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?
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, is closed?
I believe this is true. Since . So, if A is closed, then is open. If f is continuous, then the preimage of every closed set in R is closed in A. By definition.
Again true.6. If A is open, is open?
True. If f is 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, is compact?
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, is bounded?
I do not know. I believe this is false, but I can not think of a counterexample.