$2^{\aleph_0}$ and $\log$ function

Question

I am a layman when it comes to infinite sets, and I have had this question for a while. I have read that $2^x > x$ for all cardinals, including infinite cardinals. If this is the case, then it would imply that the the inverse must be true, namely $\log_2(x) < x$ for all cardinals, including infinite cardinals. If this were true, then $\log_2(\aleph_0) < \aleph_0$, and $\log_2(\log_2(\aleph_0)) < \log(\aleph_0)$, and so on. This implies that there are an infinite number of infinities between the finite cardinals and $\aleph_0$. Where am I going wrong with this argument?

Answer 1

There is no well-defined "base 2 logarithm" function on cardinal numbers. Indeed, there does not exist any cardinal $x$ such that $2^x=\aleph_0$, since if $x$ is finite then $2^x$ is finite and if $x$ is infinite then $2^x>x\geq\aleph_0$. So what you might call "$\log_2 \aleph_0$" simply does not exist.

More generally, you cannot assume that cardinal numbers have any familiar properties that real numbers have. Some properties of real numbers still hold for cardinal numbers, but most do not. Just because there is an operation called "logarithms" on real numbers does not allow you to assume such an operation also exists on cardinal numbers.

Answer 2

Look at the $\log_2$ function on the natural numbers. Not every natural number is $\log_2(n)$ for some $n$. Only numbers of the form $2^k$ even have a base $2$ logarithm.

Sure, we can "round down", but what would $\log_2(n)$ be? It would be the largest number $k$ such that $2^k\leq n$. In the case of the transfinite cardinals, this can result in a set without a largest element. Like for example $\aleph_0$, since for all $k<\aleph_0$ we have that $2^k<\aleph_0$. To mitigate that, the "round down" operation would be to take the supremum, which in this case would be $\aleph_0$ itself.

So, in summary, not every cardinal (even a finite one) is $\log_2$ of something, and if you want to have a "rounding down" version, then you would get that $\log_2\aleph_0=\aleph_0$, but this does not mean that $2^{\aleph_0}=\aleph_0$, just like how $2^{\lfloor\log_2 3\rfloor}\neq 3$.

Answer 3

We have a well defined function $\log_2(x)$ if $x$ is real. If you want to talk of $\log_2(\kappa)$ for $\kappa$ an infinite cardinal you need to define it in a way that returns a single value for each $\kappa$. The fact that set theory generally doesn't care about reals but only naturals is not a problem for this question, we can take the floor to get a function $\lfloor \log_2(n) \rfloor$ which is well defined on the naturals. For many functions we want to say that for $\alpha$ a limit ordinal $f(\alpha)=\sup_{\beta \lt \alpha}f(\beta)$. If we use that definition we get $\log_2 \aleph_0=\aleph_0$. You are going wrong in the assumption that $\log_2(x) \lt x$ implies $\lim_{x \to \aleph_0} \log_2 x \lt \lim_{x \to \aleph_0}x$, which is not true.