# Does a set R inherit proerties of its own elements?

I stumbled upon Mr Russel's and Ernst Zermelo Paradox http://en.wikipedia.org/wiki/Russell%27s_paradox. According to wiki, Let R be the set of all sets that are not members of themselves. If R is not a member of itself, then its definition dictates that it must contain itself, and if it contains itself, then it contradicts its own definition as the set of all sets that are not members of themselves. This contradiction is Russell's paradox.

I understand that if a set R contains set - elements that are not members of them selfs, but because those sets are not members of them selfs, that means that R inherits the property of its contained elements the "none membership of them selfs"?

And then why it "must" contain it self. Why does that property of R has to be inherited by its child sets. I just can't get the way the mathematical mind works around this problem.

Thank you

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Well, the Russel class doesn't inherit the properties of anything; there's no such thing. –  Malice Vidrine Dec 16 '13 at 3:50
then why it contradicts its own definition? –  themhz Dec 16 '13 at 3:56
The class doesn't contradict the definition; the definition contradicts itself. –  Malice Vidrine Dec 16 '13 at 4:07
I am still reading the above, man mathematicians are more strict with language then my grammar teacher..and I am beginning to understand why –  themhz Dec 16 '13 at 4:15
@themhz, its more like: with practice, you learn the situations in which you have to be careful with language. This isn't computer programming, where a single stray character undermines everything. –  goblin Dec 16 '13 at 7:03

Sets do not normally inherit the properties of their members, but the Russell property (of not being an element of one's self) is a very special property in this regard.

The idea is that if $\phi$ is some property, say the property of being blue, then we should be able to form the set of all the things that have that property. This is written with the rather odd notation $$\{ x : \phi(x) \}$$ and we read this “The set of all $x$ such that $x$ has property $\phi$”. For example, $$\{x : \text{x is blue}\}$$ is the set of all $x$ such that $x$ is blue, or, more concisely, the set of all blue things.

The rule for deciding if something is in one of these sets is this: Say that $S$ is the set of all things that have property $\phi$. That is, $$S = \{ x : \phi(x)\}.$$ Then if someone asks us if a particular $y$ is in $S$, we should be able to say that it is in $S$ if, and only if $y$ has property $\phi$. For example, we should be able to say that the sky is a member of the set of all blue things because the sky is blue. And more generally, a particular $y$ is an element of the set of blue things if, and only if, $y$ is blue. We can write this as $$y\in\{x : \text{x is blue}\}\quad\text{ if and only if }\quad\text{y is blue}.$$

Note that we have said nothing at all about whether the set of all blue things is itself blue. It might be, or it might not be; or the question might be silly. We can ask about it: Let $B$ be the set of all blue things. Is $B$ blue? All we know right now is that

$$B\in B\text{ if and only if B is blue}.$$ This says that that set of blue things will be a member of itself if and only if it is itself blue. There is no problem with this. Maybe $B$ is blue, maybe it isn't.
Whether the set of blue things is a blue thing is not part of our theory.

So far I hope this is all very simple. Here's a summary:

The theory is that for every property $\phi$, there is a corresponding set, called the ‘extension’ of $\phi$, which consists of exactly those things that have property $\phi$, and no other things. And a thing should belong to the extension of $\phi$ if that thing has property $\phi$, and not otherwise.

However, this very simple-seeming theory is wrong, and that is what the Russell paradox is: it is a proof that the straightforward-seeming theory of the previous paragraph is actually incoherent.

The reason is that there are properties that have no extension. And the simplest example of such a property is the Russell property of not being an element of one's self. We say $x\notin x$ if $x$ is not an element of itself. Some things have this property; some don't. My mother is not a set, so certainly she has no elements, and therefore she is not an element of herself, and so she does have the Russell property. But the set of all the things I discussed in this answer is itself one of the things I discussed in this answer, so it is an element of itself, and it does not have the Russell property.

Let's try to form the extension of the Russell property. This is the set $R$ of all things that are not elements of themselves. That is, $$R = \{ x : x\notin x\}.$$

If $R$ is in fact the extension of the property $x\notin x$, then some thing $y$ will be an element of $R$ if it has this property, namely that $y\notin y$, and not otherwise. For example, my mother does have the Russell property, so she is an element of $R$. Recalling the formula from earlier when I was talking about the set $B$ of all blue things, we had, for any $y$ in the universe: $$y\in\{x : \text{x is blue}\}\quad\text{ if and only if }\quad\text{y is blue}$$

and in this case, instead of blueness, we are talking about the Russell property of not belonging to one's self, so we should have, for every $y$:

$$y\in\{x : x\notin x\}\quad\text{ if and only if }\quad y\notin y$$

But this can't hold for every $y$, because it doesn't hold for $R$ itself:

$$R\in\{x : x\notin x\}\quad\text{ if and only if }\quad R\notin R$$

and $R = \{x : x\notin x\}$, so this is saying that $$R\in R\quad\text{ if and only if }\quad R\notin R.$$

This is flat-out false. Our theory about properties and extensions proved something false, so the theory is wrong.

We started with the ideas that:

1. Every property $\phi$ has an extension, a set of all the things with that property.
2. Each thing $y$ is an element of the extension of $\phi$ if, and only if, it has the property $\phi$

and this turns out not to be the case: for at least one property $\phi$, in particular the Russell property of not being an element of one's self, there is no such extension.

There are several different ways to fix this up, but this answers the part you asked about: sets don't normally inherit properties from their members; for example the set $B$ of all blue things is not necessarily blue; it could be, or not; either way works. But the Russell property is special because of the way we want sets to behave: the set of all things with the Russell property must either have the Russell property or not, and either way gets us into trouble.

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I have not finished reading all of your answer yet and another question popped out of my head. You said something very important that my brain seems to ignore but my mind sets caution. "Note that we have said nothing at all about whether the set of all red things is itself red; it might be, or it might not be; or the question might be silly. Whether the set of red things is a red thing is not part of our theory. So far I hope this is all very simple. " The sentence 'the set of all red things is itself red' –  themhz Dec 16 '13 at 4:27
So basically the starting point is the very statement. If I knew that "the set of all red things was green, then that would imply that the very set does not belong in it self. But if it was for instance, red then that would imply that it is a member of it self since it is red? –  themhz Dec 16 '13 at 4:28
Thank you MJD for now. I will come back tomorrow and read your answer again. I need to rest and I just can't keep focused.. –  themhz Dec 16 '13 at 4:54
I tried to address your question in more detail about whether the set of all red things is red in more detail. I also changed "red" to "blue" so that the letter $B$ in "blue" would be different from the letter $R$ I was using for the Russell set. –  MJD Dec 16 '13 at 14:22

The notation is confusing, so try to make the "levels" more explicit. Let $R$ and $R'$ both be the set whose members do not contain themselves, but we think of $R$ as the "outer" set and $R'$ as the "inner set" (that is, either $R'\in R$ or $R'\notin R$).

If $R'\in R$ then $R'$ does not contain itself. If $R'$ were any other set, there are no problems, because we would have no information about $R'$; maybe it is consistent that it does not contain itself. But we are in the special case when $R=R'$, so means that $R$ does not contain itself. In this sense, yes, $R$ has "inherited the properties of one of its members". But this is only because $R=R'$; in general no such thing is true, it's just because it is one of its members that it inherits the property.

To finish the proof: Again, if $R'$ were any other set, then this would be fine, because we would have no reason to believe that $R\in R$. But because $R'=R$, the statement $R'\in R$ is precisely the statement that $R\in R$.

Does this help?

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The origin of the paradox is clearer if you write the definition of the Russel class in a more basic form. The purported class would be a $z$ such that $\forall x(x\in z\leftrightarrow x\notin x)$. It's easy to see that an immediate instance of this is $z\in z\leftrightarrow z\notin z$.

In general, sets don't inherit anything from their members. A counterexample is $\{\mathbb{N}\}$, all members of which are infinite, but which is finite itself.

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