1. If set A is compact, show that f(A) is compact. Is the converse true?

Ans. f:M to N is continuous and A subset M is compact. Then f(A) is compact.
The converse is not necessarilly true. For Ex: F(x)=0 for every x in R(real #'s) and
k={0}. Then f^-1(k)=R is not compact.

2. If set A is connected, show that f(A) is connected. Is the converse true?

Ans. f:M to N is continuous and A subset M is connected. Then f(A) is connected.
The converse is not necessarilly true. For Ex: F(x)=x^2 and k=1.
Then f^-1(k)={-1,1} which is not connected.

3. If set B is closed, show that B inverse is closed.

Ans. f is continuous on B if f is continuous on every x sub 0 element B.
a. f is continuous on B
b. for every x sub n to x sub 0 in A. f(x sub n) approaches f(x sub 0)
c. for any u open in N, f^-1(u) is open in M
d. for any F closed in N, f^-1(F) is closed in M.

If anyone can add to this I would be greatfull.

I would say that your solutions for 1 and 2 really don't seem to be sufficient, unless you're already told that the continuous images of compact/connected sets are compact/connected, in which case this is a trivial question. Is f continuous? For these to be true it must be but you haven't stated that anywhere. Furthermore, this constitutes a topology question rather than an analysis question.

1) To show that a set is compact it is necessary to show that every family of covers for that set has a finite subcover. In regard to this question, we need to show that if C is a cover for f(A), then there exists a finite subcover of f(A).

Let C be a cover of f(A), and define

$$C^\prime = \{ f^{-1}(U) | U \in C \}$$

Now what can we say about each $U \in A$ given that f is continuous?

Furthermore, what is $C^\prime$ with respect to A? How can we use the relation between $C^\prime$ and A to find a finite subcover of the image?

2) For the sake of contradiction, assume that f(A) is not-connected. That is, it can be written as the disjoint union of two open sets. Since f is continuous, consider the pre-image of these sets. Why does this give a contradiction?

morphism