# Comparing two binary relations using Cartesian products

Let $$f$$ be a binary relation between sets $$A_0$$ and $$B_0$$ and $$g$$ be a binary relation between sets $$A_1$$ and $$B_1$$.

It can be proved that $$[f\cap(A_1\times B_1)=g]\wedge [g\cap(A_0\times B_0)=f]$$ implies $$f=g$$.

Really, it follows $$g\cap(A_0\times B_0)\cap(A_1\times B_1)=g$$. Thus $$g\subseteq A_0\times B_0$$. Consequantly $$g=f$$.

Do the following hold?

1. $$[f\cap(A_1\times B_1)\supseteq g]\wedge [g\cap(A_0\times B_0)]\supseteq f$$ implies $$f=g$$.

2. $$[f\cap(A_1\times B_1)\subseteq g]\wedge [g\cap(A_0\times B_0)]\subseteq f$$ implies $$f=g$$.

• what exactly is the difference between $\land$ and $\cap$ in this context? – Yanko Nov 22 '18 at 17:55
• @Yanko: Their meanings are standard - $\cap$ is intersection of sets, $\wedge$ is conjunction of statements (meaning 'and'). – Clive Newstead Nov 22 '18 at 17:56
• @CliveNewstead I get that, but if $f,g$ are binary relations, isn't $f\land g$ the same as $f\cap g$ where $f$ and $g$ are being viewed as subsets of the cartesian product? – Yanko Nov 22 '18 at 17:57
• @Yanko: I'd presume so, yes. What's the confusion? – Clive Newstead Nov 22 '18 at 17:58
• @Yanko: Ah, you're just parsing it wrong. Read it as $$[f \cap (A_1 \times B_1) = g] \wedge [g \cap (A_0 \times B_0) = f]$$ and so on. – Clive Newstead Nov 22 '18 at 18:01

It is true in general that if $$X \subseteq Y \cap Z$$, then $$X \subseteq Y$$ and $$X \subseteq Z$$. The hypothesis of statement (1) implies that $$f \subseteq g$$ and $$g \subseteq f$$, and so $$f = g$$, as required. So statement (1) is true.
The hypothesis of statement (2) holds for all $$f,g$$, provided that $$A_0 \cap A_1 = \varnothing$$ or $$B_0 \cap B_1 = \varnothing$$, so it doesn't follow in general that $$f=g$$. So statement (2) is false in general.
In fact, more generally, we have $$(A_0 \times B_0) \cap (A_1 \times B_1) = (A_0 \cap A_1) \times (B_0 \cap B_1)$$, and so the hypothesis of (2) holds whenever the relations $$f$$ and $$g$$ agree on $$(A_0 \cap A_1) \times (B_0 \cap B_1)$$; but then $$f$$ and $$g$$ can say what they like about the elements not in these intersections, so they need not be equal.