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$W = \{a,b,c\}, Z = \{W,\emptyset\}$

What is $Z\cap W$?

I'm not sure how to work this out as my thoughts would be $$Z = \{W,\emptyset\} = \{\{a,b,c\},\emptyset\}$$ Which means $Z \cap W = \{a,b,c,\{a,b,c\},\emptyset\}$

Or should it just be $$Z = \{W,\emptyset\} = \{a,b,c,\emptyset\}$$ Which means $Z \cap W = \{a,b,c,\emptyset\}$

Or should it just be $$Z = \{W,\emptyset\}$$ Which means $Z\cap W = \{W,\emptyset,a,b,c\}$

Any thoughts would be appreciated!

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4 Answers 4

The elements of $W$ are: $a,b$ and $c$

The elements of $Z$ are: $W$ and $\emptyset$

Can you see any element both in $W$ and $Z$? If yes, those elements are the answer. If there is no such element, then the answer is $\emptyset$.

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The intersection of two sets is the collection (set) of elements that are common to both sets.

In this case, $Z$ has two elements: $W$ and $\emptyset$. Sets can be elements of other sets. It is not true, however, that $a$, $b$ and $c$ are elements of $Z$ even though they are elements of $W$. $W$ has three elements: $a$, $b$ and $c$. Is there anything in common between $Z$ and $W$?

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$a,b$ and $c$ cannot be $W$ since a set cannot be an element of itself (axiom of regularity) The question is if any of $a,b$ or $c$ are the empty set. If so this will be the intersection, if not the intersection will be empty.

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It seems to me that we're talking about naive set theory here. In particular, I don't think you can invoke axiom of regularity. On the other hand, you can probably assume that $a,b,c$ are meant to be distinct from $W$ and $\emptyset$. –  tomasz Jun 23 at 1:53
    
Well you know how much trouble naive set theory caused. But Ok, I think Ill just leave my answer as it is, I think its clear what I am trying to say. –  Rene Schipperus Jun 23 at 1:57
    
Well, as much as naive set theory may cause trouble, merely dropping regularity is not quite so troublesome. It's a matter of convenience, really. –  tomasz Jun 23 at 11:41

Intersection = available in both the sets. So,

Z intersection W = W $ \rightarrow $ {a,b,c}

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$W \not\in W$, so $W$ cannot be in $Z \cap W$. –  Strants Jun 23 at 4:29

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