NO- by definition a set is not a proper subset of itself . ( It is a subset, but not a proper one. )
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Let A be the set {1,2,3,4} B is {1,2} and B is a proper subset of A C is {1} and C is also a proper subset of A. B and C are proper subsets of the set A because they are strictly contained in A. necessarily excludes at least one member of A. The set A is NOT a proper subset of itself.
First of all, the null set( denoted by is a subset of every set. But it being a proper set or improper set is debatable. Many mathematicians regard it as an improper set, and rightly have as when we say a set is a subset of another, the super set always contains at least one element. For eg,. Let A be the set, in roster form we take it as: A = {ϕ}, we clearly see n(A)=1 then P(A) = {ϕ,{ϕ}} We observe that at least a set must have 1 element for it to have a proper set, but if we take A = ϕ ( i.e. n(A)=0), then clearly ϕ and A itself are improper sets of A and. Hence the minimum amount of proper sets a set has is nil and improper is 2. But I have seen a few high school text books who regard null set as a proper set, which is totally false, arguable by mathematicians, clearly signifying the lethargy of authors of the book failing to update their error driven books. I assure you, that null set is an improper set of every set.
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If you have a set of 6 elements, you can make a total of 26 different subsets - including the empty set and the set itself.
no. A subset would have to allow for values in its parent which are not in its self.