MHB Why is $X \subset \mathcal{P}X$ true for a transitive set $X$?

[evinda]

Hello! (Wave)

Proposition

Let $X$ be a set. The following are equivalent:

1. $X$ is transitive
2. Each element of $X$ is a subset of $X$ $\left( \forall x(x \in X \rightarrow x \subset X) \right)$
3. $X \subset \mathcal{P}X$
4. $\bigcup X \subset X$

Could you explain me why this: $X \subset \mathcal{P}X$ holds?

When we have for example $X=\{ a, b, c \}$, then $x \in a \rightarrow x \in X$.
So, we see that $a \subset X \rightarrow a \in \mathcal{P}X$.
But why does it also hold that $X \subset \mathcal{P}X$ ? (Thinking)

 

[Evgeny.Makarov]

Sentence

Proposition, or claim.

Let $X$ be a set. The following are equivalent:

1. $X$ is transitive
2. Each element of $X$ is a subset of $X$ $\left( \forall x(x \in X \rightarrow x \subset X) \right)$
3. $X \subset \mathcal{P}X$
4. $\bigcup X \subset X$

Could you explain me why this: $X \subset \mathcal{P}X$ holds?

It doesn't in general. It only holds if any of the other conditions hold.

When we have for example $X=\{ a, b, c \}$, then $x \in a \rightarrow x \in X$.

Why do you say so? What are $a$, $b$ and $c$?

 

[evinda]

Proposition, or claim.

A ok.. (Nod)

It doesn't in general. It only holds if any of the other conditions hold.

Could you explain me why it holds if for example the first condition holds? (Thinking)

Why do you say so? What are $a$, $b$ and $c$?

I thought so because $X$ is transitive. Am I wrong? (Thinking)

 

[Evgeny.Makarov]

Let $X$ be a set. The following are equivalent:

1. $X$ is transitive
2. Each element of $X$ is a subset of $X$ $\left( \forall x(x \in X \rightarrow x \subset X) \right)$
3. $X \subset \mathcal{P}X$
4. $\bigcup X \subset X$

Could you explain me why it [$X \subset \mathcal{P}X$] holds if for example the first condition holds?

It's pretty straightforward to show that 1 implies 2 and 2 implies 3.

When we have for example $X=\{ a, b, c \}$, then $x \in a \rightarrow x \in X$.
So, we see that $a \subset X \rightarrow a \in \mathcal{P}X$.

I thought so because $X$ is transitive. Am I wrong?

This sounds like the following dialog.

"John is a murderer and must go to jail."

"Which John? Every John? And why is he a murderer?"

"I mean, if John is a murderer, he must go to jail."

"Ah..."

You introduced $X$ by saying that $X=\{ a, b, c \}$ for unknown $a$, $b$ and $c$. Who told you that $X$ is transitive? Do you say, "If $X$ is transitive", i.e., are you assuming that $X$ is transitive? Then you should say so. For each statement that you write, you should indicate its status: whether it's an assumption, a theorem that you are starting to prove or it follows from some previous statement.

Also, in this:

So, we see that $a \subset X \rightarrow a \in \mathcal{P}X$.

it is not clear whether we see that $a \subset X$ implies $a \in \mathcal{P}X$ or that $a \subset X$ holds, and therefore $a \in \mathcal{P}X$ holds. To see the difference, $n>5$ implies $n>3$ for all $n$, but it is not true that for any $n$ it is the case that $n>5$ and therefore $n>3$.

 

[evinda]

It's pretty straightforward to show that 1 implies 2 and 2 implies 3.This sounds like the following dialog.

"John is a murderer and must go to jail."

"Which John? Every John? And why is he a murderer?"

"I mean, if John is a murderer, he must go to jail."

"Ah..."

You introduced $X$ by saying that $X=\{ a, b, c \}$ for unknown $a$, $b$ and $c$. Who told you that $X$ is transitive? Do you say, "If $X$ is transitive", i.e., are you assuming that $X$ is transitive? Then you should say so. For each statement that you write, you should indicate its status: whether it's an assumption, a theorem that you are starting to prove or it follows from some previous statement.

Also, in this:

it is not clear whether we see that $a \subset X$ implies $a \in \mathcal{P}X$ or that $a \subset X$ holds, and therefore $a \in \mathcal{P}X$ holds. To see the difference, $n>5$ implies $n>3$ for all $n$, but it is not true that for any $n$ it is the case that $n>5$ and therefore $n>3$.

I am sorry... (Worried)(Tmi)

So can we show that the propositions are equivalent like that? $1 \to 2$: Let $x \in X$. Since $X$ is transitive we have that if $y \in x \rightarrow y \in X$. Therefore $x \subset X$.

$2 \to 3$: Let $x \in X$. Since we assume that $2$ holds we have that $x \subset X \rightarrow x \in \mathcal{P}X$.
Therefore $X \subset \mathcal{P}X$.

$3 \to 4$: Let $x \in \bigcup X$. Then $\exists y \in X$ such that $x \in y$. From $3$ we have that $y \in \mathcal{P}X \rightarrow y \subset X$.
Since $x \in y$ we have that $x \in X$.
Therefore $\bigcup X \subset X$.

$4 \to 1$: Let $x \in \bigcup X $. Since $\bigcup X \subset X$ we have that $x \in X$.
But $x \in \bigcup X$ means that $\exists y \in X$ such that $x \in y$.
How could we continue? (Thinking)

 

[Evgeny.Makarov]

$1 \to 2$: Let $x \in X$. Since $X$ is transitive we have that if $y \in x \rightarrow y \in X$. Therefore $x \subset X$.

$2 \to 3$: Let $x \in X$. Since we assume that $2$ holds we have that $x \subset X \rightarrow x \in \mathcal{P}X$.
Therefore $X \subset \mathcal{P}X$.

$3 \to 4$: Let $x \in \bigcup X$. Then $\exists y \in X$ such that $x \in y$. From $3$ we have that $y \in \mathcal{P}X \rightarrow y \subset X$.
Since $x \in y$ we have that $x \in X$.
Therefore $\bigcup X \subset X$.

Not bad. Sorry, I'll be using this post as evidence that I don't have to explain trivial things to you. :p

$4 \to 1$: Let $x \in \bigcup X $. Since $\bigcup X \subset X$ we have that $x \in X$.
But $x \in \bigcup X$ means that $\exists y \in X$ such that $x \in y$.
How could we continue?

You need to start with what you need to prove. $X$ is transitive if $x\in X$ and $y\in x$ imply that $y\in X$. So assume that $x\in X$ and $y\in x$. Then $y\in\bigcup X$ by the definition of generalized union, and since by assumption $\bigcup X\subseteq X$, we have $y\in X$.

 

[evinda]

Not bad. Sorry, I'll be using this post as evidence that I don't have to explain trivial things to you. :p

(Giggle)

You need to start with what you need to prove. $X$ is transitive if $x\in X$ and $y\in x$ imply that $y\in X$. So assume that $x\in X$ and $y\in x$. Then $y\in\bigcup X$ by the definition of generalized union, and since by assumption $\bigcup X\subseteq X$, we have $y\in X$.

I understand... Thanks a lot! (Happy)