Prove that if is a subset of the reals, dense around 0, then there may be no such that .
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.
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 if you want to be dense in the whole of . 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 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.
Ah! here is one. Let be an hamel basis. Consider the set of real numbers of the form
where is an increasing sequence of integers, and is of the form for some . Clearly the set of these numbers is dense, and .
Sort of awkward as you can't really picture it, but that does it.