Problem in understanding set theory I am stuck with a small question. It is below,

Let $A$ be a set and let $B = \{A,\{A\}\}$. Then, since $A$ and $\{A\}$ are elements of $B$, we have $A \in B$ and $\{A\} \in B$. It follows that $\{A\}\subseteq B$ and $\{\{A\}\} \subseteq B$. However, it is not true that $A\subseteq B$.$\hspace{302pt}\blacksquare$

I have to ask two questions from this text,


*

*What is the difference between $A$ and $\{A\}$?

*As $A$ is contained by $B$ so $A$ belongs to $B$ should be true but it is not. Why is that?


Thanks.
 A: (1) A is a set with certain elements, and {A} is a set with only one specific element, and that element is itself a set: the set A. In set theory we not only regard the elements of a set as mathematical objects, but the sets themselves as mathematical objects, and we distinguish between the two. For example, the numbers 1, 2, and 3 are all numbers whereas the set {1,2,3} is not a number but a set of numbers. Furthermore, the set {{1,2,3}} is not a set of numbers (as {1,2,3} isn't a number) but a set containing a set of numbers.
(2) A is an element of B, that is $ A \in B $. However, that doesn't mean any of the elements of $ A $ are actually straight-up elements of $ B $. For example, say A is the set {1,2,3}. Then B = {{1,2,3},{{1,2,3}}}. Neither {1,2,3} nor {{1,2,3}} are the numbers 1, 2, or 3, nor are they numbers at all (they are sets), so we can't say any of 1, 2, or 3 are in the set B. 
A: *

*$A$ is a given set, and $\{A\}$ is the set which has only one element, namely $A$.

*Be careful, it is true that $A\in B$. That is, the set $A$ is an element of the set $B$. It is not true, though, that $A\subseteq B$. Why? Well, by definition: $$A\subseteq B \iff (x\in A \Rightarrow x\in B)$$
This is not true, because there is an element of $A$ that is not an element of $B$. In fact, none of the elements of $A$ can be elements of $B$, for the elements of $B$ are only $A$ and $\{A\}$, but $A\notin A$ and $\{A\}\notin A$. See Axiom of regularity
Indeed, it is not true that $A\in A$ because "no set is an element of itself" as explained in the Wikipedia link (in Zermelo-Fraenkel theory which is a well-founded theory), and it is not true that $\{A\}\in A$, for this would imply that $A\in \{A\}\in A \in \{A\}$... forming an infinite descending sequence of sets, which is also impossible by regularity.
A: $A$ is the set $A$ itself, but $\{A\}$ is a singleton set with just one element, namely $A$. Yes, sets themselves can be elements of other sets.
You don't know that $A$ is contained by $B$, you only know that $\{A\}$ is contained by $B$, $\{A\}\subseteq B$. This is because the only element $A$ of $\{A\}$ is also an element of $B$. This is the definition of what it means to be a subset. 
A: *

*$A$ is a set containing elements (which are not explicited here, by the way) and $\{ A \}$ is a set containing one element and this element is the set $A$. The elements of $A$ are not elements of $\{ A \}$.


The relation $A \in B$ means that $A$ is an element of $B$. The relation $A \subseteq B$ means that every element of $A$ is an element of $B$. Since the only elements of $B$ are $A$ and $\{ A \}$, unless $A$ has only A or $\{ A \}$ as an element (which would make a recursive definition which is kind of weird... I don't know if this is somehow possible but I think it's not allowed under some axioms, to see how weird it could be, you would have $A = \{ A, \{A\} \}$, $A = \{A \}$ or $A = \{ \{ A \} \}$, and to be honest usually you don't consider those cases), you can't say that $A \subseteq B$.
Hope that helps,
