Power sets question about $\subseteq$ and $\in$ I have a question with regards to this question as follows: 
Let $A = \{1,2,\{1,2\}\}$. Determine whether the following statements are true or false with a brief explanation of why. 
a) $\{1,2\} \subseteq P(A)$
b) $\{1,2\} \in P(A)$
In this case, I am not sure why $a)$ false since I wrote out $P(A): \{\emptyset,\{1\},\{2\},\{\{1,2\}\},\{1,2\},\{1,\{1,2\}\},\{2,\{1,2\}\},\{1,2,\{1,2\}\}\}$
Then is it not the case that $\{1,2\} \subseteq P(A)$ ? I think I need some clarification about the difference of $\subseteq$ and $\in$. I had always thought that if it was listed without braces inside an overall brace of the entire set, then it would be written as $\subseteq$ since I thought of it as "the set containing the set". In this case, I know that $1 \in A$ and $\{1,2\} \subseteq A$ but I am not sure about the first two statements since the solutions stated that statement $a)$ is false while statement $b)$ is false.
Would someone please help me correct my understanding ? 
Thank you
 A: Keep in mind the difference between $\{1,2\}$ and $\{\{1,2\}\}$.  These are not the same as sets.  The first has two elements, the second has only one element.  Furthermore, $\{\{1\},\{2\}\}$ is different from both of these.
If it were true that $\{1,2\}$ were indeed a subset of $P(A)$ that would require both $1$ and $2$ to be elements of $P(A)$, however that is not the case.  You happen to have $\{1\}$ and $\{2\}$ are elements of $P(A)$, but that is not the same thing as $1$ and $2$ being elements of $P(A)$.  Since $\{1,2\}$ has an element that is not an element of $P(A)$ we see that it is not a subset of $P(A)$.

Additional information:
A set is a collection of elements.  To write the set, we may write a pair of braces and each item within those braces are the elements.
E.g. $\{1,2\}$ is the set containing two elements, namely the elements "$1$" and "$2$".
Some times, the elements may be sets themselves.  $\{\{1,2\}\}$ is a set containing one element.  That one element namely being the set described in the previous example.
We say an object is "an element of" a set if it is one of those items which is an element of the set.  For example, in the set $\{1,2\}$ the elements are $1$ and $2$.  $3$ is not an element of $\{1,2\}$.  Neither is $\{1\}$.  $\{1\}$ is not an element of $\{1,2\}$.  To denote this "is an element of" relation, we use the symbol "$\in$".  To denote the opposite, we use "$\notin$"
Given two sets, $A$ and $B$, we say the set $A$ is a subset of $B$ iff every element of $A$ is also an element of $B$.  Worded a different way, $A$ contains no elements that are not also elements of $B$.  Worded a different way, if we were to pick an arbitrary object, if that object happened to be an element of $A$ it must also be an element of $B$.
Given two sets, $A$ and $B$, we say $A$ is not a subset of $B$ iff there exists an element of $A$ which is not an element of $B$.
We denote these as $A\subseteq B$ ($A$ is a subset of $B$) and $A\not\subseteq B$ ($A$ is not a subset of $B$).  (Side note: some authors use $\subset$ to denote the same thing described here, while others use this specifically to denote being a proper subset, one where it is a subset but not equal)

One additional example which will hopefully clear up final confusion:
$\{1,\{2\}\}$ has elements $1$ and $\{2\}$.  It has subsets $\emptyset, \{1\},\{\{2\}\},\{1,\{2\}\}$.  It does not have subset $\{2\}$ since $2$ is not an element of $\{1,\{2\}\}$.
In the same way, your example has $\{1,2\}$ is an element of $P(A)$ but it is not a subset of $P(A)$.  $\{\{1,2\}\}$ is a subset of $P(A)$ however, but this is not the same thing.
A: Take $B=\{1,2,\{3,4\}\}$ 
$x\in B$ means $x$ is contained in $B$.   Element $x$ is a member of $B$.
Then: $1,2,\{3,4\}$ are all of the elements of $B$.   So we have: $1\in B\\2\in B\\\{3,4\}\in B$ 

$X\subseteq B$ means $X$ is a subset of $B$.   All elements of set $X$ are elements of set $B$.
$\{\}, \{1\},\{2\}, \{1,2\}, \{\{3,4\}\}, \{1,\{3,4\}\}, \{2,\{3,4\}\},\{1,2,\{3,4\}\}$ are all of the subsets of $B$.   So we have : $\{\}\subseteq B\\ \{1\}\subseteq B\\ \vdots\\\{1,2,\{3,4\}\}\subseteq B\\$
Naturally then $\big\{\{\}, \{1\},\{2\}, \{1,2\}, \{\{3,4\}\}, \{1,\{3,4\}\}, \{2,\{3,4\}\},\{1,2,\{3,4\}\}\big\}$ is the powerset of $B$; that is $\mathcal P(B)$
Then $\{1,2\}\in \mathcal P(B)$ but $\{1,2\}\nsubseteq \mathcal P(B)$.   Can you now see why?
