Answer
a) Assume $S \in S$ and show that $S\notin S$
b) Assume $S \notin S$ and show that $S\in S$
Work Step by Step
Given S={${x| x \notin x}$}
a) Let us assume that $S \in S$
Since $S \in S$, we note that the statement $S \notin S$ is false.
As $S \in S$, all the elements of S have the requirement that $x \notin x$. Since S does not satisfy the requirement, S can then not be an element of S.
$$S \notin S$$
We have then derived a contradiction as it is not possible that $S \in S$ and $S\notin S$ at the same time.
b) Let us assume that $S \notin S$
Since $S \notin S$, we note that the statement $S \in S$ is true.
However, all the elements of S have the requirement that $x \notin x$. Since S satisfy this requirement, S has to be an element of S.
$$S \in S$$
We have then derived a contradiction as it is not possible that $S \notin S$ and $S\in S$ at the same time.
