Where $\frac{1}{\infty}$ and $\frac{\infty}{\infty}$ are both undefined terms that generally lead to nonsense, I'm wondering if we can assert that...:

$$\frac{1+1+1+\cdots}{1+1+1+\cdots} = 1$$

...or even

$$\frac n {1+1+1+\cdots} = 0$$

...for any natural number $n$. In terms of infinite probabilities, it might make sense to recognize something like this because then we can differentiate:

$$\frac{1+1+1+\cdots}{2+2+2+\cdots} = \frac 1 2\left(\frac{1+1+1+\cdots}{1+1+1+\cdots}\right) = \frac 1 2.$$

I ask because I independently developed the following proof for selecting a natural number using a method where all natural numbers have an equal, however undefined, probability of being selected.


It is helpful to consider a simple observation pertaining to finite sets in order to get started. Let $A = \{1, 2\}$ and $B = \{3, 4, 5, 6\}$. Let $f(x) = 1$ if $x$ is odd and $f(x) = 2$ if $x$ is even. Then, function $f$ is a surjection from $B$ onto $A$ that is “uniform” in the sense that selecting an element $x \in B$ uniformly at random will result in the selection of $f(x) \in A$ uniformly at random as well. Note that in order to do this, we have effectively partitioned $B$ into subsets the same size as $A$ so that we could biject those subsets with $A$. Likewise, this work shows how we can partition $[ \,0, 1) \,$ into countable sets that are then bijected with the natural numbers. Consideration is then given to whether selecting an element of $[ \,0, 1) \,$ uniformly at random allows for the selection of a natural number uniformly at random as well.


Let $V^{( \,0.5, 1) \,}$ be a set containing one and only one element from each Vitali equivalence class on the interval $( \, 0.5, 1 ) \,$ (Vitali equivalence classes are equivalence classes of the real numbers that partition $\mathbb{R}$ under the relation $x \equiv y \iff ( \, \exists q \in \mathbb{Q} ) \, ( \, x - y = q ) \,$). The axiom of choice allows for such a selection.

For any real number $r$, let $d(r)$ equal the one and only one element $v \in V^{( \,0.5, 1) \,}$ such that $r - v \in \mathbb{Q}$.

Let $h : \mathbb{N} \longmapsto \mathbb{Q} [ \, 0, 1 ) \,$ (here $\longmapsto$ denotes a bijection).

Let $x$ be an element of $[ \,0, 1) \,$ selected uniformly at random.

Let $k(x) = \begin{cases} h^{-1}(x - d(x) + 0.5) && x \geq d(x) - 0.5 \\ h^{-1}((x + 1) - d(x) + 0.5) && x < d(x) - 0.5 \end{cases}$.

For each natural number $1, 2, 3, …$, let $V^1, V^2, V^3, \ldots$, respectively, be the sets such that $( \, \forall x \in V^{n} ) \, ( \,k(x) = n ) \,$. We then have $V^{( \,0.5, 1) \,} = V^{h^{-1}(0.5)}$, for example. Each $V^{n}$ will be a Vitali set by definition with the collection $\{ V^{n} : n \in \mathbb{N} \}$ forming a partition of $[ \,0, 1) \,$.

Comments on Uniformity:

A uniform distribution is a concept of translation invariance. For example, if $S$ is a measurable set, we may want the probability of $S$ to be the same as the probability of $\{y : y = z + n, z \in S \}$ for each natural number $n$. In the case of function $k$ over the domain $[ \,0, 1) \,$, however, we end up mapping each element of each non-measurable Vitali set $V^n$ to a distinct natural number $n$. Where $a, b \in \mathbb{N}$, it is easy to see that the probability of $x$ falling within $V^a$ is equal to the probability of $x$ falling in $V^b$, but we cannot rely on a Lebesgue measure as a means of establishing probability or creating any sort of cumulative distribution function on $\mathbb{N}$. The probability of selecting any given natural remains undefined.

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    $\begingroup$ It is exaggerated to say that $\infty/\infty$ generally leads to nonsense. If $f(x)\to\infty$ and $g(x)\to\infty$ as $x\to a,$ then $f(x)/g(x) \to \text{what?}$ Here's an example: $$\lim_{n\to\infty} \frac{3n^2+5n+14}{n^2+1} = 3. $$ The limit can be any real number or $+\infty$ or $-\infty$ depending on which functions $f$ and $g$ are. Things like this arise when an integral is defined as a Riemann sum and the lengths of subintervals are $(b-a)/n.$ Then $n\to\infty$ and $(b-a)/n$ is multiplied by something that approaches $\infty$, and the limit is a finite number, the value of the integral. $\endgroup$ – Michael Hardy Apr 20 '17 at 1:37
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    $\begingroup$ Until you tell us what you mean about what you are writing, the answer could be "yes," "no," or "maybe." Basically, what you've written is just a bunch of symbols that don't mean anything, without you defining or referring to some existing definition. We usually define things so that they are useful, but we can define things in pretty arbitrary ways. But here, there is no definition, and hence no meaning can be inferred. $\endgroup$ – Thomas Andrews Apr 20 '17 at 1:50
  • $\begingroup$ The context here is of probabilities involving infinite sets. The above shows perhaps why the Lebesgue measure of a Vitali set is undefined (if each had a positive real measure, then the sum of the probabilities assigned to each natural above would be infinite, or if they had a 0 measure, then the sum would be 0). We are guaranteed a natural using the above process and no natural is favored over any other in terms of its probability. The cumulative distribution function is undefined through any given $n$, except "perhaps" over $\mathbb{N}$ itself we can assert it is equal to 1. I'm fishing $\endgroup$ – AplanisTophet Apr 20 '17 at 2:57

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