7

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".

jdods
  • 6,248
  • 4
    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. – Arthur Aug 13 '16 at 23:14
  • 10
    $\infty$, 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 $\infty$. – Bernard Aug 13 '16 at 23:17
  • 2
    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. – bof Aug 13 '16 at 23:19
  • 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. – Asaf Karagila Aug 13 '16 at 23:28
  • The only thing all infinities have in common is being not finite. If $S$ is any set then $f(x)={x}$ is an injection from $S$ into the power-set $P(S)$ but there is no injection from $P(S)$ into $S.$ So in this sense $P(S)$ is bigger than $S,$ whether $S$ is finite or not. – DanielWainfleet Aug 13 '16 at 23:31
  • @Asaf Actually, I think this is one of those questions that cannot possible have too many duplicate answers. So, even if we find a duplicate, I'd like your answer to stay (basically for anyone finding this post and being reluctant to read named duplicate). – Stefan Mesken Aug 13 '16 at 23:31
  • 1
    (Possible candidates might include: http://math.stackexchange.com/questions/1807083 http://math.stackexchange.com/questions/1690711 http://math.stackexchange.com/questions/1597741 http://math.stackexchange.com/questions/532803 http://math.stackexchange.com/questions/103858 http://math.stackexchange.com/questions/90758 http://math.stackexchange.com/questions/371905 and maybe a few others. Maybe none of them. All seem related, though.) – Asaf Karagila Aug 13 '16 at 23:34
  • Definitely seems like a duplicate just after browsing those variations. I admit I didn't search before asking. Feel free to mark as duplicate. Thanks! – jdods Aug 13 '16 at 23:44

3 Answers3

24

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.

Asaf Karagila
  • 393,674
  • 1
    Okay... You say that cardinals are not real. Why would we then bother to think about them? – Stefan Mesken Aug 13 '16 at 23:29
  • 5
    Nothing is real. The Beatles said so. – Asaf Karagila Aug 13 '16 at 23:30
  • Damn, I knew it! On the positive side: Now I have a compelling argument the next time one of my relatives tells me to get a real job. – Stefan Mesken Aug 13 '16 at 23:33
  • Are you real yourself? – Bernard Aug 13 '16 at 23:35
  • 1
    @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. – Asaf Karagila Aug 13 '16 at 23:35
  • Is such banner really necessary? Who cares about it, anyways? ;) – Pedro Aug 13 '16 at 23:39
  • @Asaf Karagila, Certainly cardinals aren't reals, and your answer is the solution, I suspect. But can you quickly elaborate or point me in since direction on how the use of inequality symbols for comparison of cardinals is justified and what it means? It seems that you are confirming my intuition though, that If we were to be and to compare the length of the real line to the cardinality of the natural numbers in a meaningful way, they should be equal. – jdods Aug 13 '16 at 23:40
  • @Pedro: Isn't it obvious? – Asaf Karagila Aug 13 '16 at 23:41
  • 3
    @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. – Asaf Karagila Aug 13 '16 at 23:43
  • 3
    @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). – Asaf Karagila Aug 13 '16 at 23:45
  • That clears it up. Thanks! – jdods Aug 13 '16 at 23:46
  • On your side note: claiming that $2^{\aleph_0}$ and $\aleph_1$ are distinct cardinals is just the negation of what the OP is claiming, so also independent of ZFC. Claiming that they are defined differently is OK tough – Jens Renders Aug 14 '16 at 14:16
  • 1
    Hm, is it like saying $i<4$? – Simply Beautiful Art Aug 14 '16 at 15:27
  • @jdods I think somewhere there is something along the lines of $2^{\aleph_0}=\aleph_n$ for $n=1,2,3,\dots \omega,\dots$ – Simply Beautiful Art Aug 14 '16 at 15:28
  • @Jens: Yes, the statement that the two are equal is a meaningful statement. Just not a provable one (from ZFC). – Asaf Karagila Aug 14 '16 at 15:29
  • @SimpleArt: No, since on the complex numbers we don't have any order structure which naturally arise from the rest of the structure. As for your second comment, what??? – Asaf Karagila Aug 14 '16 at 15:30
  • @AsafKaragila I think this is it: https://en.wikipedia.org/wiki/Easton%27s_theorem – Simply Beautiful Art Aug 14 '16 at 15:32
  • @SimpleArt: That is most certainly not what you wrote in your comment. And when restricted to the continuum itself, this is actually Solovay's theorem, that the value of the continuum is consistently anything whose cofinality is uncountable. – Asaf Karagila Aug 14 '16 at 15:35
  • I would be happy to hear some verbal reasoning for the downvote, by the way. – Asaf Karagila Aug 14 '16 at 20:31
  • @AsafKaragila where did I claim it is not meaningfull to claim that they are equal? You just make a mistake in your language: you try to explain that the equality is something that is independent from ZFC but in that same sentence you claim that they are distinct, ie not equal, which is just the negation so just as unprovable – Jens Renders Aug 15 '16 at 10:37
  • @Jens: distinct in the sense that each has its own particular definition, and these definitions are entirely unrelated. One is the cardinality of the power set of the natural numbers, another is the cardinality of all the countable ordinals. These are two distinct notions, entirely independent of one another, and it is unprovable from ZFC whether or not the two cardinals are equal. I agree that my formulation was somewhat ambiguous. – Asaf Karagila Aug 15 '16 at 11:34
  • @AsafKaragila I understand what you meant, but what you mean is not what you say, you say "distinct cardinals" while you mean "distinct definitions so not neccecarily the same cardinals". Saying that they are "distinct cardinals" is unambiguously claiming the negation of CH – Jens Renders Aug 15 '16 at 12:11
1

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 $-\infty < H < \infty$.

  • 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). – Asaf Karagila Aug 14 '16 at 13:07
  • @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? – jdods Aug 14 '16 at 14:34
  • @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! – Asaf Karagila Aug 14 '16 at 15:56
  • @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? – jdods Aug 14 '16 at 16:07
  • @jdods: You can do almost anything. The question is whether or not the structure you're putting on something has some meaningful interaction with other properties. For example, you can enumerate the rationals, does this enumeration have any meaningful relation to the algebraic structure of the rationals? Not at all. – Asaf Karagila Aug 14 '16 at 16:08
  • @AsafKaragila, ok, so then we could define a relation so that we get to choose one: $\infty=\aleph_0$, $\infty<\aleph_0$, or $\infty>\aleph_0$ to our hearts content, and it could be consistent with all the rest of mathematics, but it would probably not be a useful construction. If that is the case, then that is the answer I was looking for. Similarly, we could define $i<4$, and be consistent, but it may very well be useless. – jdods Aug 14 '16 at 16:19
  • 1
    @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. – Asaf Karagila Aug 14 '16 at 16:36
1

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!

jdods
  • 6,248
  • 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 . – Han de Bruijn Aug 14 '16 at 19:50
  • 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. – Asaf Karagila Aug 14 '16 at 20:22
  • 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. – Asaf Karagila Aug 14 '16 at 20:23
  • 1
    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. – Asaf Karagila Aug 14 '16 at 22:33
  • @AsafKaragila, Whether we take the limit over reals of naturals it doesn't matter. I meant just a sum of real numbers $1+1+\cdots$ where for each integer, we add a $1$. So we are adding $\aleph_0$ ones. I'm likely misunderstanding a technical point you are making that I would simply have to study set theory to understand though. – jdods Aug 14 '16 at 23:01
  • Yes, I am making a slightly technical point here. Although it is not set theoretic. Consider the function $f(x)=\begin{cases} x & x\in\Bbb N\ 0 & x\notin\Bbb N\end{cases}$, this function will satisfy that $\lim_{n\to\infty}f(n)=\infty$ while $\lim_{x\to\infty}f(x)$ does not exist (setting the context of natural vs. reals as I did before), moreover $\int_{\Bbb R} f(x)\mathrm d,x=0$. So in one context the limit is infinite, where in the other it is "essentially $0$". The situation is the same with summation (which is similar to integration). Context is key. – Asaf Karagila Aug 14 '16 at 23:10
  • @AsafKaragila, but I don't see how that bears on comparing $\infty$ to $\aleph_0$. I wouldn't sum $\mu([x,x+1])$ over $\mathbb R$ because it isn't disjoint. – jdods Aug 14 '16 at 23:17
  • 2
    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. – Asaf Karagila Aug 14 '16 at 23:19
  • @AsafKaragila, Thanks for the clarification. I do understand that I'm asking to compare two completely different types things for which no comparison exists. I really appreciate your patience and help here! The point I'm attempting to make is that $\mathbb R$ can be measured/covered by $\aleph_0$ finite identical length measuring sticks. I now understand that comparing infinities of fundamentally distinct contexts is not fruitful, mathematically. It's a fun thought exercise though! – jdods Aug 14 '16 at 23:54
  • @Asaf Karagila I think you are too radical. $\infty$ represents any infinity, $\aleph_0$ represents any countable infinity. I of course object calling them "exact", because the both are classes of infinities, like "positive" is a class of real numbers. What infractts me is calling either $\infty$ or $\aleph_0$ numbers. It is like calling "non-zero" a number. – Anixx Feb 18 '21 at 02:37
  • @Anixx: I don't see the point of these comments that you're leaving me, where you record your objection to standard set theoretic ideas. Please stop. – Asaf Karagila Feb 18 '21 at 08:56
  • I think it's important to distinguish between rigorous formal mathematics and less rigorous informal exploration. I enjoy both approaches. As far as comparing objects of a completely different formal type, such as "$\lim_{x\to\infty}x$" (as a real function limiting procedure) and "$\aleph_0$" (the cardinal number) the matter is settled as far as established formal math goes. It's still fun to push the "philosophical boundary" but that may or may not be fruitful. I still appreciate any thoughts on it though. – jdods Feb 18 '21 at 19:29