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AnalysisHW1 125A

# AnalysisHW1 125A - MAT 125a HW1 Solutions 5(a(i First...

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MAT 125a, HW1 Solutions 5 (a) (i) First assume x B ∩ {∪ i I X i } . Since x B ∩ {∪ i I X i } we have that x B and j I such that x X j . Since x B and x X j we have x B X j ⊂ ∪ i I ( B X i ). So x B ∩ {∪ i I X i } implies x ∈ ∪ i I ( B X i ) hence B ∩ {∪ i I X i } ⊂ ∪ i I ( B X i ). (ii) Next assume x ∈ ∪ i I ( B X i ). Since x ∈ ∪ i I ( B X i ) j I such that x B X j . Since x B X j we have that x B and x X j hence x B and x ∈ ∪ i I X i which implies that x B ∩ {∪ i I X i } . Therefore i I ( B X i ) B ∩ {∪ i I X i } . Therefore by (i) and (ii) we have i I ( B X i ) = B ∩ {∪ i I X i } . (b) (i) First assume x ( i I X i ) { . Since x ( i I X i ) { we have that x S and x 6∈ ∪ i I X i which implies that x S and x 6∈ X i i I . Since x S and i I s 6∈ X i we have that x X { i i I and hence x ∈ ∩ i I X { i . Therefore x ( i I X i ) { ⊂ ∩ i I X { i .

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AnalysisHW1 125A - MAT 125a HW1 Solutions 5(a(i First...

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