1. ## Transitive Set Problem

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

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

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

If $A$ is a transitive set, then by definition: whenever $x \in A$, and $y \in x$, then $y \in A$. That is it. Now, if $A$ is nonempty, as is the case by your assumption, then there is some element $x \in A$. If $x$ is the empty set, then the case satisfies our proof. On the other hand, let us suppose $x$ is nonempty. Then there is some $y \in x$. By the definition of transitivity, $y \in x \Rightarrow y \in A$. The same reasoning applied to $x$ can be applied to $y$ now--i.e., either y is empty or it contains something that is also in $A$. 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 $A \neq \phi$ , using this axiom,

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

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

$B \subseteq A$

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

and

$B\cap A =\phi$ means

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

Now lets assume that

$B \neq \phi$

$\exists x \in B$

$\Rightarrow x \in B$

since

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

and

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

we have

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

which is contradiction. so the assumption that

$B \neq \phi$ is wrong.

$\Rightarrow B=\phi$

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

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

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

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