**PROPOSITION 1: ** **For a real number x there exists a sequence $ x_1, x_2, x_3,…$ of integers such that**

$ \hspace{4cm} x=x_1 +\frac{x_2}{2!}+\frac{x_3}{3!} + \cdots + \frac{x_n}{n!} + \cdots, \hspace{2cm} (*) $

**where $x_1$ can be any integer, but for $ n \geq 2$, $x_n \in \{ 0,1,…,n-1 \}.$ Furthermore, if we require that the partial sums be strictly smaller than x, then such a representation is unique.**

**Remark:** One cannot help recalling decimal or binary expansion of numbers. Notice that $\frac{n}{n!}=\frac{1}{(n-1)!}$ (drops back to previous digit), so the bound on $x_n$ is logical.

**Proof:**

Choose the biggest integer $x_1$ *strictly* smaller than $x$. If $x_1+\frac{1}{2!}$ is strictly less than $x$ then choose $x_2=1$, otherwise, choose $x_2=0$. Assume we have picked $x_1, x_2, … , x_n$, then we’ll choose $x_{n+1}$ to be the largest of $\{ 0, 1, 2, … , n \}$ so that

$ \hspace{6cm} x_1+\frac{x_2}{2!}+ … +\frac {x_n}{n!} + \frac{x_{n+1}}{(n+1)!} < x .$

We’ll prove that this inductive choice of $ \{ x_n \}_{n=1}^\infty $ satisfies the expansion $(*)$.

**Claim**: For every ,

$ \hspace{6cm} 0 < x- (x_1+\frac{x_2}{2!}+ \cdots +\frac{x_n}{n!}) \leq \frac{1}{n!} \hspace{2cm} (ES) $

Proof of the claim: If $x_n \neq n-1$, this is an immediate consequence of the choice of $x_n$: by optimality of $x_n$ we have

$ \hspace{5.5cm} x_1+\frac{x_2}{2!}+ \cdots +\frac {x_n}{n!} < x \leq x_1+\frac{x_2}{2!}+ \cdots +\frac {x_n + 1}{n!} $

Subtracting $ x_1+\frac{x_2}{2!}+ \cdots +\frac {x_n}{n!} $ from each term yields (ES).

If $x_n=n-1$, the maximum possible, then we have the identity

$ \hspace{1cm} x_1+ \cdots +\frac {x_n}{n!} + \frac{1}{n!} = x_1+\cdots + \frac{x_{n-1}}{(n-1)!}+\frac {n-1}{n!} + \frac{1}{n!} = x_1+\cdots +\frac {x_{n-1}}{(n-1)!} + \frac{1}{(n-1)!} $

To obtain (ES) it suffices to show that this quantity is bigger than or equal to x. Comparing the first and last expressions, we see that we have reduced case n to (n-1). Thus, either we work backwards to reach n=1, considering cases for $x_{n-1}$ for this current step and so on, or we switch to an inductive proof, to attain (ES). Case n=1 is obviously true.

The uniqueness part’s proof: Assume that some has two different representations:

$\hspace{3cm} x_1 +\frac{x_2}{2!}+\frac{x_3}{3!} + \cdots + \frac{x_n}{n!} + \cdots = y_1 +\frac{y_2}{2!}+\frac{y_3}{3!} + \cdots + \frac{y_n}{n!} + \cdots $

We’ll prove that one of them is a finite sum. Assume that $k$ is the first index where $x_k \neq y_k$, and, without loss of generality, that $x_k > y_k \ $ . Then,

$ \hspace{6cm} \frac{x_k – y_k}{k!} = \displaystyle\sum _{n=k+1}^\infty \frac{y_n – x_n}{n!} \hspace{2cm} (EQ)$

Notice that while $\frac{x_k – y_k}{k!} \geq \frac{1}{k!}$,

$\hspace{2cm} \left|\displaystyle\sum _{k+1}^\infty \frac{y_n – x_n}{n!}\right| \leq\displaystyle\sum _{n=k+1}^\infty \frac{|y_n – x_n|}{n!} \leq\displaystyle\sum_{n=k+1}^\infty \frac{n-1}{n!} =\displaystyle\sum_{n=k+1}^\infty \left(\frac{1}{(n-1)!}-\frac{1}{n!}\right) = \frac{1}{k!} .$

Thus, the only way for (EQ) to hold is to have $x_k=y_k+1$ and for all $n>k$, $x_n= 0$ and $y_n=n-1$ ; an analogous situation to $1.73=1.7299999…$ in decimal base.

Note that out of the two representations only one is strictly increasing to its limit, proving the uniqueness claim.

**The Backstory: **My independent discovery of this expansion was triggered by my search for a compact set in $R$ with no isolated points (every point’s every neighborhood contains other points of the set as well) and *no* rationals. This was a question my favorite analysis professor, Dr. Rezaee, had asked me to think about.

For months my approach had been to start with $[0,1]$ and then try to remove successively more and more subsets. But I had failed to land on the right set. Then there was this morning that I was sitting in a different class when suddenly I recalled an exercise from Tom M. Apostol’s *Mathematical Analysis* book:

**Exercise: **The number $x=x_1 +\frac{x_2}{2!}+\frac{x_3}{3!} + \cdots + \frac{x_n}{n!} + \cdots $ (as in proposition 1) is rational if and only if there exists an $N \in \mathbb{N} $ such that

$\hspace{6cm} n>N \implies x_n=n-1.$

“I know so many irrational numbers!” I said to myself and there I was with a set.

Solution to the exercise: Suppose the condition holds. Then, as shown by uniqueness part’s proof above, the sum is equal to a finite sum, each of whose terms are rational.

For the other direction, suppose $x$ is rational then, for relatively prime $p \in \mathbb{Z},q\in \mathbb{N}$ we have

$\hspace{6cm} x=\frac{p}{q}=\displaystyle\sum_{i=1}^{\infty} \frac{x_i}{i!}$

Multiplying the sides by $q!$ yields

$ \hspace{6cm} q!\displaystyle\sum_{q+1}^{\infty} \frac{x_i}{i!} = p(q-1)!-q!\displaystyle\sum_{i=1}^{q} \frac{x_i}{i!}. $

The right hand side is an integer. If for even only one index $ i>q$ the equality $ x_i = i-1$ failed to hold, then we would have

$ \hspace{6cm} 0 < q!\displaystyle\sum_{q+1}^{\infty} \frac{x_i}{i!} <q!\displaystyle\sum_{q+1}^{\infty} \frac{i-1}{i!} =1 $

Which contradicts it being an integer. (The strict positivity is due to strictly increasing assumption on the series.)

**A Perfect Set**

**Proposition: 2 The set**

$\hspace{6cm} S=\left\{\left.\displaystyle\sum_{i =4}^\infty \ \frac{x_i}{i!} \right| \ x_i \in {1,3}\right\}$

**is a “perfect set” (closed, and each point is a limit point) without rationals.**

Proof: The idea is hidden in the arguments we have already made. The point is that when we change one digit by 1, the tail has to go full speed to catch up. Since here we have restricted to choices 1 and 3, when we change a digit then the new number is by a significant distance away from any member of $S$.

Take $y \in R \backslash S$. Let’s restrict to $y$’s of the form

$\hspace{6cm} y=\displaystyle\sum_{i=4}^{\infty} \frac{y_i}{i!} .$

Other cases where $y$ has earlier digits are just as easy, but we want to avoid complications in notation! Since $y$ is not in $S$, there is an index $j$ such that $y_j \notin \{1,3\}.$ For any given $x \in S$, the representations of $x$ and $y$ will differ somewhere earlier than $j$, say at $k$’th component, $4 \leq k\leq j$. Therefore,

$\hspace{2cm} \left|x-y\right|=\left|\displaystyle\sum_{i=k}^{\infty} \frac{x_i-y_i}{i!}\right|\geq \frac{\left|x_k-y_k\right|}{k!}-\displaystyle\sum_{i=k+1}^{\infty} \frac{\left|x_i-y_i\right|}{i!} \geq \frac{1}{k!}-\left|\displaystyle\sum_{i=k+1}^{\infty} \frac{i-2}{i!}\right|. $

It follows that

$\hspace{6cm} |x-y| \geq \frac{1}{(k+1)!}\geq \frac{1}{(j+1)!}$

The index $j$ depends on $y$ only, thus we proved that within radius $1/(j+1)!$ of $y$ there are no points of $S$, or, equivalently, $S^c$ is open, $S$ is closed.

Now, pick any $x\in S$. Given $\epsilon >0$, in order to find another point in $S$, $\epsilon$-close to $x$, move far enough in the representation of $x$, and switch $x_N$ to $1$ if it is $3$, or to 3 if it is 1. Keep all other digits the same. The new number is again is $S$, closer than $\epsilon$ much to $x$ provided that $\frac{2}{N!}<\epsilon$. Thus, every point in $S$ is a limit point.

**Questions:**

Do you know of other perfect sets without rationals (in the usual topology of $\mathbb{R}$)?

What else can be done with the factorials representation of numbers?!

We have proved that $e$ is an irrational number! Do you see where?