Recent content by Shaggydog4242

I think I got it right after I posted that. C is closed in A implies that A-C is open in X. Then by subspace topology A-C=A\bigcapK compliment for some closed K in X. A\bigcapK compliment implies C = A - (A-C) = A-K compliment = A\bigcapK! unless I messed up somewhere...

Thanks again for the help! Yeah, i meant x \inX. I know that if C is open in A then automatically C = A\bigcapK for some closed set K of X but I don't see how I can go from assuming A-C is open in A to C being open in A.

Homework Statement 1. Let A= {a,b,c}. Calculate the subspace topology on A induced by the topology T= { empty set, X,{a},{c,d},{b,c,e},{a,c,d},{a,b,c,e},{b,c,d,e},{c}, {a,c}} on X={a,b,c,d,e}.Homework Equations Given a topological space (X, T) and a subset S of X...

Thanks for replying! In this case C would be in A. I'm not sure why I would be allowed to set K=C if K can be any closed subset in X while C has to be closed in A.

Homework Statement Hi, This is my first post. I had a question regarding open/closed sets and subspace topology. Let A be a subset of a topological space X and give A the subspace topology. Prove that if a set C is closed then C= A intersect K for some closed subset K of X. Homework...