Prove that if $\displaystyle S$ is a subset of the reals, dense around 0, then there may be no $\displaystyle s \in S$ such that $\displaystyle s/2 \in S$.

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- Jun 16th 2009, 07:24 PMBruno J.S dense in 0
Prove that if $\displaystyle S$ is a subset of the reals, dense around 0, then there may be no $\displaystyle s \in S$ such that $\displaystyle s/2 \in S$.

- Jun 17th 2009, 06:16 AMalunw
Let S be the set of numbers that can be expressed as a finite "decimal" fractions in some base with the last non-zero digit taking some fixed value. If you choose the base and the digit appropriately you can easily find sets with the required property. For example the set of finite base 10 decimals ending in 6 is dense round 0 since any real number near zero can be approximated as closely as you like. But if the decimal expansion of x ends in 6 then that of x/2 must end with either 3 or 8 so is not in the set. Another example is base 3 "decimals" ending with 2, in which case x/2 either ends in 1 or is non terminating.

- Jun 19th 2009, 11:17 AMBruno J.
Very good.

Here is my solution : it is easy to see that rationals of the form $\displaystyle \frac{a^2}{b^2}$ are dense in $\displaystyle \mathbb{R}$. But clearly $\displaystyle \frac{a^2}{2b^2}$ is not of such form, QED. - Jun 19th 2009, 02:09 PMalunwAnother answer
Yes, that is a nice solution because you can generalise it more easily to sets not containing x/n. But it took me a minute or two to convince myself that your set is dense, and you need to have $\displaystyle \pm a^2/b^2$ if you want to be dense in the whole of $\displaystyle R$. I suppose you just consider a sequence of rational approximations to the square root of the number you want to approximate with a number from your set. But then you could also do the same trick using numbers of the form $\displaystyle \pm a^n/b^n$ for any n, and it is quite surprising that they are dense. Another answer I thought of is the set of rationals of the form p/q with p and q both odd.

- Jun 19th 2009, 03:02 PMputnam120
- Jun 19th 2009, 04:16 PMalunw
How about?

Prove that there exist dense subsets $\displaystyle S$ of $\displaystyle \mathbb{R}$ such that $\displaystyle \forall x\in S \ \frac x 2\in\mathbb{R}-S$

As you can see we have already found plenty of constructions for such sets. But so far they are all countable. I wonder if it would be possible to find an uncountable set with this property. - Jun 20th 2009, 12:45 AMBruno J.
- Jun 20th 2009, 10:40 AMBruno J.
Ah! here is one. Let $\displaystyle \{U_n\}$ be an hamel basis. Consider the set $\displaystyle S$ of real numbers of the form

$\displaystyle r = \sum_{j=1}^na_jU_{s_j}$

where $\displaystyle s_n$ is an increasing sequence of integers, and $\displaystyle a_1$ is of the form $\displaystyle x^2$ for some $\displaystyle x \in \mathbb{Q}$. Clearly the set of these numbers is dense, and $\displaystyle r/2 \notin S\: \: \forall r \in S$.

Sort of awkward as you can't really picture it, but that does it.