Russell Paradox
# Russell Paradox
Imagine that there is a set $R$ ;
$R=$ { $x\text{ | } x\notin x$ }
Now we want to decide whether $R$ is the element of $R$.
Thus, just consider 2 cases;
1. $ R \in R $
2. $ R \notin R $
$ 1 $ $ \Rightarrow $ R is an element of set R .
so $ R \notin R $ but $ \bot $ 1.
$ 2 $ $ \Rightarrow $ R satisfies the condition of R .
so $ R \in R $ but $ \bot $ 2.
Both 1 & 2 is $ \bot $ , this is inconsequence .
# Predicate
We prepare "Predicate" for all Sets,
so that $ A = $ { $ x \ | \ P(x) $ }
where $ P(x) $ is defined as `if x is the element of Set A then True else False`.
Now, consider the above set $R$ with predicate $P$
$ R = $ { $ x \ | \ x \notin x \wedge P(x) $ }
And test the 2 cases;
1. $ R \in R $
2. $ R \notin R $
$ 1 $ $ \Rightarrow $ R is an element of set R.
So $ R \notin R $ but $ \bot $ 1.
$ 2 $ $ \Rightarrow \ \lnot P(x) $ .
So $ R \notin R \wedge \lnot P(x) $
This $R$ is not the element of $R$ .
So now, the case $ 2 $ ," $R$ is not the element of $R$ ", is True.
This is the birth of Russel's Type Thoery.