# Does $1-\frac{1}{2^\omega}$ equal 1? What about $1-\frac{1}{2^{\aleph_0}}$?

(Correct if I'm wrong on any of this.)

Recently, I've been learning about transfinite ordinals and cardinals. For some cases, I understand the difference between ordinals and cardinals, for instance that $$|\{1,2,3,\ldots,\omega\}| = \aleph_0$$ I get that ordinals index a set, while cardinal numbers are the sizes of sets. For some cases, however, I'm still uncertain about which to use.

An equation that would seem to me to work is $$1-\frac{1}{2^\omega} = 1$$

After all, $$\lim_{x\to\infty}{1-\frac{1}{2^x}}=1$$

However, I'm not certain that $\omega$, an ordinal number, is the right infinity to be using. Does this work: $$1-\frac{1}{2^{\aleph_0}}=1$$

Or perhaps, am I making a fallacy and does neither work?

• What does division by an ordinal or a cardinal mean? If you can come up with a consistent answer to that, then your question would basically answer itself. – Arthur May 29 '16 at 20:16
• $$\Huge\textbf{CARDINALS}\\ \Huge\textbf{ARE}\\ \Huge\textbf{NOT}\\ \Huge\textbf{REAL}\\\Huge\textbf{NUMBERS}\\ \Huge\textbf{!!!!!!!!!!!!!!!!!!!!!!}$$ – Asaf Karagila May 29 '16 at 20:17
• @AsafKaragila How does that help me? I'm new to this, so I don't know why that affects my question. – ostrichofevil May 29 '16 at 20:20
• You're new to this, great. When I teach my students about cardinals, I always write in huge letters on the board the same thing. – Asaf Karagila May 29 '16 at 20:21
• It stops you because multiplication of real numbers and multiplication of cardinals are two completely different things, so it shouldn't really be surprising that one gives rise to a well-functioning division and the other doesn't. Or at least the possibility that that might be the case shouldn't come as a surprise. – Arthur May 29 '16 at 20:26

Cardinals measure the number of elements in a set. Since there is no such thing as half an element of a set, there is absolutely no meaning to a reciprocal of a cardinal number. Certainly not an infinite cardinal.

Similarly, ordinal numbers measure the order type of an ordered set. Specifically, a well-ordered set. Since in this case there is no meaning to a reciprocal either, both of the equations $1-\frac1{2^{\aleph_0}}=1$ and $1-\frac1{2^\omega}$ are absolutely meaningless (also, note that cardinal and ordinal arithmetic are very different, $2^\omega=\omega$, whereas $2^{\aleph_0}>\aleph_0$).

It is tempting to say that in the surreal numbers one can find meaning to a reciprocal of an ordinal, but that would be false. Because the surreal numbers form a field, and as such their arithmetic is incompatible with the ordinal arithmetic, and with the cardinal arithmetic as well.

In short, there is no meaning to $\frac1{2^x}$ when $x$ is a cardinal or an ordinal, and there is even less meaning to $\lim_{x\to\infty}\frac1{2^x}$ when $x$ is either an ordinal or a cardinal, not because there is no meaning for limits. There is a meaning of limits in either context of cardinals and ordinals. But because (1) there is no meaning for reciprocals, and (2) there is no meaning for the $\infty$ symbol: do you mean the natural numbers, do you mean the entire ordinals/cardinals? The meaning of $\infty$ is unclear, and just today I was scolded by my colleague for using the $\infty$ symbol in a rather ambiguous set theoretic context.

• As an undergraduate, I tried to come up with a suitable definition for reciprocals of countably infinite ordinals. I ended up with some possible definitions - all of them where already known and didn't fit my vision of how such an object should behave. It still seems to be, that there should be a natural candidate, but the more I think about it, the less certain I am which properties they should have... – Stefan Mesken May 29 '16 at 23:50
• @StefanMesken Do you have any references re: "already known"? Curious what you came up w/ as well! – meowzz Nov 22 '18 at 3:54

There are lots of different things in mathematics called "infinity" or "infinite". Infinite cardinals are different from infinite ordinals, and the $+\infty$ encountered in calculus is a different thing from those, and the $\infty$ that is neither $+\infty$ nor $-\infty$ but is approached by going in either direction on the line is slightly different from those (that's the one that is often appropriate when one writes $\lim\limits_{x\to0}\dfrac 1 x=\infty$) and the infinite value of Dirac's delta function is an altogether different thing from all of the above, and the infinite number of Abraham Robinson's non-standard analysis or of Euler's way of doing calculus is different from all of the above, and Knuth's "surreal numbers" (relevant to strategy in games of skill) are different from all of the foregoing. And there are yet others.

Some arithmetic operations are not appropriate for some of these "infinities", just as, for example, one cannot do matrix multiplication when the number of columns on the left fails to match the number of rows on the right.

Accordingly, one does not evaluate things like $\dfrac 6 {\aleph_2}$, etc.

• There are hardly any thing in mathematics which is more cringe worthy than $$\int_{\Bbb R}f({\aleph_{52}}^\omega)\mathrm{d}\omega.$$ – Asaf Karagila May 29 '16 at 20:30
• @AsafKaragila : Here is a suggestion : $$\int_\mathbb{R} f\left( \left(1-\frac{1}{2^{\aleph_0}}\right) \cdot \aleph_{52}^\omega \right) \; d\omega.$$ – M.G May 29 '16 at 20:38
• @Asaf What about $\int_{X} \int_{n < \omega} M_{f(n)}^{\#} dn df$?, where $M_n^{\#}$ is the sharp for the minimal model with $n$ Woodin cardinals and $X \subseteq ^\omega\omega$. Given a careful coding and suitable $X$, this yields an actual real (well...sort of). – Stefan Mesken May 30 '16 at 0:01