elements of sets, subsets relations give an example (if possible) such that:
a)$x\in y$ and $x\not\subseteq y$
here I take $x=\{1,2\}$ and $y=\mathcal{P(x)}=\{\emptyset, \{1\},\{2\},\{1,2\}\}$ then $x\in y$ but $x\not\subseteq y$ as e.q $1\not\in y$
b)$x\subseteq y$ and $x\not\in y$ here I cannot find any counter example...By definition of subset any element in x must be in y...
is it true?
c)$x\in y$ and $x\subseteq y$ 
Here I take $x=\emptyset$ and $y=\mathcal{P(x)}=\{\emptyset, \{\emptyset\}\}$ then $x\in y$ and $x\subseteq y$
could someone please check?
 A: $(a)$ and $(c)$ are alright. For $(b)$ let $x=\{1,2\}$ and $y=\{1,2,3\}$, Then $x\subseteq y$ but $x\notin y$. In contrast an example for $x\subseteq y$ but $x\in y$ would be when $x=\{1,2\}$ and $y=\{1,2,\{1,2\}\}$.
A: Your answers for (a) and (c) are fine. For (b), remember that $\varnothing\subseteq y$ for any set $y$ whatsoever, so take $x=\varnothing$ and let $y$ be any set that does not have $\varnothing$ as an element, e.g., $\{\{\varnothing\}\}$.
Added: For (b) you might also consider the fact that a finite set $y$ has $2^{|y|}$ subsets and $|y|$ elements, and $n<2^n$ for all $n\in\Bbb N$. Thus, every finite $y$ will give you an example: it has more subsets than elements, so at least one of those subsets cannot be an element. In fact you can always take $x=y$: $y\subseteq y$, but $y\notin y$. If you’re working in $\mathsf{ZF}$ this is true for all $y$ as a consequence of the axiom of regularity (also called foundation), but you can check it explicitly with any simple finite example that you might construct, like your answers to (a) and (c).
