Answer:
Step-by-step explanation:
REcall that given sets S,T if we want to prove that 
, then we need to prove that for all x that is in S, it is in T.
a) Let (a,b) be a non empty interval and 
. Then a<x <b. Let 
Consider 
, then
and
.
Then 
. Hence, (a,b) is open.
Consider the complement of [a,b] (i.e 
).
Then, it is beyond the scope of this answer that
.
Suppose that 
and without loss of generality, suppose that x < a (The same technique applies when x>b). Take 
and consider . Then
Then y \in (-\infty,a). Applying the same argument when 
we find that [a,b] is closed.
c) Let I be an arbitrary set of indexes and consider the family of open sets ![\{A_i\}_{i\in I}. Let [tex]B = \bigcup_{i\in I}A_i](https://tex.z-dn.net/?f=%5C%7BA_i%5C%7D_%7Bi%5Cin%20I%7D.%20Let%20%5Btex%5DB%20%3D%20%5Cbigcup_%7Bi%5Cin%20I%7DA_i)
. Let 
. Then, by detinition there exists an index 
such that 
. Since 
is open, there exists a positive epsilon such that 
. Hence, B is open.
d). Consider the following family of open intervals 
. Let 
. It can be easily proven that
![B =[a,b]](https://tex.z-dn.net/?f=B%20%3D%5Ba%2Cb%5D)
. Then, the intersection of open intervals doesn't need to be an open interval.
b) Note that for every 
and for every 
we have that 
. This means that 
is open, and by definition, 
is closed.
Note that the definition of an open set is the following:
if for every 
, there exists a real number 
such that 
. This means that if a set is not open, there exists an element x in the set S such that for a especific value of epsilon, the subset (x-epsilon, x + epsilon) is not a proper subset of S. Suppose that S is the empty set, and suppose that S is not open. This would imply, by the definition, that there exists an element in S that contradicts the definition of an open set. But, since S is the empty set, it is a contradiction that it has an element. Hence, it must be true that S (i.e the empty set) is open. Hence is also closed, by definition. If you want to prove that this are the only sets that satisfy this property, you must prove that is a connected set (this is a topic in topology)
