I'm having a bit of difficulty seeing whether a function or its inverse preserves whether a set's image/pre-image remains open of closed. There are 5 parts for which I must either prove or give a counter-example. I cannot find an example for3.and can't think of how to prove4.I put all of them down though - just in case I messed anything up.

Setup:

Let be continuous. Also, let and

1.If is open, then is open.

False: If for some constant , then its range is closed no matter the pre-image.

2.If is closed, then is closed.

False: Using and , then .

3.If is bounded, then is bounded.

False:ButI'm having difficulty thinking of an example.

4.If is open, then is open.

True:Butagain, I'm not sure where to turn for the proof.

5.If is closed, then is closed.

True: If I had the above, I'd use its complement to show this one holds, but since I don't... Let and let be a sequence of elements from that converges to . Since is continuous, converges to . Since is closed, is in C. So, is in .