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Consider R² with the usual topology and A,B subset or equal R² then prove intersection of int(A intersection B)=int(A) intersection int(B)?
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Consider R² with the usual topology and A,B subset or equal R² then pr...
Proof:

1. Inclusion of Intesections:
- Let x be an element in int(A ∩ B), then there exists an open set U containing x such that U is a subset of A ∩ B.
- This implies that U is a subset of both A and B, hence x is in int(A) and int(B).

2. Inclusion of Intersection of Ints:
- Now, let x be an element in int(A) ∩ int(B), then x is in both int(A) and int(B).
- This means there exist open sets U and V containing x such that U is a subset of A and V is a subset of B.
- Since U ∩ V is an open set containing x and U ∩ V is a subset of A ∩ B, x is in int(A ∩ B).

Conclusion:
- From the above arguments, we have shown that int(A ∩ B) is a subset of int(A) ∩ int(B) and int(A) ∩ int(B) is a subset of int(A ∩ B).
- Therefore, we can conclude that int(A ∩ B) = int(A) ∩ int(B).
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Consider R² with the usual topology and A,B subset or equal R² then prove intersection of int(A intersection B)=int(A) intersection int(B)?
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