MHB We cannot define an order over all the sets

[evinda]

Hi! (Mmm)

This:
$$\subseteq_{\mathcal{P}A}=\{ <X,Y> \in (\mathcal{P}A)^2: X \subset Y\}$$
is a partial order of the power sets $\mathcal{P}A$.

But, we have to take attention to the following fact:

We cannot define an order over all the sets because if $R=\{ <X,Y>: X \subset Y \}$ is a set, then if $X=\varnothing$ we have that for each set $Y$, $\varnothing \subset Y$. Therefore $rng(R)=\{ Y: Y \text{ set } \}$ is a set.

Contradiction.

Could you explain me what is meant? I haven't understood it.. (Worried)

 

[Evgeny.Makarov]

You can define a partial order by inclusion on subsets of a given set, but not on all sets. This is because if $R$ is a set of pairs, then $\operatorname{rng}(R)$ is also a set. But for $R=\{\langle X,Y\rangle\mid X,Y\text{ are sets}, X\subseteq Y\}$, $\operatorname{rng}(R)$ includes all sets; therefore, $\operatorname{rng}(R)$ is not a set (recall that the collection of all sets is not a set), which means that $R$ is not a set.

 

[evinda]

You can define a partial order by inclusion on subsets of a given set, but not on all sets. This is because if $R$ is a set of pairs, then $\operatorname{rng}(R)$ is also a set. But for $R=\{\langle X,Y\rangle\mid X,Y\text{ are sets}, X\subseteq Y\}$, $\operatorname{rng}(R)$ includes all sets; therefore, $\operatorname{rng}(R)$ is not a set (recall that the collection of all sets is not a set), which means that $R$ is not a set.

The range is this:

$$\operatorname{rng}(R)=\{ Y: \exists X (XRY) \}=\{ Y: \exists X (X \subseteq Y)\}$$

right? If so, could you explain me why $\operatorname{rng}(R)$ includes all the sets? (Thinking)

 

[Evgeny.Makarov]

This is explained in post #1.

 

[evinda]

This is explained in post #1.

Oh, yes.. you are right! Is $X=\varnothing$ the only set for which we can say for sure that $rng(R)$ and therefore $R$ is not a set? (Thinking)

 

[Evgeny.Makarov]

Is $X=\varnothing$ the only set for which we can say for sure that $rng(R)$ and therefore $R$ is not a set?

This reminds me a joke. A visitor to Odessa, a city in Ukraine famous for its peculiar humor, asks a local: "Excuse me, if I go in this direction, will the railway station be there?" "Young man", says the local, "The station will be there even you don't go in that direction".

$\operatorname{rng}(R)$ is either a set or it isn't. This fact does not depend on the choice of $X$, though some choices make it possible to prove it.

 

[evinda]

This reminds me a joke. A visitor to Odessa, a city in Ukraine famous for its peculiar humor, asks a local: "Excuse me, if I go in this direction, will the railway station be there?" "Young man", says the local, "The station will be there even you don't go in that direction".

$\operatorname{rng}(R)$ is either a set or it isn't. This fact does not depend on the choice of $X$, though some choices make it possible to prove it.

(Tmi) (Blush) Yes, you are right.. $\operatorname{rng}(R)$ is always a set.
By picking $X=\varnothing$, we found that $\operatorname{rng}(R)$ is not a set, so we found a contradiction...
So, is $\varnothing$ the only choice for $X$ in order to find a contradiction? (Thinking)

 

[Evgeny.Makarov]

$\operatorname{rng}(R)$ is always a set.

Not always, but if $R$ is a set, which is not the case for $R=\{\langle X,Y\rangle\mid X,Y\text{ are sets}, X\subseteq Y\}$.

So, is $\varnothing$ the only choice for $X$ in order to find a contradiction?

For an arbitrary $X$ it is the case that for all $Y$, $X\subseteq Y$ implies $Y\in\operatorname{rng}(R)$. For any given $X$, the collection
\[
\{Y\mid Y\text{ is a set},X\subseteq Y\}\text{ is not a set.}\qquad(*)
\]
So any $X$ works for proving that $\operatorname{rng}(R)$ is not a set. But according to your threads, (*) has been shown in your course for $X=\emptyset$.

 

[evinda]

Not always, but if $R$ is a set, which is not the case for $R=\{\langle X,Y\rangle\mid X,Y\text{ are sets}, X\subseteq Y\}$.

For an arbitrary $X$ it is the case that for all $Y$, $X\subseteq Y$ implies $Y\in\operatorname{rng}(R)$. For any given $X$, the collection
\[
\{Y\mid Y\text{ is a set},X\subseteq Y\}\text{ is not a set.}\qquad(*)
\]
So any $X$ works for proving that $\operatorname{rng}(R)$ is not a set. But according to your threads, (*) has been shown in your course for $X=\emptyset$.

When we pick $X=\varnothing$ it holds because it is known that $\varnothing \subset A$ for any set $A$.
But could you explain me further why we could pick any $X$ ? (Thinking)

 

[evinda]

For example if we pick $X=\{1,2,3\}$ how could we justify that $\{ Y| Y \text{ is a set, } \{1,2,3\} \subseteq Y \}$ is not a set? (Thinking)

 

[Evgeny.Makarov]

Let $\Bbb U_X=\{Y\mid Y\text{ is a set and }X\subseteq Y\}$. Then any set $Z$ can be viewed as $Y\setminus X'$ for some $Y\in\Bbb U_X$ and some $X'\subseteq X$. But if $\Bbb U_X$ is a set, then so is $\{Y\setminus X'\mid Y\in \Bbb U_X,X'\subseteq X\}$.