Nature of infinity

Question

If there are an infinite number of whole numbers, and an infinite number of decimals in between any two whole numbers, and an infinite number of decimals in between any two decimals, does that mean that there are infinite infinities? And an infinite number of those infinities? And an infinite number of those infinities? And…(infinitely times. And that infinitely times. And that infinitely times. And…) …

Answer 1

(Making my comment into an answer.) There are "infinite infinities", but not for the reason you give.

Discussing "infinities" requires certain definitions, but it really boils down to: two sets $X, Y$ have the same cardinality (i.e. correspond to the same "infinity") if there exists a bijection between them. In this case, we write $|X|=|Y|$.

If $X=[a, b]$ is the set of numbers between $a$ and $b$, and similarly $Y=[c, d]$ is the set of numbers between $c$ and $d$, then the map $f:X\to Y$ defined by $f(x) = (x-a)\frac{d-c}{b-a}+c$ is a bijection. Hence, all the sets you defined have the same cardinality.

These sets have infact the same cardinality as the real numbers. To see this, note that they all have cardinality equal to that of $[-\pi/2, \pi/2]$. Then prove that the closed interval $[-\pi/2, \pi/2]$ has the same cardinality as the open interval $(-\pi/2, \pi/2)$, which is tricky. Then the map $g:(-\pi/2, \pi/2)\to\mathbb{R}$ defined by $g(x) = \tan^{-1}(x)$ is a bijection. Combining these three bijections $$[a, b]\to [-\pi/2, \pi/2]\to (-\pi/2, \pi/2)\to\mathbb{R}$$ gives a bijection $[a, b]\to\mathbb{R}$, and the result follows.

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There are "infinite infinities", and the standard proof of this uses powersets. In particular, the powerset $\mathcal{P}(X)$ of $X$ has strictly larger cardinality than $X$, i.e. there exists an injection but no bijection $X\to\mathcal{P}(X)$. Taking powersets of powersets, etc., gives the result.