# Suppose $R$ is a relation on $A$. Let $S$ be the transitive closure of $R$. Then $Dom(S)\subset Dom(R)$

$$Proof$$. Suppose $$x\in Dom(S)$$ and $$x\notin Dom(R)$$. Then there is some $$y\in A$$ such that $$xSy$$. Since $$S$$ is the transitive closure of $$R$$, there exist some set of pairs in $$R$$ such that $$x$$ and $$y$$ can be connected. But $$x\notin Dom(R)$$, so this connection cannot be made. Therefore it must be that $$x\in Dom(R)$$.

Im not 100% confident that this proof is valid. Is the wording used ok? If $$S$$ is the transitive closure of $$R$$ and $$xSy$$, then if you understand what transitive closure means, you would know that there are some $$xRz$$ and $$zRb$$ and $$bRy$$ to make $$xSy$$, but I feel like the way I explain it in the proof would be considered bad.

• I don't see any issue with the proof. Can you explain what you find concerning? Your explanation is also okay, other than being worded as if there can be only two intermedaries between $x$ and $y$ (also some confusing abbreviation of what you are actually trying to so - take the extra effort to speak clearly "there exist $z$ and $b$ such that ..."). But that is a easy fix. In both cases, the point is that the chain of $R$-relationships from $x$ to $y$ has to start with $x$, which requires $x$ to be in the domain of $R$. Jun 10 at 4:34
• @PaulSinclair someone else told me that "it is definitely too vague" Jun 10 at 11:47
• Well, I disagree. It does expect you to understand how the transitive closure is constructed, which might be inappropriate in circumstances where the concept is completely new with little to no development, but apparently that was not the case for you as you supply that detail. I presume that there was enough development of the concept in its context for you to recognize it. With that concept, is obvious that if $x \notin \text{Dom}(R)$, the first step of that chain cannot exist. Jun 10 at 16:42