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I'm slightly confused as to how

$$\{\emptyset,\{\emptyset,\emptyset\}\} = \{\{\emptyset\},\emptyset,\{\emptyset\}\}$$

are equivalent. I thought two sets were equivalent if and only if "$A$" and "$B$" have exactly the same elements. In this case, we have one element which is in both sets but then two elements aren't in other! Can someone please explain where I am going wrong in my definition?

Thanks a bunch!

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    $\begingroup$ There is a trick here. For sets, $\{A, A, B\} = \{A, B\} = \{A, B, B\}$, because each set has exactly the same elements, namely $A$ and $B$. It might seem that $\{A, A, B\}\not\subset\{A, B\}$, but in fact the opposite is true: $\{A, A, B\}\subset\{A, B\}$ because every element of $\{A, A, B\}$ is also an element of $\{A, B\}$. $\endgroup$ – MJD Mar 18 '14 at 18:27
  • $\begingroup$ Using the rules given by MJD, prove that both sets are equal to $\{\emptyset , \{\emptyset\}\}.$ $\endgroup$ – Dan Christensen Mar 19 '14 at 21:45
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Two sets are equal if every element of one set is also an element of the other and vice-versa.

So you must convince yourself that $\{\emptyset,\emptyset\}=\{\emptyset\}$ and the exercise becomes very simple...

By the way it happens that natural numbers are usually defined by $0=\emptyset$, $n+1=\{n\} \cup n$ so that the set in your question is actually the number $2$.

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When you have a set like $\{a,a\}$ then its same as $\{a\}$ so LHS is precisely $\{\phi, \{\phi\}\}$. Similarly, RHS becomes $\{\phi, \{\phi\}\}$ as we are identifying the inner $\{\phi\}$.

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  • $\begingroup$ Please note phi and empty set are not the same symbol. $\endgroup$ – CiaPan Dec 19 '16 at 15:17
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A set is a collection of distinct objects. Every element of a set must be unique; no two members may be identical.

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may be it is better to use that $\{ \emptyset, \emptyset\} = \{\emptyset\} \cup \{\emptyset\}$

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