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I wonder if we can we say $x\in \{\{\{x\}\}\}$?

In one viewpoint the only element of $\{\{\{x\}\}\}$ is $\{\{x\}\}$. In the other viewpoint $x$ is in $\{\{\{x\}\}\}$, for example all people in Madrid are in Spain.

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  • $\begingroup$ You either have too many set braces, or neither of the viewpoints are correct (the second one is, as already pointed out, never correct). $\endgroup$ – Tobias Kildetoft Dec 12 '14 at 9:20
  • $\begingroup$ If we allow this, then you are allowing a set to be a member of itself. There is a famous paradox here. Let S be the set of all sets that are NOT elements of themselves, now we call a set good if it is a member of itself, and bad if it is not. Is S Good? Well then S is in S and this is not allowed by definition of S YIKES!, is S bad? then S is not in itself, and therefor must be in S by definition YIKES AGAIN! YIKES YIKES YIKES! $\endgroup$ – H_B Dec 12 '14 at 9:49
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    $\begingroup$ @H_B The part that causes that paradox is not so much the sets being elements of themselves but having the set of all sets to start with. $\endgroup$ – Tobias Kildetoft Dec 12 '14 at 10:06
  • $\begingroup$ ohh, Sorry I am new to set theory myself, and this is how it was explained to me. Do you think I should delete it? $\endgroup$ – H_B Dec 12 '14 at 10:11
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    $\begingroup$ Not everyone in Madrid is a city in Spain though. $\endgroup$ – T.J. Gaffney Dec 12 '14 at 10:20
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If $x\in\{\{\{x\}\}\}$, then $x=\{\{x\}\}$, and we have a cycle $x\in\{x\}\in x$. In Zermelo-Fraenkel set theory, this behavior is forbidden by the foundation axiom. So there does not exist any set $x$ such that $x\in\{\{\{x\}\}\}$, no matter how complicated $x$ may be!

(One can study less-popular set theories without the foundation axiom, but I don't know anything about those.)

In your example, you can use the subset relation to model the relationship between Madrid and Spain, instead of the set membership relation. The set of people in Madrid is a subset of the set of people in Spain. If we identify a place with the set of people in that place, then we can simply say that Madrid is a subset of Spain. In that case, we wouldn't say that Madrid is also an element of Spain, because Madrid isn't a person.

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$$x\in\{\{\{x\}\}\}\iff x = \{\{x\}\}$$ The latter is obviously not true for many sets.

Your example of "all people in Madrid are in Spain" is not good here, because, if $\{x\}$ is madrid, then spain would be some set which includes both $\{x\}$ and some other elements

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As many here explained, this certainly don't have to be true, and assuming the axiom of regularity this can be disproved quite easily.

I will remark that it is consistent that the axiom of regularity fails, and that this situation does happen. Of course it would mean that $x=\{\{x\}\}$, but it is a plausible scenario (which, for example, follows from certain "anti-foundation" axioms).

But perhaps one thing missing from the answers here is the mention of a transitive closure. Given a set $X$ we define the transitive closure of $X$ as the set of all elements which can be "reached" by taking elements of elements and so on. The transitive closure of $X=\{\{\varnothing\}\}$ would be to take $X\cup\{\varnothing\}\cup\varnothing=\{\varnothing,\{\varnothing\}\}$, for example.

So while $x$ need not be an element of $\{\{\{x\}\}\}$, it is an element of the transitive closure of $\{\{\{x\}\}\}$.

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While there are some variants of set theory that allow, e.g., that a set is an element of itself ("Quine atoms") or such deeper level weirdnesses as in your question, the usual axiomatization of set theory (Zermelo-Frenkel with or without the Axiom of Choice) contains the Axiom of Foundation (or Regularity)

Axiom. For every nonempty set $a$, there exists $b\in a$ with $b\cap a=\emptyset$.

In your example, $a=\{\{\{x\}\}\}$, the only possibility for $b$ is $b=\{\{x\}\}$, and then $a\cap b=\emptyset$ which merely gives us that the only element of $a$ is not in $a$, i.e., $\{x\}\notin \{\{\{x\}\}\}$. In principle, this would still allow $x\in a$ - after all, $x\ne\{x\}$.

However, we can use a better $a$. The only way for $x\in\{\{\{x\}\}\}$ is that $x=\{\{x\}\}$ holds. Assume that this is the case. Now let $$a=\{\,x,\{x\}\,\} = \{\,\{\{x\}\},\{x\}\,\}.$$ By the Axiom of Foundation there exists $b\in a$ with $b\cap a=\emptyset$. Then either $b=\{x\}$ or $b=x=\{\{x\}\}$. But in the first case $x\in b\cap a$ and in the second case $\{x\}\in b\cap a$, contradiction. Hence the assumption that $x=\{\{x\}\}$ (or equivalently $x\in\{\{\{x\}\}\}$) could hold for some $x$ must be wrong.

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If Mariano Rajoy is in Madrid, and Mariano Rajoy is in Spain and Madrid is in Spain, then expressing those facts as memberships of sets would look like

Madrid = { Rajoy, ... }

Spain = { Madrid, Rajoy, ... }

so this doesn't resemble your example at all.

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there are a lot of good answers, but I want to add:

In the set $A=\{Chris,Culter,5,Adam,\{\{x\}\},525\}$, the elements are, by definition, all things between $\{$ and $\}$. so they are exactly:
Chris , Culter, 5, Adam, {{x}} , 525
so in $B=$$\{${{x}}$\}$ the only element between $\{$ and $\}$ is...?

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The answer to your question is no. Note that the set $ \{ \{ \{ x \} \} \} $ has one element, which is $ \{ \{ x \} \} $, not $x$. To say $ x \in \{ \{ \{ x \} \} \} $ means that $x$ is an element of $ \{ \{ \{ x \} \} \} $, implying that $ x = \{ \{ x \} \} $, which is not true if $x$ is a number.

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  • $\begingroup$ Perhaps $x = \{\{x\}\}$. Without the axiom of regularity, you can't rule this out. $\endgroup$ – MJD Dec 12 '14 at 15:16
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No, if it were true it would break many things, for example the set theoretic construction of the natural numbers:

$$ 0 = \emptyset \\ 1 = \{\emptyset\}\\ 2 = \{\{\emptyset\}\}\\ \vdots$$

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    $\begingroup$ I thought $2 = \{\emptyset, \{\emptyset\}\}$ $\endgroup$ – H_B Dec 12 '14 at 9:45
  • $\begingroup$ The definition $2=\{\{\emptyset\}\}$ doesn't seem widely used, for the simple reason that this set has just one element, while it's preferable that $2$ has two elements. $\endgroup$ – egreg Dec 12 '14 at 10:43
  • $\begingroup$ @H_B You can define it that way too, but the way I explained is related to the question. $\endgroup$ – Latyov Dec 12 '14 at 12:31
  • $\begingroup$ This isn't really correct; Zermelo's construction of the natural numbers works just fine in non-well-founded set theory, as does von Neumann's. $\endgroup$ – goblin Dec 12 '14 at 12:33
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    $\begingroup$ I'm not claiming that you are claiming that Zermelo's (i.e. your) construction is the only way of getting the natural numbers in ZFC. My point really is really that the answer to "is $x∈\{\{\{x\}\}\}$" is only a "resounding no" in ZFC. In ZFC minus Regularity, the answer to "is $x∈\{\{\{x\}\}\}$" may be "yes" for some $x$. In ZFC minus regularity plus an anti-foundation axiom, the answer to "is $x∈\{\{\{x\}\}\}$" will necessarily be "yes" for certain sets $x$ (called non-well-founded sets.). Furthermore, Zermelo's (i.e. your) construction of the natural numbers works fine without regularity. $\endgroup$ – goblin Dec 12 '14 at 12:43

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