Is there any way to save this "proof" that $\aleph_0=\aleph$?

Question

I came up with this idea of proving that $\aleph_0=\aleph$. I know this is not true at all, but maybe there is more to it than I can see.

we start with the inequality $\aleph_0 \leq 2^{\aleph_0}=\aleph$. We should have $2^{\aleph_0} \leq e^{\aleph_0}$, where $e^{\aleph_0}$ is defined by the following power series:

$$1+\frac{\aleph_0}{1}+\frac{\aleph_0}{2}+\frac{\aleph_0}{6}+\dots $$ The sequence of partial sums of this series is $1,\aleph_0,\aleph_0,\aleph_0,\dots$, therefore its limit should be $\aleph_0$ as well, no?

And if $e^{\aleph_0}$ is really $\aleph_0$, we have "proven" that only one infinity exists.

Is this complete BS, or maybe some hope exists for this idea?

NB another possible definition of $e^{\aleph_0}$ as $\lim_{n \to \infty} \left(1+\frac{\aleph_0}{n} \right)^n$ yields the same result.

Answer 1

Last I checked, $\aleph_0$ was not a real number. It was not subjected to the convergence laws applied to real functions.

It is very important to remember that $\infty$ in calculus is a notion which represent a point "larger than any real number", but it is completely distinct from any infinity that is found in set theory. In particular we cannot replace $\aleph_0$ by $\infty$ and use all these machinery from calculus to manipulate infinite cardinals.

Moreover, the continuum function is not continuous. That is $$\lim_n 2^n\neq2^{\lim_n n},$$ at least for cardinal numbers. It is true for real numbers (both are $\infty$) and it is true for ordinal numbers (both equal $\omega$, but ordinal exponentiation is different than cardinal exponentiation).

Answer 2

$e^1$ is not a natural number. Neither is $e^2$. Nor $e^3$. These numbers, being reals, are not cardinal numbers - they do not count the number of elements of a set.

So even if you could define $e^{\aleph_0}$ in some fashion, why would you think it would count anything, given that $e^1,e^2,\dots,e^n,\dots$ don't count anything?

Also, why would you think that $e^{\aleph_0}>2^{\aleph_0}$? Wedefine $2^{\aleph_0}$ one way, and $e^{\aleph_0}$ is defined another way. We use the same notation for $e^x$ and $2^x$ when $x$ is real because they have a similarity. But the proof of that similarity doesn't extend to arbitrary cardinalities.

There is no way to salvage this proof. The very nature of the definition of $e^x$ doesn't admit to a useful infinite cardinal extension.

Answer 3

Here's another challenge for you to try and reconcile, one with a similar 'sense' to your original question but with all of the quantities well-defined: it is in fact the case (for suitable definitions of everything here) that $|\omega|=|\omega^\omega|$. The 'trick' is that the latter is using the ordinal definition of $\omega^\omega$, where $\omega^\omega$ is the union of the sets $\{\omega, \omega^2, \omega^3, \ldots\}$ and for instance $\omega^2$ is defined as $\{0, 1, 2, \ldots, \omega, \omega+1, \ldots, \omega\cdot2, \ldots, \omega\cdot3, \ldots\}$; that is, $\omega^\omega$ is essentially the limit of the quantities $\omega^n$ as $n\to\infty$. But 'obviously' it must be the case that $\omega^\omega\leq 2^\omega$, and we 'know' that $2^\omega=\mathfrak{c}\gt\omega$, so what's going on here?

The important part of the puzzle here, as Asaf hints, is that ordinal exponentiation is distinct from cardinal exponentiation; when we talk about cardinal exponentiation $2^{\aleph_0}$ we're referring to the cardinality of the set of all functions from $\omega$ to 2, $|{}^\omega2|$. This is a very different beast from $\omega^\omega$, which refers to an ordinal quantity that's the limit of a sequence of other ordinal quantities.

In essence, $\omega^\omega$ represents a sequence, whereas $2^{\aleph_0}$ only represents (the cardinality of) a set — and the key distinction between the two is the additional structure that the sequence (the ordinality) offers. For instance, it should be obvious how to 'well-order' all of the elements of $\omega^\omega$, almost by definition; each element $\alpha$ has a clear successor $\alpha+1$, there's the least element $0=\{\}$, and with a little thinking you may be able to convince yourself that there are no infinite descending chains.

By contrast, it's not immediately obvious how to put any sort of order structure on the set ${}^\omega2$ of functions from $\omega\to2$ (or equivalently, sets of integers), and in fact it's consistent that there is no way of putting any such order structure on the set.