Question

1. Prove that for any set S S R, S is closed if and only if Se is open. Notice the book has a proof of this, but it uses a different notation for set complements and a different definition of neighborhood. You may consult it, but you must write your proof using the definition for interior point I presented in class (also in the notes on blackboard). If you copy the proof from the book you will not receive credit. Here are some hints: You must write two subproofs “ifs is closed, then Sc is open', and “IfSc is open, then s is closed." Recall that a set being closed requires that all of its accumulation points are in the set. To prove this you need to pick an arbitrary accumulation point of your set. Recall that a set being open requires that all elements of the set are interior points. To prove this you need to pick an arbitrary element of the set. a. b. c.

Answer 1

Statement for any set s subset equal R s is closed if and only if sc is open proof Let S is closed we have to show that s^l is open Let x belongs to s^c rightarrow not belongs to S since x is not an accumulation point of s, there is a neibhour should n or x such that y N intersection S = phi rightarrow N sub set S^&C That is there exist a neighbhour N of X which completely contained in s^c rightarrow x is an interior point od s^c since x is arbitrary aall elements of s^c are interior points rightarrow s^c is open we have to show that s is closed let x belongs to s and N be any neighbour hood of x rightarrow x belongs to x bnot belongs to s^c since x is not an interior point of sc there is no neighbourhood x contains elements of s