Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I redeaded the following definitions:

--1) let $A$ a set, $A$ is transitive set if $\forall B \in A (B \subseteq A )$

--2) let $A$ a set, $A$ is transitive set if $\forall x \in A (x \subseteq A )$

I think that 2) is correct, in fact if $A:=\{\emptyset, \{\emptyset\},a,f\}$ then in case 1) $A$ is transitive set but in case 2) $A$ is not transitive set because $a,f \nsubseteq A$... is correct?

Thanks in advance!

share|cite|improve this question
The two definitions are exactly equivalent; they just use different names for the elements of $A$. – user61527 Aug 20 '13 at 13:47
@T.Bongers, I think that first definition is only for sets of sets.. or not? – mle Aug 20 '13 at 13:49
No, they just have different letters. – user61527 Aug 20 '13 at 13:50
In Zermelo-Fraenkel Set Theory, all objects are sets. – João Júnior Aug 20 '13 at 13:50
@JoãoJúnior, okok is true.. but if I am in ZF with ur-element? just out of curiosity... – mle Aug 20 '13 at 13:53
up vote 1 down vote accepted

The two definitions are equivalent. They are also equivalent to this one:

A set $A$ is transitive if $\forall\heartsuit\in A(\heartsuit\subseteq A)$.

The only difference is what we're calling the variable ranging over $A$.

As for your particular example, it depends on what $a$ and $f$ are. In the case that $a=\bigl\{\emptyset,\{\emptyset\}\bigr\}$ and $f=a\cup\{a\},$ then $A$ is indeed transitive. It also works if $f=\bigl\{\emptyset,\{\emptyset\}\bigr\}$ and $a=f\cup\{f\}.$ It also works if $a$ and $f$ are elements of $$\Bigl\{\emptyset,\{\emptyset\},\bigl\{\emptyset,\{\emptyset\}\bigr\}\Bigr\}.$$ Otherwise, you're correct that $A$ is not transitive, regardless of what $a$ and $f$ might be (sets or ur-elements).

share|cite|improve this answer
okok.. thanks...!! :) ;) – mle Aug 20 '13 at 13:55
As a side observation, one can readily prove that if $A$ is a transitive set, then there are no ur-elements in $A$ (i.e.: any set with an ur-element in it is non-transitive). – Cameron Buie Aug 20 '13 at 14:00
I can edit the definition for this case "$A$ is transitive set if $A - \{B \in A|B \subseteq A\}=\emptyset$.. is correct? – mle Aug 20 '13 at 14:10
That is another perfectly fine definition, and works whether or not we have ur-elements in our theory. – Cameron Buie Aug 20 '13 at 14:11
okok thanks soo much!! :) – mle Aug 20 '13 at 14:16

Your Answer


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.