkel set theory), so it must be established as an axiom. It is, of course, provable that $2^{\aleph_0}\geq \aleph_1$, but not much more than that can be said. $\endgroup$
    – Arthur
    Aug 13, 2016 at 23:14
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    $\begingroup$ $\infty$, in the language of limits, is a purely metaphoric way of expressing $n$ gets larger and larger, without bound. At least in classical calculus, there is no object such as $\infty$. $\endgroup$
    – Bernard
    Aug 13, 2016 at 23:17
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    $\begingroup$ If $\aleph_0$ is identified with the ordinal number $\omega$ and the natural number $n$ with the corresponding finite ordinal number, then $$\lim_{n\to\infty}n=\omega,$$ i.e., the infinite sequence of finite ordinals converges to the least infinite ordinal. $\endgroup$
    – bof
    Aug 13, 2016 at 23:19
  • $\begingroup$ By the way, I sort of feel like I'm repeating past answers in my answer below. If anyone wants to suggest a duplicate, I'll be happy to remove my answer later on. $\endgroup$
    – Asaf Karagila ♦
    Aug 13, 2016 at 23:28
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    $\begingroup$ (Possible candidates might include: math.stackexchange.com/questions/1807083 math.stackexchange.com/questions/1690711 math.stackexchange.com/questions/1597741 math.stackexchange.com/questions/532803 math.stackexchange.com/questions/103858 math.stackexchange.com/questions/90758 math.stackexchange.com/questions/371905 and maybe a few others. Maybe none of them. All seem related, though.) $\endgroup$
    – Asaf Karagila ♦
    Aug 13, 2016 at 23:34

3 Answers 3

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The reason the answer is negative is that $$\huge\underline{\underline{\color{red}{\textbf{Cardinals are not real numbers.}}}}$$

What do I mean by that? For finite cardinals we can nicely match the natural numbers with the ordinals, the finite cardinals, the iterated sums of the unity of the real numbers, or the rationals, or the complex numbers, or whatever.

But once infinitary operations are involved (via limits or otherwise) we are no longer playing by the same rules.

It is true that $\lim_{n\to\omega}n=\aleph_0$ if you consider this sequence as a sequence of cardinals. But using $\infty$ means that you clearly don't think about these as cardinals, but rather as real numbers or something related. And these are two entirely distinct systems. The role of $\infty$ in analysis is entirely different than the role of $\aleph_0$ as a cardinal, or $\omega$ as an ordinal.

The above mixing that finite cardinals allow is to do between these systems is whence all these mistakes come from. And you're not alone in making them. Many people do, which is why I usually write the above line in huge letters, with several underlines, when I teach this stuff to my students. I want it to be comically rememberable to them, so they never again make this mistake.

On a side note, $2^{\aleph_0}$ and $\aleph_1$ are two distinct cardinals with two distinct definitions. Positing their equality is known as the continuum hypothesis, which the standard axioms of set theory can neither prove nor disprove.

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    $\begingroup$ Okay... You say that cardinals are not real. Why would we then bother to think about them? $\endgroup$
    – Stefan Mesken
    Aug 13, 2016 at 23:29
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    $\begingroup$ Nothing is real. The Beatles said so. $\endgroup$
    – Asaf Karagila ♦
    Aug 13, 2016 at 23:30
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    $\begingroup$ @Bernard: Probably not. You should ask people who met me. I am certainly part irrational and part imaginary. So even if I'm not real, I'm certainly complex. The question is whether or not I have a real part, too. $\endgroup$
    – Asaf Karagila ♦
    Aug 13, 2016 at 23:35
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    $\begingroup$ @jdods: The $\leq$ is merely a symbol used to denote, most of the time, a partial order. It doesn't have any strict association with the real numbers, once you leave the realm of real analysis. Cardinals have a natural order, which is in fact a linear order, and actually a well-order (at least if we assume the axiom of choice). So we can use this order to define a myriad of things, from a topology (that's how we can talk about limits) to just a natural notion of "size", which should come equipped with some sort of ordering. $\endgroup$
    – Asaf Karagila ♦
    Aug 13, 2016 at 23:43
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    $\begingroup$ @jdods: Which is why you don't see $\infty\leq\aleph_0$ anywhere in mathematics. Since $\infty$ is not a cardinal and $\aleph_0$ does not lie in the extended real numbers line, the $\leq$ used for the two systems are unrelated (although both extend the usual ordering of the natural numbers, that much is true). $\endgroup$
    – Asaf Karagila ♦
    Aug 13, 2016 at 23:45
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There is an ordered set of extended natural numbers $\mathbb{N} \cup \{ \infty \}$.

The ordered class of cardinal numbers has an initial segment $\mathbb{N} \cup \{ \aleph_0 \}$.

These two ordered sets happen to be isomorphic. This fact is pretty much the entirety of the relationship between $\infty$ and $\aleph_0$.

However, there is something else along these lines that may be interesting. If you consider the hyperreal numbers of nonstandard analysis, the hyperreals contain a lot of infinite numbers $H$. However, every hyperreal (including the infinite ones) satisfies $-\infty < H < \infty$.

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  • $\begingroup$ I think the OP is looking at this from a real-analysis point of view (as evident by the use of [real-numbers] in the original tagging scheme). $\endgroup$
    – Asaf Karagila ♦
    Aug 14, 2016 at 13:07
  • $\begingroup$ @Hurkyl, I've only vaguely read about the hyperreals. Your first 3 sentences summarize a good perspective here. Thanks! Is there no rigorous way to compare the "magnitude/size" of "infinite linear extent" vs "countably infinite many distinct things" (i.e. length vs counting)? They are both infinite and there may be several ways to compare them. Can we place a well-order on the set including $n$-dimensional Lebesgue measure of all subsets of $\mathbb R^n$ for all $n$ and all cardinals? $\endgroup$
    – jdods
    Aug 14, 2016 at 14:34
  • $\begingroup$ @jdods: Admittedly, I couldn't make sense of your last question, no matter how long I stared into it. My suggestion is to study some set theory, and answer that yourself! $\endgroup$
    – Asaf Karagila ♦
    Aug 14, 2016 at 15:56
  • $\begingroup$ @AsafKaragila, fair enough. To be more clear: can we put a well order on $\{...,\infty,\aleph_0,...\}$ where the set includes both cardinal and non-cardinal versions of infinity? $\endgroup$
    – jdods
    Aug 14, 2016 at 16:07
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    $\begingroup$ @jdods: This is just like defining a linear ordering on the complex numbers. Yes, we can do this. In many ways. But we cannot do it in a way which coheres with the field structure. $\endgroup$
    – Asaf Karagila ♦
    Aug 14, 2016 at 16:36
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To summarize information I've gotten from the existing answers and discussion in comments:

  • Cardinals and real numbers are not comparable with the standard relations for real numbers nor those for cardinals (e.g. the usual $=$, $<$, etc., $<$ for real numbers is not the same $<$ as for cardinals, etc.).
  • $\infty$ is not a cardinal either and so isn't comparable to cardinals
  • One could probably define any arbitrary relation they want between $\aleph_0$ and $\infty$ and it would be of no consequence to mathematics.

===

Now, how about the following:

Let $\infty$ represent the length of the real line. We have that $1=\mu\left([n,n+1]\right)$ the length of each segment between consecutive integers for $\mu$ the standard length measure.

Thus the length of the real line is $\infty=\displaystyle\sum_{i\in\mathbb Z}1$.

Since there are exactly $\aleph_0$ unit length intervals for consecutive integers (and exactly $\aleph_0$ consecutive intervals of any finite length, of course), then to get the length of the real line, we just count these unit intervals, hence the length of the real line would be $\aleph_0$ if we were to allow $\aleph_0$ to represent a spatial magnitude.

So the only reasonable/natural comparison would be $\infty=\aleph_0$ if one were to make a comparison. NOTE: The $=$ used here is not the equals sign used to show identity of real numbers! Nor is it the equals sign used to equate cardinals!

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  • $\begingroup$ Oh, I'm struggling with this (called trolling in some circles) for more than 40 years now! You might be interested in : Double Think about Numerosity . $\endgroup$
    – Han de Bruijn
    Aug 14, 2016 at 19:50
  • $\begingroup$ The problem with the sum is that summation makes sense when summing over a partial order, and each summand is a partial order. In the case of the index being $\Bbb Z$ and the summand being $1$, you just get $\Bbb Z$ again. This is not a cardinal, this is an order type. But comparing order type is not very useful, because even linear orders can embed into one another without being isomorphic. So you're losing the antisymmetry that we want to have of the notion of comparison. $\endgroup$
    – Asaf Karagila ♦
    Aug 14, 2016 at 20:22
  • $\begingroup$ Also, since "the only reasonable comparison" is to equate $\infty$ and $\aleph_0$, the answer is that there is no reasonable comparison. Because if you think about this in terms of real numbers, there is no difference between $\lim_{n\to\infty}n$ and $\lim_{x\to\infty}x$, where $n$ means we go over the natural numbers and $x$ means we go over the real numbers. So why not use the real numbers instead? Well, because it's a different cardinality, and that really means that this notion is doomed from the beginning. $\endgroup$
    – Asaf Karagila ♦
    Aug 14, 2016 at 20:23
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    $\begingroup$ Also, while we do allow some overloading of many mathematical symbols. Equality is not really one of them. I cannot recall a situation where the equality symbol was actually overloaded. $\endgroup$
    – Asaf Karagila ♦
    Aug 14, 2016 at 22:33
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    $\begingroup$ No, the point is that summation of "something" over "something else" depends a lot on the context in which the summation takes place. In the context of real numbers, the sum $\sum_{i\in\Bbb Z}1$ is just $\infty$; in the context of cardinals it is $\aleph_0$; and in the context of order theory it is $\Bbb Z$. All of these are very different from one another. And you're trying to mix the contexts, but they don't quite mix, much like oil and water. Or people who use emacs and people who use vim. $\endgroup$
    – Asaf Karagila ♦
    Aug 14, 2016 at 23:19

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