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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$B\cap D=\{6\}$
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?
user105916
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yes i think so, so in set D you ignore the first 6 and the 7 – user106594 Nov 09 '13 at 15:22
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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 09 '13 at 15:26
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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 $\{6,7\}$
mm-aops
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well, what are the common elements of $B$ and $D$ based on what I said? – mm-aops Nov 09 '13 at 15:12
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$7$ 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 09 '13 at 15:16
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yup, the answer is in fact ${6}$ because $6$ is the only common element, hence the set of common elements is the set ${6}$ – mm-aops Nov 09 '13 at 15:21