Is there any way to save this “proof” that $\aleph_0=\aleph$? [closed]

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.

closed as not a real question by egreg, Potato, Chris Godsil, Sasha, Shuhao CaoJun 22 '13 at 4:47

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• There is no such thing as division of cardinal numbers. – Thomas Andrews Jun 17 '13 at 19:25
• I don't think the Taylor series for $e^x$ is valid at $x=+\infty$. – gt6989b Jun 17 '13 at 19:26
• In particular, the value of $e^n$ is not a natural number when $n$ is a natural number, so there is no way to extend the idea of $e^x$ to cardinals in a useful fashion. – Thomas Andrews Jun 17 '13 at 19:27
• Of course, it may not even be $\aleph_1$ if the continuum hypothesis is false - better to just call it $\mathfrak{c}$... – Steven Stadnicki Jun 17 '13 at 22:02
• @Hans, Panda: Actually $\aleph$ is $2^{\aleph_0}$ in Cantor's original notation. $\aleph_1$ is a whole other thing. – Asaf Karagila Jun 17 '13 at 22:07

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).

$e^1$ 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.

• As the OP I would be disappointed by this answer which just says that you must not compare apples and oranges. I'd like to learn a bit more about the subtle ways in which mistaken analogies may lead to wrong results. – Hans-Peter Stricker Jun 17 '13 at 20:05
• @HansStricker I think the fact that $e^n$ is never a counting number when $n$ is a counting number is key. Why would you expect that $e^x$ when $x$ is an infinite cardinal would ever be something that has a cardinal meaning when $e^n$ never does? Sometimes, apples and oranges is the heart of the confusion. I leave it open to him to try to prove that this $e^X$ still satisfies $e^X>2^X$, but it requires proof, and we know it isn't true... – Thomas Andrews Jun 17 '13 at 20:12
• @Hans: But it's true. You can't compare apples, pears, and oranges. And those exactly reals, ordinals, and cardinals. As arithmetical systems (especially with respects to infinitary operations, e.g. limits) those are pairwise different. – Asaf Karagila Jun 17 '13 at 22:37

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.