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I have a problem that I don't under because I don't know what the answer is because there are two lots of brackets in D B n D when B = {6,7,8,9} when D = {{6,7},6,4}

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marked as duplicate by Asaf Karagila, Dan Rust, Lord_Farin, SHOBHIT GAUTAM, Cameron Buie Nov 9 '13 at 16:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

$B\cap D=\{6\}$

Because in B are the elements 6,7,8 and 9. To D belongs the elements $\{6,7\}$,6 and 4.

Note that in the set D is the additional set $\{6,7\}$ as an element. Do not misunderstand that with the set $D'=\{6,7,6,4\}=\{6,7,4\}$

The intersect of two sets contains all elements which are in both sets. And this is only the 6.

Does that make sense to you?

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yes i think so, so in set D you ignore the first 6 and the 7 – user106594 Nov 9 '13 at 15:22
Ignore is maybe a wrong word for it. It is simply that the set \{6,7\} is an element of D. You have to view this as one object. You could write $D=\{A,6,4\}$ where $A=\{6,7\}$. And the element A is not in B. – user105916 Nov 9 '13 at 15:26

it means that the element of a set is again a set, hence in your case the elements of $D$ are numbers $6, 4$ and a set consisting of $6$ and $7$hence $\{6,7\}$

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whats the answer then – user106594 Nov 9 '13 at 15:10
well, what are the common elements of $B$ and $D$ based on what I said? – mm-aops Nov 9 '13 at 15:12
the common elements are 6,7 – user106594 Nov 9 '13 at 15:14
$7$ is not an element of $D$, a set $\{6,7\}$ is an element of $D$ but that's a totaly different object – mm-aops Nov 9 '13 at 15:16
so the answer is 6 then – user106594 Nov 9 '13 at 15:17

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