# Proving $\{P(a)\}$ (where $a$ is transcendental) is dense

So, I have read this problem, and it's bugged me since:

Let $a \in (1,2)$ be a transcendental number

1) Let $Y = \{ P(a) : P \in \mathbb{Z}[X] \}$, show that Y is dense in $\mathbb{R}$.

2) Let $X = \{ P(a) : P \in \mathbb{Z}[X], P \text{ has coefficients in } \{0,1,-1\}\}$, show that X is dense in $\mathbb{R}$.

I have been able to prove 1), by showing that $Y$ is a non-cyclic subgroup of $\mathbb{R}$ and therefore is dense (a well-known result which I can prove too).

2) had a hint attached:

Hint: Start by proving that $0$ is in the closure of $X \backslash \{0\}$

I was able to prove this too:

• I took $Z = \{ \sum_{k=0}^{n-1} b_k a^k / \forall k, b_k \in \{0, 1\} \}$.
• We have $\forall z \in Z, 0 \leq z \leq \frac{a^n - 1}{a-1} < \frac{a^n}{a-1}$.
• $Z$ has $2^n$ elements, so if we cut $[ 0, \frac{a^n}{a-1} )$ in the $2^n - 1$ intervals $I_k = [ \frac{(a/2)^n}{a-1} \times k, \frac{(a/2)^n}{a-1} \times (k+1) )$, then there is a $k$ such that $I_k \cap Z = \{z_1, z_2\}$ with $z_1 \neq z_2$.
• Therefore, $| z_1 - z_2 | \in X \cap [0, \frac{(a/2)^n}{a-1})$. Since $0 < a < 2$, $(\frac{a}{2})^n \rightarrow 0$ and we have that X is dense around 0.

But now I'm stuck. I've thought of trying to approximate any real by an element of the form $a^n \varepsilon$ with $\varepsilon \in X$ close to 0, but it doesn't work. Do you have any hints?

Edit: It's not homework, it's in preparation of an upcoming oral exam.

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$1)$ also follows from the fact that for $\alpha$ irrational the set $\{ m \alpha+n | m,n \in {\mathbb Z}$ is dense in ${\mathbb R}$. Thus you only need linear polynomials. – N. S. May 28 '11 at 17:25

## 2 Answers

For every $n$ you can find an infinite sequence $x_1,x_2,\dots\in X$ such that $2^{-(n+1)}<x_i<2^{-n}$ (for every $i$) and such that, for every $k$, $\pm (x_1+ x_2+\dots+ x_k)$ is still in $X$. The sequence can be found inductively: if you have $x_1,\dots,x_k$, choose a $0<P(a)\in X$ small enough and set $x_{k+1}=a^pP(a)$, with $p$ such that $2^{-(n+1)}<x_{k+1}<2^{-n}$. You need to choose $P(a)$ so small that no $a^q$, $q\geq p$, is present in any of $x_i$, $i\leq k$ (every $x_i$ is a polynomial in $a$). Such a $P(a)\in X$ exists as $0$ is in the closure of $X\backslash\{0\}$.

As a consequence, you can approximate any real number by an element of $X$ with error $<2^{-n}$ (using a suitable $\pm (x_1+ x_2+\dots+ x_k)\in X$). Hence $X$ is dense.

(there should be a simpler proof)

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The answer is wrong. I'll try and find a good solution. :)

For the second part, it is easy to see that $x\neq y \in X\setminus\{0\} \Rightarrow \frac{x+y}{2},\frac{x-y}{2} \in X$. Now, all you need is to prove that there are arbitrarily big and small elements in $X$, and you can do this by picking the polynomials $P=X^n,\ P=-X^n,\ n \geq 1$.

The idea is that if you have a set $A$ which contains two elements $a<b$ and you have the property that $x,y \in A \Rightarrow \frac{x+y}{2} \in A$, then you can construct a dense subset of the interval $(a,b)$ which is contained in $A$ by picking at each step the midpoint of the interval formed by two known elements of $(a,b)$(which we know it is in $A$).

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@zulon: See the modifications made to the solution. Now it should work. – Beni Bogosel May 28 '11 at 17:46
It still does not work. Take $X$ and $X^2$, the mean of these two is $\frac{X+X^2}{2}$ which hasn't coefficients in $\{0,1,-1\}$. – Najib Idrissi May 28 '11 at 17:58