3
$\begingroup$

Let $A\subset \mathbb R$. Is it true that $$ \dim(A+A)\le 2\dim A $$

for some dimensions – say, lower box for the LHS and upper box for the RHS.

$\endgroup$
1

1 Answer 1

5
$\begingroup$

$A+A$ is a Lipschitz image of the set $A\times A\subset \mathbb{R}^2$ under the map $(x,y)\to x+y$. If $A$ is covered by $N$ balls of radius $\varepsilon$, then $A\times A$ is covered by $N^2$ balls of radius, say, $\sqrt{2}\varepsilon$, thus the box dimension (lower or upper) of $A\times A$ does not exceed twice that of $A$, and the Lipschitz map does not increase the dimension.

$\endgroup$
2
  • $\begingroup$ This seems to contradict the answer to the linked question. $\endgroup$ Sep 24, 2022 at 22:02
  • 2
    $\begingroup$ @ChristianRemling why? It is about other dimension $\endgroup$ Sep 24, 2022 at 22:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.