
Russ P. answered 10/14/14
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Rose,
Because it is an even multiple of 3, we can multiple this by 2 and then relate it to the +- integers as follows. Then the set members are functions of n and we don't have to worry about even or odd, just +- integers since you didn't state that x was positive.
S(x) = {x | where x = 6n and n = 0, +-1, +-2, +-3, ... to +- infinity}
Note that x=zero is even because it is divisible by 2; but no odd number can be divisible by 2, and 0 clearly is, so zero can't be odd. However, some might keep it out of the set anyway, but I'll leave it in based on this argument.
(a) 4 elements of S(x) = {-6, 6, 12, 18}
(b) The set S(x) is infinite because n is infinite and so S can have an infinite number of elements.
(c) x = -9 is not an even multiple of 3 or a multiple of our 6 formulation so it is not in S(x). I can't type the symbol here, but you would use the with a slash thru it to denote "not a member of the defined set".

Byron S.
10/14/14