Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Fix some binary relation $f$.

Does there necessarily exist a set $C$ such that $(x\times x)\cap f\ne \varnothing \Leftrightarrow x\cap C\ne \varnothing$ for all sets $x$?

share|improve this question
add comment

1 Answer

up vote 3 down vote accepted

Not necessarily.

Say $f=\{(a,b)\}$, with $a\neq b$, and assume such a set $C$ exists.

In particular, $\{a,b\}\cap C\neq\varnothing$, since $\{a,b\}\times\{a,b\}\cap f\neq\varnothing$. Hence either $a\in C$ or $b\in C$ (or both). If $a\in C$, then $\{a\}\cap C\neq\varnothing$, but $\{a\}\times\{a\}\cap f=\varnothing$. If $b\in C$, then $\{b\}\cap C\neq\varnothing$ but $\{b\}\times\{b\}\cap f=\varnothing$. So no such $C$ exists.

By the above argument, if $f$ contains a pair $(a,b)$ such that (i) $a\neq b$; and (ii) $(a,a)\notin f$; and (iii) $(b,b)\notin f$; then no such set $C$ can exist.

Conversely, suppose that for all $a,b$, if $(a,b)\in f$ then either $(a,a)$ or $(b,b)$ lies in $f$. I claim that then the set $C$ does exist, by letting $C$ consist precisely of those elements $r$ for which $(r,r)\in f$.

Indeed, suppose that $x$ is a set such that $x\times x\cap f\neq\varnothing$. Then there exist $a,b\in x$ such that $(a,b)\in f$; by assumption, either $(a,a)\in f$, and hence $a\in C$ so $x\cap C\neq\varnothing$; or else $(b,b)\in f$ and hence $b\in C$, so $x\cap C=\varnothing$.

Conversely, if $a\in x\cap C$, then by definition of $C$ we have $(a,a)\in f$, so $x\times x\cap f\neq \varnothing$.

In summary:

A set $C$ as given exists if and only if $f$ satisfies the condition $$(a,b)\in f\implies (a,a)\in f\text{ or }(b,b)\in f.$$

If that is the case, then $C=\{a\mid (a,a)\in f\}$ is the desired set; one can justify that $C$ is a set since it is contained in $\mathrm{dom}(f)\cup\mathrm{range}(f)$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.