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I got the following sets:

A= { { 1 , 3 , 5 , 7 , 9 } , { 1 , 2 , 3 } , 1 , 2 , 3 , 4 , 5 }

B= { 1 , 3 , 5 , 7 , 9 }

I'm a bit confused. What would A ∩ B be?

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$A$ is an unusual set in that some of its elements are numbers and others are sets of numbers. This kind of sets are rarely used in mathematics -- except sometimes in exercises that confuse beginning students ...

$\{1,3,5,7,9\}$ is one of the elements of $A$. This set is not an element of $B$ -- it happens to be $B$ itself, but that doesn't make it an element of $B$ -- and therefore it is not an element of $A\cap B$.

$\{1,2,3\}$ is another element of $A$ that is not an element of $B$; it is not an element of $A\cap B$ either.

On the other hand $7$ is not an element of $A$. It's an element of an element of $A$, but that's a different thing.

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    $\begingroup$ I disagree with your first paragraph. Yes, it's a bit unusual to "mix types", but it's not used mainly to "confuse students". It is used to teach students how to carefully follow definitions, and pushes them to better understand the concept of elementhood. $\endgroup$ – Asaf Karagila Aug 17 at 2:25
  • $\begingroup$ @AsafKaragila: I've toned it down a slight bit. $\endgroup$ – Henning Makholm Aug 17 at 2:27
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    $\begingroup$ I got it! :D The answer is A ∩ B = (1,3,5) , because those are the ones who are actual elements in both. $\endgroup$ – Gerardo Hernández Aug 17 at 2:28
  • $\begingroup$ @GerardoHernández: Correct! $\endgroup$ – Henning Makholm Aug 17 at 2:28
  • $\begingroup$ Thanks for the help guys! :D $\endgroup$ – Gerardo Hernández Aug 17 at 2:28
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When sets are elements, they are elements. The elements within them are irrelevant.

We know $B = \{1,3,5,7,9\}$. Let $L = \{1,2,3\}$.

Then $A =\{\{1,3,5,7,9\}, \{1,2,3\},1,2,3,4,5\}= \{B,L,1,2,3,4,5\}$

So what is $A\cap B$? It is, as always, the elements that $A$ and $B$ common.

The first element of $A$, is the element $B$. Is $B$ an element of $B$? No, it is not. The elements of $B$ are $1,3,5,7,9$ and not a single one of them is the same thing as $B$.

That's all we are concerned with. We don't need to think about "well, the elements of $B$ are within $B$ so...". $B$, the entire set itself, is the first element. Is the entire set itself an actual element of $B$? No, it is not. That's it. $B \not \in B$. Period.

The second element of $A$, is the element $L$. Is $L$ an element of $B$? No, it is not. The elements of $B$ are $1,3,5,7,9$ and not a single one of them is the same thing as $L$.

The third element of $A$, is the element $1$. Is $1$ an element of $B$? Yees, it is. The elements of $B$ are $1,3,5,7,9$ and one of them is the same thing as $1$. So $1\in A\cap B$

The fourth element of $A$, is the element $2$. Is $2$ an element of $B$? No, it is not. The elements of $B$ are $1,3,5,7,9$ and not a single one of them is the same thing as $2$.

The fifth element of $A$, is the element $3$. Is $3$ an element of $B$? Yees, it is. The elements of $B$ are $1,3,5,7,9$ and one of them is the same thing as $3$. So $3\in A\cap B$

The sixth element of $A$, is the element $4$. Is $4$ an element of $B$? No, it is not. The elements of $B$ are $1,3,5,7,9$ and not a single one of them is the same thing as $4$.

The last element of $A$, is the element $5$. Is $5$ an element of $B$? Yees, it is. The elements of $B$ are $1,3,5,7,9$ and one of them is the same thing as $5$. So $5\in A\cap B$

So $A\cap B =\{1,3,5\}$

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  • $\begingroup$ If we are working with ZFC I would have to consider that. I took this to be a more elementary course. It's the room I read and I'm sticking with it. $\endgroup$ – fleablood Aug 17 at 4:22
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To answer your question, $A\cap B=\{1,3,5\} $.

And, for extra credit, appreciate the distinction between these two true statements:

$$\{1,2,3,4,5\}\subset A $$ $$\{1,3,5,7,9\}\in A $$

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  • $\begingroup$ Great way to clear it out! Thank You! $\endgroup$ – Gerardo Hernández Aug 17 at 3:07

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