Set theory questions about $ϵ$ and concatenation I have two questions I'm wondering about. Take a look at my answers and tell me if I did anything wrong or any suggestions/hints would be appreciated thanks!
Q1 (T/F):
Let A be an alphabet, e represents an empty STRING, 0 represents an empty SET
(a) e $∈$ 0
(b) $e^{R}$ $=$ $(ee)^{4}$
(c) e $∈$ {{e}, x, y}
(d) 0 $=$ e
(e) 0 $∈$ A*
(f) e $∈$ A*
(g) 0 is not the same as {0}
(h) e $∈$ A
(i) 0 $⊆$ A*
Q2:
A={a,b} and B={0,1}
a) compute concatenation of $A^{2}$ and $B^{2}$
b) $(AB)^{*}$$=$$(BA)^{*}$ is this true?
My answers:
Q1:
a. T
b. T because if you concat 2 empty strings it will still be empty and e^R is empty still
c. F
d. T
e. F  A* contains empty STRING not empty SET?
f. T
g. T
h. F
i. T
Q2:
a) I think this is a tricky question, not to sure about it...
$A^{2}$ $=$ aabb
$B^{2}$ $=$ 0011
b) I would say this is false. Because the concatenation of AB is a different string compared to the concatenation of BA so $(AB)^{*}$ would contain ALL combinations of AB and $(BA)^{*}$ would contain ALL combinations of BA which would not be equal.
 A: In the first question (a) is wrong: $0$ has no elements, so in particular $e\notin 0$. It’s not clear how (d) and (e) are to be answered. If you’re to understand $e$ and $0$ as distinct objects, then your ansower to (d) is wrong, but your answer to (e) is right. If, on the other hand, you’re to consider them simply different names for the same object, then your answer to (d) is right, but your answer to (e) is wrong — unless you’re expected to know that even though they denote the same object, they are properly used only in different contexts, in which case your answers to (d) and (e) are both right. The other answers are correct.
In the second question, $A^2=\{xy:x,y\in A\}$: it’s the set of all words that you can get by concatenating two elements of $A$. Thus, $A^2=\{aa,ab,ba,bb\}$. From this you should be able to correct your answer for $B^2$ as well. You need to be a bit more explicit in your answer to part (b). You’re right that $(AB)^*\ne(BA)^*$, but you should explain this more clearly with a specific example of a word that is in one of the sets but not the other. For instance, $a0\in(AB)^*$, but $a0\notin(BA)^*$, because every non-empty word in $(BA)^*$ begins with a member of $B$, i.e., with $0$ or $1$, not with $a$.
