-1
$\begingroup$

Any binary relation over any set (finite or infinite) must has a transitive closure. Moreover, every binary relation must has a minimal transitive closure. Who proved this well-known result in abstract algebra first?

Many abstract algebra texts include this.

$\endgroup$
  • $\begingroup$ What's there to prove? Who was the first to prove that any finite set of integers has a maximum? $\endgroup$ – Gerry Myerson May 3 at 3:36
  • $\begingroup$ @GerryMyerson The proof can be very short - about four lines. Of course you can claim that it is trivial, but this is kind of subjective. $\endgroup$ – High GPA May 3 at 4:54
  • $\begingroup$ @GerryMyerson The example you posed has a named theorem (in the infinite/continuous case) - Weierstrass theorem. The complete proof can be of a few pages $\endgroup$ – High GPA May 3 at 4:57
  • $\begingroup$ This statement is very obvious so I doubt it would have been stated and proved explicitly as a theorem by whoever first used it. $\endgroup$ – Eric Wofsey May 5 at 4:45
2
$\begingroup$

TRANSITIVE. (Of a binary relation) Bertrand Russell wrote in "On the Notion of Order," Mind, 10, (1901), p. 32. "When ARB and BRC imply ARC, I call R transitive .... This term was used in this sense by De Morgan ... It is now generally adopted." De Morgan wrote in "On the Symbols of Logic, the Theory of the Syllogism, and in particular of the Copula," Transactions of the Cambridge Philosophical Society, 9, (1850) p. 104: "The first is what I shall call transitiveness, symbolized in X—Y—Z = X—Z; meaning that if X stand in the relation denoted by — to Y, and Y to Z, X therefore stands in that relation to Z." (OED)

Source: http://jeff560.tripod.com/t.html

$\endgroup$
  • $\begingroup$ Thank you very much!! To me it sounds like only for transitive. Is it also for the definition of transitive closure? $\endgroup$ – High GPA May 3 at 11:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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