3

I am having a really difficult time proving part two of this problem in Baby Rudin.

If$\,\,A\subset\mathbb{R^k}$ and $B\subset\mathbb{R^k}$, define $A+B$ to be the set of all sums $x+y$ with $x\in A$ and $y\in B$.

(b) Let $\alpha$ be an irrational real number. Let $C_1$ be the set of all integers, let $C_2$ be the set of all $n\alpha$ with $n\in C_1$. Show that $C_1$ and $C_2$ are closed subsets of $\mathbb{R^1}$ whose sum $C_1+C_2$ is not closed, by showing that $C_1+C_2$ is a countable dense subset.

I understand that both $C_1$ and $C_2$ are closed, and I understand why $C_1+C_2$ might be dense, but I am just having no luck figuring out how to tackle the problem.

Here are two possible ideas I think might work, but don't know how to make work:

  1. Prove that for every element $z\notin C_1+C_2$ and every $\epsilon$, there exist an element $c\in C_1+C_2$ such that $|z-c|<\epsilon$.

  2. Assuming $z\notin C_1+C_2$, prove that if $ d_z:\mathbb{R^1}\to \mathbb{R^1}$ where $d_z(x)=|z-x|,$ then $\inf (f(C_1+C_2))=0$

Any ideas as to what might be the best way to tackle this problem?

Community
  • 1
user160110
  • 2,700
  • 17
  • 27

1 Answers1

0

Hint regarding your first idea: you only need one such $z, c$ pair for each $\varepsilon$.

Henrik supports the community
  • 6,831