This is false.

No, consider the constant function.2.If A is open, is F(A) open?

I believe this is true but I am not sure.

Yes, this is true. This is a well-known result.3.If A is compact, is F(A) compact?

True.

False again.4.If A is bounded, is F(A) bounded?

False. Let and . Since is bounded, but is unbounded.

With pre-images this is true.5.If A is closed, isclosed?

I believe this is true. Since . So, if A is closed, then is open. Iffis continuous, then the preimage of every closed set in R is closed in A. By definition.

Again true.6.If A is open, isopen?

True. Iffis continuous, then the preimage of every open set in R is open in A by definition.

This is false since the next one is false.7.If A is compact, iscompact?

False, let A = [0, 1] which is compact because it is closed and bounded. Then let f(x) = 1/2, the the image which is not compact.

Consider defined by . Then is bounded but its preimage is unbounded.8.If A is bounded, isbounded?

I do not know. I believe this is false, but I can not think of a counterexample.