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.


1 Answer 1


$A+A$ 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.

  • $\begingroup$ 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

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