Universal set of subsets A and B The question is:
Two subsets given:
$A = \{ Z, H, O, V, N, I, R \}$;
$B = \{ I, G, O, R \}$
The aim is to find universal set of this subsets. I tried to use definition of "universal set" and here are my suggestions:


*

*Universal set is array of UNIQUE characters of subsets:
$U = \{ Z, H, O, V, N, I, R, G \}$

*Universal set is ALL characters of subsets:
$U = \{ Z, H, O, V, N, I, R, I, G, O, R \}$

*Universal set is all alphabetical characters:
$U =\{ A, \dots Z \}$
Which one is true?
Thanks in advance!
 A: The universal set $U$ could be either $(1): U = A\cup B$ or $(3)$ (in which case $A\cup B\subsetneq U)$: 
Without more information, we cannot conclude which, if either. The Universal Set is simply the set which contains all elements in the domain. Without the domain clearly defined, we cannot conclude just how large $U$ is; we can only conclude, by the definition of "subset", that if $A \subseteq U$ and $B\subseteq U$, then $A\cup B \subseteq U$.
The second "option" you list is simply the same set as described by $(1)$: A set of elements is a set with each element counting once and only once. So, for example, $\{1, 1, 2, 3, 3\} = \{1, 2, 3\}$.
A: According to set theory, both 1. and 2. are the same set: $\{X,Y,X\}=\{X,Y\}$.  Repetition and order does not matter.  This set is equal to $\{G,H,I,N,O,R,V,Z\}$ and can be notated as $A\cup B$ (union of sets $A$ and $B$).
Any universal set must contain $A$ and must contain $B$ so it must contain its union: $A\cup B\subseteq U$.  So, whatever you take as you universal set is up to you as longer as $G\in U$, $H\in U$, $I\in U$, $N\in U$, $O\in U$, $R\in U$, $V\in U$ and $Z\in U$.  Your alternative 3. is such a solution.
