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I am a little consufed in basic set theory

Let $A = \{\{x\},y,z\}$, $B= \{z,\{y\},x\}$.

Are they equal?

I am clear about that order does not matter, but can we prove they are equal or not?

I am little confused in $\{x\}$ and $x$, how to compare them?

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Well, $\{x\}\in A$. Is $\{x\}\in B$? Do you think $\{x\}=x$? – Noomi Holloway Dec 28 '12 at 10:34
All these words are written here below just for two braces in each sides. How wonderful mathematics is. :) – Babak S. Dec 28 '12 at 11:25

The question is a bit subtle. Without knowing, what $x,y,z$ are, we have to take into account the possibility that e.g. $z=\{x\}$, in which case $\{x\}$ is an element of both $A$ and $B$, in spite of the difference between $\{x\}$ occuring in $A$ and $x\ne\{x\}$ ocurring in $B$. In fact, we can have $A=B$, for example if $x=y$. In concreto, letting $x=y=z=\emptyset$, we find that both $A$ and $B$ are the two-element set $\{\{\emptyset\},\emptyset\}$. But in general (i.e. for "typical" $x,y,z$) we have $A\ne B$. However, a proof of $A\ne B$ would require to exhibit an object that is element of one set and not of the other and this may not be easy without additional assumptions about $x,y,z$. (And as my above remark shows, $x\ne y$ must be among these assumptions). Indeed, if $A=B$ then $x\in A$ (because $x\in B$), hence $x=\{x\}$ or $x=y$ or $x=z$. The first case is impossible (with most set theories); the second case corresponds to my remark above; thus let us assume the third case $x=z$. Similarly, we conclude $y\in B$ and from that $y=\{y\}$ or $y=x$ or $y=z$, from which only the case $y=z$ remains interesting. But if $x=z$ and $y=z$ then we are again arriving at $x=y$. Therefore,

$$ \{\{x\},y,z\}=\{x,\{y\},z\}\iff x=y$$

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The sets are not equal: $x\ne\{x\}$, and $y\ne\{y\}$. You might think of $x$ as some object and $\{x\}$ as a box with that object inside it: not the same thing.

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Some confusion may arise because of the notation. $\{x\}$ doesn't actually name a set. It may help to label it $S=\{x\}$. Then we could write $\forall a (a\in S\leftrightarrow a=x)$. (In words: For all a, a is an element of set S if and only if $a = x$.) It would then be more obvious that $S$ and $x$ are not the same things.

Alternatively, we could look at it as $\{x\}=\{a|a=x\}.$

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Remember $$A=B\iff (\forall x)\ (x\in A\iff x\in B)$$ Clearly $\left\{x\right\}\in A$. But is it $\left\{x\right\}\in B$? No, as $\left\{x\right\}\neq x$.

In fact, by the Axiom of Foundation, $$(\exists y\in \left\{x\right\})(\forall z\in \left\{x\right\})\ z\notin y$$ which can be translated to $$(\forall z\in \left\{x\right\})\ z\notin x$$ or even $$x\notin x$$ This implies $\left\{x\right\}\neq x$ (why?)

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These sets are nor equal. Actually, $x$ and $\{x\}$ are different things: $x$ is an element, while $\{x\}$ is a set that have only one element (it is called a singleton). Thus, your two sets $A$ and $B$ are completely different.

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