1. ## Transitive Set Problem

Prove that if $\displaystyle A$ is transitive and $\displaystyle A \ne \phi$, then $\displaystyle \phi \in A$.

I can only say that if $\displaystyle A$ is transitive, then $\displaystyle \forall a(a \in A \longrightarrow a \subseteq A)$. How would I have to do in order to continue from here?

2. ## Re: Transitive Set Problem (Help)

If $\displaystyle A$ is a transitive set, then by definition: whenever $\displaystyle x \in A$, and $\displaystyle y \in x$, then $\displaystyle y \in A$. That is it. Now, if $\displaystyle A$ is nonempty, as is the case by your assumption, then there is some element $\displaystyle x \in A$. If $\displaystyle x$ is the empty set, then the case satisfies our proof. On the other hand, let us suppose $\displaystyle x$ is nonempty. Then there is some $\displaystyle y \in x$. By the definition of transitivity, $\displaystyle y \in x \Rightarrow y \in A$. The same reasoning applied to $\displaystyle x$ can be applied to $\displaystyle y$ now--i.e., either y is empty or it contains something that is also in $\displaystyle A$. It might be fruitful to index these elements as a sequence. Thus, what we end up having to prove is that this member sequence will "bottom out" at some point. Now, how to go about this proof, I am unsure, and maybe I've approached it in an inefficient or less fruitful manner.

3. ## Re: Transitive Set Problem (Help)

Hi

we can use the axiom of regularity , which is one of the axioms of
Zermelo–Fraenkel set theory. Look up here

Axiom of regularity - Wikipedia, the free encyclopedia

axiom says

"Every non-empty set A contains an element B which is disjoint from A"

Since $\displaystyle A \neq \phi$ , using this axiom,

$\displaystyle B \in A \;\;\mbox{and}\; B\cap A =\phi$

$\displaystyle \because B \in A$ and A is transitive , so

$\displaystyle B \subseteq A$

$\displaystyle \forall x[x \in B\Rightarrow x \in A ]$

and

$\displaystyle B\cap A =\phi$ means

$\displaystyle \forall x[x \in B \Rightarrow x \notin A]$

Now lets assume that

$\displaystyle B \neq \phi$

$\displaystyle \exists x \in B$

$\displaystyle \Rightarrow x \in B$

since

$\displaystyle \forall x[x \in B\Rightarrow x \in A ]$

and

$\displaystyle \forall x[x \in B \Rightarrow x \notin A]$

we have

$\displaystyle x \in A \wedge x \notin A$

which is contradiction. so the assumption that

$\displaystyle B \neq \phi$ is wrong.

$\displaystyle \Rightarrow B=\phi$

$\displaystyle \because B \in A \Rightarrow \phi \in A$

hope this helps.... $\displaystyle \smile \heartsuit$

4. ## Re: Transitive Set Problem (Help)

just curious about the solution given by me..... can somebody confirm my solution here ...... thanks