Is the concept A + ∅ = ∅ correct? A is any set of non-negative integers. I
think that if it is so, it contradicts the condition on the cardinality of sum sets
that |A| + |B| − 1 ≤ |A + B| ≤ |A| |B|. Please give your expert opinions.
Dear Sudev,
Consider the following paper; pp.112-113.
J.E. Olson, On the Sum of Two Sets in a Group, Journal of Number Theory, 18, 110-120, 1984.
THEOREM 4 (Kemperman). Let C=A+B be the sum of two finite sets A and B in G. If some element c₀ = a₀ + b₀, appears exactly once in A+B then |C| ≥ |A|+|B| - 1.
In your above problem, the set B is empty set, the condition
c₀=a₀+b₀ can not holds, so we can not apply the inequality in this case.
A + B is the set of all sums a + b with a in A and b in B. If there is no such sum, as in your question, then A + B is empty. In particular A + empty = empty.
No it is not possible. Since |A|≤ |A + B|, that is, if A is a non-empty set with |A|=n>0, then n= |A|≤ |A + ∅|=|∅|=0, which is impossible.
Hi Sudev !! consider A U ∅ = ∅ implies n(A U ∅) = n(A).n(∅) = 0 = n(∅) = ∅ compare cardinality of sets and equalently assigning set theory from measurability of lebesge then it is true A + ∅ = ∅. No doubt it is right to take the sumset A + ∅ = ∅ if you think in this way .....Dr Nimmagadda Venkata Nagendram
This notation does not make much sense unless you define the necessary operations. However if you replace + by U {union }, it would be meaningful.
Regards ,
Kulthoum
I agree with Mihai and Cenap: Kemperman's theorem assumes non-empty sets. However, the equality A+\emptyset = \emptyset is meaningful and quite true (vacuously, as it were). (At first glance, I thought you were using the old notation for set union, so was a little taken aback.)
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