Cl(U ^ A) = cl( U ^ cl(A) )

[Unassuming]

Homework Statement

Prove that cl(U^A) = cl(U^cl(A)).

Homework Equations

Theorems...

1.) cl(U^A) $$\subseteq$$ cl(U)^cl(A)
2.) cl(A)=cl(cl(A))

The Attempt at a Solution

My proof so far,

Let x be in cl(U^A). From #1, then x is in cl(U)^cl(A). From #2, x is in cl(U)^cl(cl(A)).


I am stuck there. I feel like I am one hair away from finishing but cannot.

EDIT: I realized I need to use neighborhoods for this proof. I am currently working on it and I think I might have something.

 

[mighty2000]

Will update soon.

Proof:

Assume x is in cl(U^A). This means that every neighborhood of x intersects with U^A.

Let N be a neighborhood of x. Then N intersects with both U and A. Since N is arbitrary, this implies that every neighborhood of x intersects with both U and A.

Now, consider cl(U)^cl(A). By definition, cl(U) is the intersection of all closed sets containing U, and cl(A) is the intersection of all closed sets containing A. This means that any neighborhood of x must intersect with both cl(U) and cl(A).

Since x is in both cl(U) and cl(A), this means that every neighborhood of x intersects with both cl(U) and cl(A). Therefore, x is in cl(U)^cl(A).

Now, let x be in cl(U)^cl(A). This means that every neighborhood of x intersects with both cl(U) and cl(A).

Let N be a neighborhood of x. Then N intersects with both U and A. Since N is arbitrary, this implies that every neighborhood of x intersects with U^A.

Therefore, x is in cl(U^A).

Since x is in both cl(U^A) and cl(U)^cl(A), this means that cl(U^A)=cl(U)^cl(A).