#### Figaro

Sets which doesn't contain themselves are called normal sets while sets that contain themselves are called abnormal. Let $N$ be a set of all normal sets. Prove that $N$ is normal if and only if $N$ is abnormal.
Proof. $~~\rightarrow ~~$ Suppose $N$ is normal such that $N \not\in N$ but since $N$ contains all normal sets, it should include itself since it is a normal set by assumption. Thus $N \in N$
$\leftarrow~~$ The argument is the same as before. QED

Can anyone help me check my proof? Also, is it correct to think that the crucial ingredient/key phrase here is "N contains all the normal sets"?

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#### PeroK

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You haven't quite proved precisely what was asked. You have correct logic, but you haven't clearly shown the result.

And you do need to prove the converse properly.

#### andrewkirk

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$\leftarrow~~$ The argument is the same as before. QED
It's not the same. The argument runs in reverse, for the reverse direction.

The argument can be made the same for both directions by connecting the propositions "$N$ is normal" and "$N$ is abnormal" by a chain of one or more intermediate propositions, all of which are connected by 'if and only if' relations ($\leftrightarrow$ symbol). Your argument is not currently expressed that way but, with a little thought, could be adapted to be like that.

#### Figaro

$\leftarrow$ Suppose $N$ is abnormal such that $N \in N$, but since $N$ contains all normal sets it is the case that $N \not\in N$ since $N$ is abnormal by assumption.

Is this the correct argument?

#### PeroK

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$\leftarrow$ Suppose $N$ is abnormal such that $N \in N$, but since $N$ contains all normal sets it is the case that $N \not\in N$ since $N$ is abnormal by assumption.

Is this the correct argument?
Because the point of logic is to absolutely precise, let me analyse what you have done from a very rigorous perspective:

1) "Suppose $N$ is abnormal such that $N \in N$"

Usually "such that" will introduce some additional property or constraint, but isn't used to restate the definition or draw a conclusion. Better is:

"Suppose $N$ is abnormal, then $N \in N$"

2) "but since $N$ contains all normal sets"

It's not that $N$ contains all the normal sets, it's that $N$ contains only normal sets that is key.

3) "it is the case that $N \not\in N$ since $N$ is abnormal by assumption."

This loses the thread somewhat and you appear to be aiming for a contradiction, but you never quite finish the argument. Note that what you were to show, precisely, is that:

$N$ abnormal $\Rightarrow \ N$ normal

Note that there is only a contradiction if you think being normal and abnormal are incompatible, which is technically a different question!

Does this help?

#### Figaro

Because the point of logic is to absolutely precise, let me analyse what you have done from a very rigorous perspective:

1) "Suppose $N$ is abnormal such that $N \in N$"

Usually "such that" will introduce some additional property or constraint, but isn't used to restate the definition or draw a conclusion. Better is:

"Suppose $N$ is abnormal, then $N \in N$"

2) "but since $N$ contains all normal sets"

It's not that $N$ contains all the normal sets, it's that $N$ contains only normal sets that is key.

3) "it is the case that $N \not\in N$ since $N$ is abnormal by assumption."

This loses the thread somewhat and you appear to be aiming for a contradiction, but you never quite finish the argument. Note that what you were to show, precisely, is that:

$N$ abnormal $\Rightarrow \ N$ normal

Note that there is only a contradiction if you think being normal and abnormal are incompatible, which is technically a different question!

Does this help?
So everything is ok and the only problem is my usage of words in 1) and 2) and that in 3) I should specify clearly that since I concluded that $N \not\in N$, that N is normal?

#### PeroK

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So everything is ok and the only problem is my usage of words in 1) and 2) and that in 3) I should specify clearly that since I concluded that $N \not\in N$, that N is normal?
Why do you need to say $N \not\in N$?

$N \in N$ (by hypothesis), $N$ contains only normal sets, therefore $N$ is normal. Isn't that simpler?

#### FactChecker

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$\leftarrow$ Suppose $N$ is abnormal such that $N \in N$, but since $N$ contains all normal sets it is the case that $N \not\in N$ since $N$ is abnormal by assumption.

Is this the correct argument?
That is correct. Just a little word-smithing: but since $N$ contains only normal sets

• Figaro

#### Figaro

Why do you need to say $N \not\in N$?

$N \in N$ (by hypothesis), $N$ contains only normal sets, therefore $N$ is normal. Isn't that simpler?
Yes, I should do that. Thanks!

#### sysprog

The self-referentiality of the designation "set of all sets not members of themselves" co-mingles first-order and second-order predication -- the resultant paradox can be resolved by modification of the designation to "the set, other than and not including itself, of all sets not members of themselves".

#### Bashyboy

Sets which doesn't contain themselves are called normal sets while sets that contain themselves are called abnormal. Let $N$ be a set of all normal sets. Prove that $N$ is normal if and only if $N$ is abnormal.
Proof. $~~\rightarrow ~~$ Suppose $N$ is normal such that $N \not\in N$ but since $N$ contains all normal sets, it should include itself since it is a normal set by assumption. Thus $N \in N$
$\leftarrow~~$ The argument is the same as before. QED

Can anyone help me check my proof? Also, is it correct to think that the crucial ingredient/key phrase here is "N contains all the normal sets"?
Where did you get this exercise? I am interested in a book that covers this sort of material.