What is $x\notin x$ supposed to mean?

Question

In Russell's Paradox the expression $x\notin x$ is used to define a set

$$R=\{x\mid x\notin x\}$$

which supposedly leads to a contradiction

$$R\in R\Leftrightarrow R\notin R$$

However, what is $x\notin x$ even supposed to mean? Since any nontrivial element is at least an element of itself, the set $R$ defined as above can be nothing but the empty set $R=\{\}$. But the empty set does not represent an object that could be compared to itself, so the subsequent comparison $R\in R\Leftrightarrow R\notin R$ is meaningless.

To me it seems that the derived contradiction is a petty and meaningless pathology, so I fail to see why Russell's Paradox was enough to break "naive set theory".

Did I misunderstand something, or is there some important subtlety I missed? What do you think?

EDIT:

After some discussion in the comments, it seems that I should be thinking of $x\notin x$ as a set statement:

$x\notin x$ being all sets $x$ that are not infinitely nested sets and in particular do not contain themselves.

But from this perspective, $R=\{x|x\notin x\}$ is a huge set of pretty much every thinkable object, excluding infintely nested self repeating sets.

That makes me wonder how $R\in R$ could then possibly follow, since for a set to be contained as an element in itself, it should be infinitely nested, no?

EDIT2:

Well, thinking about it a bit more, the definition

$$R=\{x\mid x\notin x\}$$

states that $R\in R$ is not true, since otherwise we could find at least one $x\in R$ (in particular $x=R$) such that $x\notin x$ would be wrong. That would create a different set $R'$ which would be defined differently.

Seems to me, the dilemma whether one should consider the set of "all sets that do not contain themselves" as element of itself is super pathological, since adding it would invalidate its definition. But the definition of the set may not be broken while its properties are being investigated. This suggests an unambiguous resolution: $R\notin R$ per definition and for any $P\in P$ per definition $R\neq P$. $R$ is missing one element -- itself, addition of which would break the definition which is why it must be omitted.

Instead of throwing out a whole set theory, I would have concluded from this that sets sometimes do not contain an exhausting number of elements but instead a maximum number of elements consistent with its definition.

Answer 1

You define $\in$ simply as a relation between sets, which are some objects that you're interested in. Just like $<$ is a relation between numbers, e.g. $2<3$, we have a relation $\in$ for which we can say $x\in y$ where $x$ and $y$ are sets.

If we take two things that are not numbers, then $<$ does not relate between them. For example $\square < \triangle$ makes no sense. Similarly, if we take two things that are not sets, e.g. $\square\in\triangle$, then the relation makes no sense.

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What Russell's paradox shows, is that if we can define a set $R$ containing all sets $x$ such that $x\notin x$, then we get a contradiction: if $R\in R$, then $R\notin R$, because the latter is the requirement all elements of $R$ have to satisfy. On the other hand, if $R\notin R$, then $R$ satisfies its own requirement, so we should have $R\in R$.

This is a contradiction. We cannot have both a statement ($R\in R$) and its negation ($R\notin R$) implying each other. Therefore, if set theory is consistent, it must be impossible to define a set such as $R$.

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What does hold true, is that $R$ is a collection of sets, also called a class. Any set is a class, but not every class is a set. A class that is not a set is called a proper class.

In case you assume the Axiom of Regularity, $R$ contains every imaginable set, since $x\notin x$ will be true for any set. We do not get the contradiction that $R\notin R$, since $R$ is a proper class, and thus the relation $\in$ is not defined on $R$ (remember that $\in$ is only defined on sets).

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Perhaps this does not address your confusion, but just in case, there is a difference between $\in$ (being an element of) and $\subseteq$ (being a subset of). We say that $x\in y$ if $y$ is a set that contains the set $x$ as a member, and we say $x\subseteq y$ if $y$ is a set that contains all the sets that are contained in $x$.

For example $\varnothing$ is a set that contains no elements: for any $x$ we have $x\notin \varnothing$. On the other hand, there is some $x$ such that $x\subseteq \varnothing$, namely $\varnothing\subseteq\varnothing$ (since both sets contain the same elements, which is none). Similarly $1\in\{1,2\}$ and $\{1\}\subseteq\{1,2\}$, while $1\not\subseteq\{1,2\}$ and $\{1\}\notin\{1,2\}$. N.B.: this depends on how the sets $1$ and $2$ are defined. Here I assume they are some urelements that do not break down into smaller sets.

So make sure you're not confused between these notions.

Answer 2

Since any nontrivial element is at least an element of itself

No, it's not.

$\text{Babar the elephant} \ne \{\text{Babar the elephant}\}$ and $\text{Babar the elephant} \not \in \text{Babar the elephant}$ and $\{\text{Babar the elephant}\}\not \in \{\text{Babar the elephant}\}$.

$\text{Babar the elephant}$ is ... an elephant. It is not a set. It does not have any memebers. And it is not a member of itself because it is not a set.

$\{\text{Babar the elephant}\}$ is a set that contains an elephant as a member. It is not an elephant. $\text{Babar the elephant}\in \{\text{Babar the elephant}\}$ because $\text{Babar the elephant}$ is in the set $\{\text{Babar the elephant}\}$ but the set itself.... is not in the set.

$\{\text{Babar the elephant}\} \not \in \{\text{Babar the elephant}\}$ because $\{\text{Babar the elephant}\}$ is not $\text{Babar the elephant}$ and $\text{Babar the elephant}$ is the only thing in $\{\text{Babar the elephant}\}$.

....

Every nontrivial element is an element of the set that contains it. But the set that contains it is a different thing than itself.

$\{$ all even numbers$\}$ is not an even number so $\{$ all even numbers $\}\not \in \{$ all even numbers}$.

And, if we don't use the safe guards of ZFC (or something similar), we can have:

$x = \{$ all set's that have seven items$\}$.

So $\{$snow white's dwarfs $\} \in x$. And $\{$the deadly sins$\}\in x$.

How many sets have seven elements? Well, more than seven... we have snow white's dwarfs, and the deadly sins, and the classical virtues, the days of the week, "the words in this phrase right here", the magnificent cowboys, the samarai in the Kurosawa film, and hills of Rome, are eight of them and there are more.

So how many elements does $x$ have. Well, not seven. We just counted eight and we can clearly come up with many many more.

So $x$ doesn't have seven items. so $x \not \in x$.

.....

And without the safeguards of $ZFC$ we can have $x = \{$ all sets$\}$ and $x \in x$ because $x$ is a set. And $x=\{$ all sets with more than seven elements$\}$ and $x \in x$.

And so on.

But what if $x=\{$ sets that do not have themselves as members $\}$.

If $x$ has itself as a member than $x$ is a set that doesn't have itself as a member.

If $x$ doesn't have itself as a member then $x \in x$ and it does have itself as a member.

So a paradox. Apparently.

.....

So either 1) not everything definable (such as the set of all sets that don't have itself as a member) is a set, or

2) It's not always the case that an element actually is in or out of a set.

3) We can have instance of elements that are not exclusively one condition or not.

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1) Breaks naive set theory if everything describable can't be a set. (But a more sophisticated system where not everything "describable" is "collectable" will work.)

2) Breaks all set theory, naive or not.

3) Breaks all mathematics.