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Are $A = \{\{a\},\{b,c\},\{d\},\{e\}\}$ and $B = \{a,b,c,d,e\} $the same set? Justify your answer.

I think I get that set $A$ has elements in the form of sets: $\{a\}, \{b,c\}, \{d\}$ and $\{e\}$. Is that the same as the elements in $B$?

Could someone explain if these two sets are the same or not?

Thank you!

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    $\begingroup$ $a \neq \{a\}$. The first set has $\{a\}$ as an element. The second set does not. The "depth" here matters. $\endgroup$
    – JMoravitz
    Sep 7, 2019 at 12:25
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    $\begingroup$ No. $A$ has 4 elements, but $B$ has 5. $\endgroup$
    – Joe
    Sep 7, 2019 at 12:25
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    $\begingroup$ @Joe Provided that $a,b,c,d,e$ are distinct. $\endgroup$
    – drhab
    Sep 7, 2019 at 12:26
  • $\begingroup$ I suppose that to be pedantic, just saying that $\{a\}$ is an element of the first while $a$ is an element of the second isn't good enough., since what if $b=\{a\}$... Joe's proof is the shortest and cleanest $\endgroup$
    – JMoravitz
    Sep 7, 2019 at 12:27
  • $\begingroup$ @drhab and JMoravitz, both great points. I was actually answering (whether or not this what was meant in the question), “Are the sets necessarily equal, given how they are defined?”, as opposed to “Can these sets be equal?”, which I think would require axioms from ZF to prove. So technically I should have said “$A$ can only have $\leq 4$ elements as defined, whereas $B$ can have 5.” $\endgroup$
    – Joe
    Sep 7, 2019 at 12:32

3 Answers 3

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Your question is in general shape ("are $A$ and $B$ the same set?") and the answer is: "no".

To justify this it is enough to show that $5$ distinct sets exists and to label them as $a,b,c,d,e$.

Then - as commented by @Joe - the set $A=\{\{a\},\{b,c\},\{d\},\{e\}\}$ has exactly $4$ elements and the set $B=\{a,b,c,d,e\}$ has exactly $5$ elements, so the sets are definitely not the same.

Another (specified) question would be: do sets $a,b,c,d,e$ exist such that $\{\{a\},\{b,c\},\{d\},\{e\}\}=\{a,b,c,d,e\}$?

Then - if the axiom of regularity is accepted - again the answer is "no", but a proof of this is much less easy.

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Two sets $A,B$ are equal iff for all $x$, $x\in A$ iff $x\in B$.

In your case $A$ elements $\{a\},\{b,c\},\{d\},\{e\}$ and $B$ has elements $a,b,c,d,e$. But $a\ne \{a\}$ (but $a\in\{a\}$), the set $\{b,c\}$ has the elements $b,c$ and $d\ne\{d\}$, $e\ne\{e\}$. So they are far from being equal.

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    $\begingroup$ So an element a does not equal the set whose only element is a? $\endgroup$ Sep 7, 2019 at 12:29
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    $\begingroup$ @FrodoBaggins correct. $a\neq \{a\}$, just like $\emptyset \neq \{\emptyset\}$. This second example should be clearer since $\emptyset$ is a set with zero elements while $\{\emptyset\}$ is a set with one element, that element happening to be the emptyset. $\endgroup$
    – JMoravitz
    Sep 7, 2019 at 12:31
  • $\begingroup$ Right, but elements can be sets too. $\endgroup$
    – Wuestenfux
    Sep 7, 2019 at 12:31
  • $\begingroup$ Detail: $\{d,e\}$ is not an element of $A=\{\{a\},\{b,c\},\{d\},\{e\}\}$. You probably misread. $\endgroup$
    – drhab
    Sep 7, 2019 at 12:38
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    $\begingroup$ I think it's worth noting that $\bigcup A=B$, or in other words that $A$ is a partition of $B$. So they are related to each other in an interesting and important way. $\endgroup$
    – user694818
    Sep 7, 2019 at 14:10
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Two Reasons why it is not true.

1) set A = contains 5 elements and Set B contains 4. For example, if you have a Set A = {{x,y,g,p}} set A only has one element which is the set of x,y,g, and p. Not every element in the set.

2) If you compares the elements inside, a $\neq$ {a} . The set {a} also includes the empty set $\phi$. You will learn more on this when you study powers of sets.

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