I would really appreciate it if someone can check the proof to a claim i made in a bigger proof..here is the whole thing and below it is the part i want you to check:

 

Let X be a space, Y subset of X, and Z subset of Y. Give Y the subspace topology the following hold:

(1) Suppose Y is closed in X. Then Z closed in Y implies Z closed in X.

(2)Suppose Y is open in X. Then Z open in Y implies Z open in X.

 

 

Let us start with (2). Denote by T the topology on X. Then the topology

on Y is {U \cap Y | U in T}, where \cap = intersection. Thus if

Z is open in Y then Z = U \cap Y for some U which is open

in X. But now U and Y are both open in X, so their intersection

is open in X. This establishes (2).

 

(1) is similar; just note that if Z is closed in Y then Y-Z is

open in Y; i.e. Y-Z = U \cap Y for some U open in X. that this implies Z = (X-U) \cap Y. Thus Z is closed in X.

 

Now the proof to the claim, the one I need you to check is:

Z = (X\U) /\ Y

I showed 2 inclusions:

let x be in Z, since Z is subset of Y then x is in Y, since Y is a closed set in X then Y is a subset of (X\U)..so Z is a subset of (X\U) /\ Y

Now let x be in (X\U) /\ Y, then x is in X and x is not in U and x is in Y, then we ca say that x is in Y\ U and since Z is a closed subset in Y, then x is in Z too. So (X\U) /\ Y is a subset of Z, so one can conclude that Z = (X\U) /\ Y. Please check my proof and correct me if i am wrong. Thanks

by the way..for the intersection sign i used /\ and some other times \cap. Also for Y minus Z i used Y-Z or Y\Z.