If we have N sets, { A 1 , … , A N }, and we form a set S by taking the sum of each element in the set with each element in the other sets, what can we say about the mode of S?Intuitively, I would like to think that we can simply take the sum of the modes, i.e: Mode ⁡ ( S ) = ∑ n = 1 N Mode ⁡ ( A n )However, this seems unlikely, especially as we would expect that Mode ⁡ ( A n ) could potentially be a set of values, rather than a single value.So I was wondering if we'd be able to relax this condition to state that Mode ⁡ ( S ) ⊆ ∑ n = 1 N Mode ⁡ ( A n ), where we define Mode ⁡ ( A ) + Mode ⁡ ( B ) as the set formed by taking the sum of each element in Mode ⁡ ( A ) with each element in Mode ⁡ ( B ), formally: Mode ⁡ ( S ) ⊆ ∑ n = 1 N Mode ⁡ ( A n ) = { ∑ n = 1 N x n : x i ∈ A i } This seems to be true, but I was wondering if we could say anything stronger?

Question

If we have   N sets,   {      A      1    ,  …  ,      A      N    }, and we form a set   S by taking the sum of each element in the set with each element in the other sets, what can we say about the mode of   S?Intuitively, I would like to think that we can simply take the sum of the modes, i.e:  Mode  ⁡  (  S  )  =      ∑          n      =      1          N    Mode  ⁡  (      A      n    )However, this seems unlikely, especially as we would expect that   Mode  ⁡  (      A      n    ) could potentially be a set of values, rather than a single value.So I was wondering if we&#039;d be able to relax this condition to state that   Mode  ⁡  (  S  )  ⊆      ∑          n      =      1          N    Mode  ⁡  (      A      n    ), where we define   Mode  ⁡  (  A  )  +  Mode  ⁡  (  B  ) as the set formed by taking the sum of each element in   Mode  ⁡  (  A  ) with each element in   Mode  ⁡  (  B  ), formally:  Mode  ⁡  (  S  )  ⊆      ∑          n      =      1          N    Mode  ⁡  (      A      n    )  =      {          ∑              n        =        1                N              x          n        :          x          i        ∈          A          i        }  This seems to be true, but I was wondering if we could say anything stronger?

Darrell Valencia · Accepted Answer

Consider the multisets   A  =  {    {  0  ,  2  ,  2  ,  3  }    } and   B  =  {    {  0  ,  2  ,  2  ,  3  ,  5  }    }, both with mode   2. Then  A  +  B  =  {    {  0  ,  2  ,  2  ,  2  ,  2  ,  3  ,  3  ,  4  ,  4  ,  4  ,  4  ,  5  ,  5  ,  5  ,  5  ,  5  ,  6  ,  7  ,  7  ,  8  }    }    ,whose mode is   5.

If we have N sets, <mo fence="false" stretchy="false">{ A 1 </msub>

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