Variant on Russell's paradox: show $B = \varnothing$

Question

Let $X$ be a set and $R$ a relationship on $X$.

Define $N = \{x \in X\mid(x, x) \notin R\}$. Let $$B =\{b \in X\mid(\forall n \in N)(b\,R\,n) \land (\forall n \notin N)(\neg b\,R\,n)\}\;.$$ Show that $B = \varnothing$.

This is a version of “Russell’s paradox”. Only there is nothing paradoxical about it in this formulation. The set $B$ is defined as a subset of $X$ in terms of how it behaves in the relation $R$, and should not necessarily be nonempty.

I'm not sure how to even start off this question so any help is appreciated.

Answer 1

Suppose that $b \in B$. Now we can try and determine if $b \in N$.

Can you complete the following arguments?

• If $b \in N$, then by definition of $B$ $\ldots$
• If $b \notin N$, then by definition of $B$ $\ldots$

What does this mean for $b$? Can it exist?

Answer 2

Suppose $y\in B$. Apply definition of $B$ for $y$. Apply definition of $N$ for $y$. Suppose $y\in N$. Obtain two successive contradictions, and hence $\neg\exists y\in B$.