Since x ∈ U(j∈N)∩(n∈N)Aj,n, there exists j0∈N such that x ∈ ∩(n∈N)Aj0,n and so x ∈ Aj0,n for all n∈N. Therefore, x ∈ U(j∈N)Aj,n for all n∈N and so x ∈ ∩(n∈N)U(j∈N)Aj,n.
We conclude that U(j∈N)∩(n∈N)Aj,n ⊆ ∩(n∈N)U(j∈N)Aj,n.
To better understand why this makes sense, consider a number of students taking the same multiple-choice test, where n is the question number on the test and j is the student number. Think of Aj,n as the set of all possible test results such that the jth student gets the nth question right. Then essentially we just proved the fairly obvious statement that if at least one student gets a perfect paper, then each test question is answered correctly by at least one student. (Of course, the converse is false. Each question could be answered correctly by at least one student without anyone getting a perfect paper.)
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Suppose that x ∈ U(j∈N)∩(n∈N)Aj,n.
We need to prove that x ∈ ∩(n∈N)U(j∈N)Aj,n.
Since x ∈ U(j∈N)∩(n∈N)Aj,n, there exists j0∈N such that x ∈ ∩(n∈N)Aj0,n and so x ∈ Aj0,n for all n∈N. Therefore, x ∈ U(j∈N)Aj,n for all n∈N and so x ∈ ∩(n∈N)U(j∈N)Aj,n.
We conclude that U(j∈N)∩(n∈N)Aj,n ⊆ ∩(n∈N)U(j∈N)Aj,n.
To better understand why this makes sense, consider a number of students taking the same multiple-choice test, where n is the question number on the test and j is the student number. Think of Aj,n as the set of all possible test results such that the jth student gets the nth question right. Then essentially we just proved the fairly obvious statement that if at least one student gets a perfect paper, then each test question is answered correctly by at least one student. (Of course, the converse is false. Each question could be answered correctly by at least one student without anyone getting a perfect paper.)
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