problem in set theory QN1 by logical argument verify that {a} is not open for any real number "a".
i guess the set is not open since there is no open interval about "a"
instead the set is said to be closed since any of its complement must be open.
am i correct?
QN2 GIVE TWO COUNTER-EXAMPLES WHICH SHOWS THAT THE IMAGE OF INTERSECTION OF TWO SETS IS NOT EQUAL TO THE INTERSECTION OF THEIR IMAGES 
 A: If you are working with the usual topology on the reals, then the complement of {a} is the union of two open intervals and so open. If {a} were open, its complement would need to be closed, which it is not (a is a limit point of the complement, but not included in the complement). So, your argument for question 1 seems OK. 
For question 2, you could always take the cheap way out by letting S and T be nonempty disjoint subsets of the reals and f a constant function. The image of the intersection of S and T is then empty, as is the image of f on the intersection. The intersections of the images is whatever the constant value of f is. Another possibility is to pick a function like $f(x) = x^{2}$ and take S=[-1,0] and T=[0,1].
A: Hint for Q2:
Must the image of a point in the intersection of two sets be in the intersection of the images of the two sets?
Must a point in the intersection of the images of two sets be the image of a point in the intersection of the two sets?
If the answers to these questions are different, then can you construct a counter-example? 
