Elements and Subsets This is my first discrete math course so this questions might seem simple but I would like clarification.
Say I have a set $A=\{1,2,3\}$ and a set $D=\{1,2,3\}$. Is it true that set $A$ is both an element and a subset of set $D$?
Now say $D=\{\{1,2,3\}\}$. Is it true that set $A$ is no longer an element of $D$ because $D$ is a set which contains a set that contains $1,2,3$?
A third case. Say I have set $D=\{x,16,\{1,2,3\}\}$. Is set $A$ an element and proper subset of $D$? 
 A: 
Say I have a set $A = \{1,2,3\}$ and a set $D = \{1,2,3\}$. Is it true that set $A$ is both an element and a subset of set $D$?

No.  $A$ is a subset of $D$, but not an element.  $A$ is also a subset of a set like $\{1,2,3,4\}$.

Now say $D = \{\{1,2,3\}\}$. Is it true that set $A$ is no longer an element of $D$ because $D$ is a set which contains a set that contains $1,2,3$?

Now $A$ is an element of $D$ but not a subset.  While $D$ in the previous example was a set that had three members, in this example $D$ is now a set that has only one member - that member being the set $\{1,2,3\}$.  This time, in opposition to the previous example, $A$ would not be an element of $\{\{1,2,3,4\}\}$, because the set needs to be exact.

A third case. Say I have set $D = \{x,16,\{1,2,3\}\}$. Is set $A$ an element and proper subset of $D$?

Once again, $A$ is an element of $D$ but not a subset.  An example of a set that has $A$ as both a member and a subset would be $\{1, 2, 3, \{1, 2, 3\}\}$
A: In the first case, $A = D$, which is different from $A \subseteq D$ (true) or $A \in D$ (false). In the second case, $A \in D$, while $A \ne D$ and $A \not\subseteq D$.  
A: In the first case $A$ is a subset of $D$, since every element of $A$ lies in $D$. The converse also holds, so you have $A = D$ in this case, but $A$ is not an element of $D$. 
In the second case, $A$ is an element of $D$. For the last case, $A$ is again an element , but not a subset.
A: In your first case $A$ is not a subset of $D$, and not an Element of $D$ since $A=D$,
in your second case, $A$ is an element of $D$, here the only element of $D$,
in your third case $A$ is an element of $D$, and since every Element ist also a subset, it is a subset of $D$.
A: If $A = \{1,2,3\}$ and $D = \{1,2,3\}$ then $A$ is a subset of $D$, but isn't a proper subset. Because
$$\nexists a \in D | a \notin A$$
$$A \subseteq D \: \& \: A \subseteq D \Longrightarrow A=D$$
If $A = \{1,2,3\}$ and $D = \{\{1,2,3\}\}$ then A is a element of D. Because $D=\{A\}$. So A is an element in D. So $A$ is not a subset of $D$. 
$$A \in D$$
If $D = \{x,16,\{1,2,3\}\}$ then $D = \{x,16,A\}$, so $A$ also is an element in $D$. 
$$A \in D$$
For better understanding, assume that input is in this format:
$$M = \{x,16,N\}$$
Now $N$ is an element in $M$ and $P=\{N\}$ and $Q=\{x,N\}$  are two subset of $M$.
A: 
"Is it true that set A is both an element and a subset of set D?"

No.  
The elements of $D$ are $1,2,$ and $3$.  
None of those things are the same thing that $A$ is (which is the set $\{1,2,3\}$).  So $A$ is not an element of $D$.
However $A \subset D$.
The elements of $A$ are $1,2$ and $3$.  And every one of those elements is an element of $D$.  So that $A\subset D$.
(As a rule for any set $X$ we have $X \subset X$ because every element of $X$ is an element of $X$.  And in this case $A = D$.)

Now say D={{1,2,3}}. Is it true that set A is no longer an element of D because D is a set which contains a set that contains 1,2,3?

The exact opposite!
Now $D$ is a set with $1$ element.  That element is $\{1,2,3\}$.  That element is the same thing that $A$ is.  So $A$ is an element of $D$.
But now it is no longer true that $A \subset D$.
The elements of $A$ are $1,2,3$ and none of those elements are an element of $D$ (whose only element is $\{1,2,3\}$).  So none of the elements of $A$ are elements of $D$. So it is not true that all of the elements of $A$ are element of $D$ (Not only are they not all; none of them are). So $A\not \subset D$.
