Why is 1 divided by aleph null undefined? [duplicate]

Question

So recently I have been thinking about infinity, and one of the things that I thought of was if you were able to get a defined value for the reciprocal of a transfinite (cardinal) number. So, I plugged $\frac{1}{א_0}$ into WolframAlpha and it said the following: img1 Why is this the case? Shouldn't this be similar to this case? $$\lim_{x\to\infty} \frac{1}{x} =0$$ Aren't $\infty$ and $א_0$ equal to the same value in this context? What am I missing here?

Answer 1

For one thing, in $\lim\limits_{x\to\infty}f(x)=L$ the symbol $\infty$ does not represent a number of any kind, cardinal or real, and the expression says something very different from $\lim\limits_{x\to c}f(x)=L$. The latter is a statement about two numbers, $L,c$, and a function $f$, that says values of $f$ get really close to $L$ when its inputs are really close to $c$. The former is a statement only about $L$ and $f$, that says the bigger the $x$, the closer $f(x)$ is to $L$. Note that it is not a statement about values of $x$ getting closer to some value: "bigger" does not mean "closer to $\infty$" (since "closer" only has a defined meaning between real numbers in this context).

There is, of course, such a thing as $\aleph_0$, but it is emphatically not a real number; it belongs to an entirely different class of numbers, with an entirely different ordering relation, while convergence to a limit is defined only in terms of real numbers and their relationships. So real division by $\aleph_0$ makes no sense.