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This is one of my weird "isolated problems" that I just think up by myself.

Question $1:\ $ Is it true that if $A$ and $B$ are nowhere dense subsets of $[0,1]$ then the Minkowski sum $A+B = \{a+b:\ a\in A, b\in B\}$ is nowhere dense in $[0,2]$?

Question $2:\ $ Is it true that if $A$ and $B$ are nowhere dense subsets of $[0,1]$ then the "multiplication product" $AB = \{ab:\ a\in A, b\in B\}$ is nowhere dense in $[0,1]$?

For the first question to have negative result, we need to show that, if $\ I\ $ is an interval inside $\ [0,2],\ $ then $\ \exists x\in I\ $ such that $\ x\ $ is not a limit point of $A+B.$ But I don't know how to go about doing this.

Alternatively, maybe the first question has positive result, which might arise if our sets $A$ and $B$ were Cantor sets, although my intuition on Cantor sets is limited. In particular, I find it hard to picture the properties of $A+B$ if $A$ and $B$ are Cantor sets.

kabenyuk
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Adam Rubinson
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    If $C\subseteq [0,1]$ is the ternary Cantor set then $C+C=[0,2]$ – Alessandro Codenotti Sep 11 '21 at 13:48
  • @AlessandroCodenotti how do you prove that? – Adam Rubinson Sep 11 '21 at 14:07
  • The second question is not clearly worded. Maybe we are talking about such a set $ AB={ab\mid a\in A,b\in B} $? – kabenyuk Sep 11 '21 at 14:28
  • @kabenyuk yes. Question amended. – Adam Rubinson Sep 11 '21 at 14:40
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    @AdamRubinson the easiest way is probably to think about $C$ as the set of reals in $[0,1]$ whose ternary expansion contains no $2$, then given any number $r\in[0,2]$ you can explicitely write $r$ as the sum of two elements of $C$ by writing $r$ in base $3$ and "splitting every $2$ in a $1+1$" – Alessandro Codenotti Sep 11 '21 at 15:29
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    @AlessandroCodenotti Isn't $C$ the the set of reals in $[0,1]$ whose ternary expansion contains no $1$ (rather than "no $2$" as you said)? It's the middle interval you take out every time in the construction process – Ewan Delanoy Sep 11 '21 at 15:34
  • @AlessandroCodenotti My criticism doesn't alter the fact that your argument works fine with the set $C'$ of reals in $[0,1]$ whose ternary expansion contains no $2$. It's just that this $C'$ is not the usual ternary Cantor set, it's a variant of it. – Ewan Delanoy Sep 11 '21 at 15:50
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    Woops I wasn't thinking. You can still do an explicit argument based on ternary expansions, takes a little more work. Anyway someone already wrote the details here https://math.stackexchange.com/questions/309080/cantor-set-cantor-set-0-2 – Alessandro Codenotti Sep 11 '21 at 16:08
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    To solve the second question, you can use the equality $exp(C)\cdot exp(C)=exp([0,2])$ – kabenyuk Sep 11 '21 at 16:09

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