Do different infinities have different base representations?

Question

I've been wondering about different types of infinity; e.g., $\aleph_0,\aleph_1$, e.t.c.; where $\aleph_0$ represents the smallest infinity, the countable infinity (e.g., the cardinality of the naturals); and $\aleph_1$ represents the next smallest infinity (i.e., the smallest non countable infinity).

My idea is as follows. Please bear with me. Can we think of infinities as being representable in particular number bases, so that we have $$\aleph_0 = \cdots111\;(\mathrm{base\;}1),$$ $$\aleph_1 = \cdots111\;(\mathrm{base\;}2),$$ $$\aleph_2 = \cdots222\;(\mathrm{base\;}3),$$ $$\aleph_3 = \cdots333\;(\mathrm{base\;}4),$$ and so on.

It feels to me as though this type of thinking might provide an alternative view point of what the different types of infinity are (particularly as you go beyond $\aleph_1$).

Think about it like this. The size of a base $1$ numeral is identical to the number of $1's$ in its representation. For the base one infinity (i.e., $\aleph_0$), it has a countably infinite number of $1's$ in its representation; therefore, the base $1$ infinity is a countable infinity (i.e., $\aleph_0$). Similarly, the size of a base $2$ numeral (with all characters equal to $1$) is given by $2^{n}-1$, where $n$ is the number of characters in its representation. Since the base $2$ infinity has a countably infinite number of characters, it is given by $$\aleph_1 = 2^{\aleph_0} -1 = 2^{\aleph_0 },$$ which is a well know result already.

Of course, we can continue with this thought process so that we have $$\aleph_2 = 3^{\aleph_0}\;\mathrm{and}\;\aleph_3 = 4^{\aleph_0}.$$
In general, we would have $$\aleph_n = (n+1)^{\aleph_{0}}.$$

I'm extremely interested to know what others think of this idea. Might it suggest that there are only $\aleph_0$ different types of infinity? Does it say anything about the continuum hypothesis? What about non integer bases?

Note - Although I do have a mathematics degree, I am an engineer. So please be kind with your comments :-)

Answer 1

No, it doesn't work that way.

In standard notation, $\kappa^\lambda$ (where $\kappa$ and $\lambda$ are cardinal numbers) means the number of different functions from a set with $\lambda$ elements to a set with $\kappa$ elements.

So in particular, for example, $7^{\aleph_0}$ is the number of different functions from $\mathbb N$ (the canonical set-with-$\aleph_0$-elements) to $\{0,1,2,3,4,5,6\}$, or in other words the number of different (countably) infinite sequences of "digits" from the set $\{0,1,2,3,4,5,6\}$.

In some sense this matches your intuition that $7^{\aleph_0}$ "should be" the number "$\ldots 6666$ in base $7$". I'm not sure that you can find a principled way to give a rigorous meaning to that notation, though.

Unfortunately, your project breaks down after that point. It turns out, though it is not entirely obvious, that with the above definition $n^{\aleph_0}$ is the same cardinal number for every $n$ between $2$ and $\aleph_0$, inclusive.

In other words there exists a bijection $2^{\aleph_0} \leftrightarrow 7^{\aleph_0}$. One way to see it is that we can let $2^{\aleph_0}$ correspond (with some exceptions that are too few to matter) to the binary fraction representations of all the real numbers between $0$ and $1$, and the base-7 representation of those same real numbers give you a bijection to $7^{\aleph_0}$.

So the cardinals you have defined are all the same cardinal, and not a way to construct $\aleph_1$, $\aleph_2$, and so forth, or anything like that.

Answer 2

Unfortunately this idea won't do us any good because what you've defined isn't the set of aleph numbers. As someone pointed out in the comments, $$ 2^{\aleph_0} = 3^{\aleph_0}.$$ In fact we can go further and note that $\lambda^{\aleph_0} = \kappa^{\aleph_0}$ for any $\kappa, \lambda \leq \aleph_0$. Your supposed "base $n$" expansions of the aleph numbers will reduce into a sum as follows:

$$ \dots aaa \text{ (base $a + 1$)} = \sum_{1 \leq i\leq \aleph_0}a^i = a^{\aleph_0}$$

which as mentioned is always the same value regardless of the choice of $a$.

Now, a heuristic justification for where you went wrong is that the presence of $\dots$ in any mathematical expression is a sign of the writer being nonrigourous and hoping the reader will understand his or her meaning. The caveat is that one can often inadvertently be wrong and not realize it. This is especially the case in set theory, where things we believe we understand behave in ways we often don't expect.

Answer 3

I guess a little bit of cardinal arithmetic won't hurt as an answer :

If there is an injection between the set $X$ and $Y$, we write $|X| \le |Y|$. By Cantor-Bernstein's theorem, if $|X| \le |Y|$ and $|Y| \le |X|$, then there is a bijection between $X$ and $Y$ ; in this case we write $|X| = |Y|$. (Think of $|X| = |Y|$ as only making sense as a whole if you don't want to think of the individual object $|X|$, which is the class of sets in bijection to $X$, but is not a set, so you cannot work in ZFC... anyway.)

Given two sets $X$ and $Y$, say that $|X| + |Y| = |X \cup Y|$ (I will always mean disjoint union here), $|X| \cdot |Y| = |X \times Y|$ and $|X|^{|Y|} = |X^Y|$ where $X^Y$ is the set of all functions from $Y$ to $X$. Then even though these "quantities" satisfy many elementary operations, such as $$ |X| \cdot (|Y| + |Z|) = (|X| \cdot |Y|) + (|X| \cdot |Z|), $$ (which follows from the bijection $X \times (Y \cup Z) \simeq (X \times Y) \cup (X \times Z)$ ), it also satisfies some funny identities, such as if $X$ and $Y$ are infinite sets (i.e. not in bijection with $\varnothing$ or $\{1,\cdots,n\}$ for any $n \in \mathbb N$ ; this is important since the definition of a finite set is tricky in ZF), then $$ |X| + |Y| = |X| \cdot |Y| = \max \{|X|,|Y|\}, $$ where with the maximum we mean that if $|X|$ doesn't give you equality then $|Y|$ does. Perhaps most importantly, $$ (|X| |Y|)^{|Z|} = |X|^{|Z|} \cdots |Y|^{|Z|}, $$ which when combined with the previous ones, lets you prove plenty of funny identities, such as $2^{|\mathbb N|} = 2^{|\mathbb N|} 2^{|\mathbb N|} = 4^{|\mathbb N|}$, hence by Cantor-Bernstein also equal to $3^{|\mathbb N|}$. So yeah, you need to watch out when counting cardinals. You can use this trick again and show that $2^{|\mathbb N|} = n^{|\mathbb N|}$ for any $n \in \mathbb N$ too.

Hope that helps,