Let$A$ be an open, dense set in $mathbb R^n$. Prove that $A + A = mathbb R^n$

If $A$ is a dense open set, then $A-frac x2$ and $frac x2-A$ are dense open sets, so their intersection is a dense open set, and in particular it's nonempty. Choose a point $yin(A-frac x2)cap(frac x2-A)$; then $frac x2+yin A$ and $frac x2-yin A$, so $x=(frac x2+y)+(frac x2-y)in A+A$.

More generally, if $A$ is a nonempty open set in $mathbb R^n$ and $B$ is a dense subset of $mathbb R^n$, then $A+B=mathbb R^n$.

Proof. Consider any point $tinmathbb R^n$; we have to show that $tin A+B$.

Since the mapping $xmapsto t-x$ is a homeomorphism, $t-A$ is a nonempty open set. Since $B$ is dense, $Bcap(t-A)neemptyset$. Choose a point $bin Bcap(t-A)$. Then $bin B$, and $b=t-a$ for some $ain A$, so $t=a+bin A+B$.