Examples of A intersection B being an element of A

[Stephen Tashi]

What are some examples where the intersection of two sets is a member of one of the sets?

Let $A,B,C,D$ be sets whose elements are sets of integers.$$A = \{\emptyset, \{1,2\},\{3\} \}$$

$$B = \{\{4,5,6\},\{7,8\} \}$$

$$C = \{ \emptyset, \{7,8\} \}$$

$$D = \{ \{3\}, \{4,5,6\} \}$$Then $A \cap B = \emptyset$ and taking the empty set to be unique (as mentioned in the recent thread https://www.physicsforums.com/threa...ms-specify-that-the-empty-set-is-open.773047/ ) this is the same empty set that is an element of $A$ so have that $A \cap B \in A$On the other hand $A \cap C = \{\emptyset \}$, which is to say that $A \cap C$ is a set with one element and that element is the empty set. This set is not the empty set, so we can't say that $A \cap C \in A$.$A \cap D = \{\{3\}\}$. My interpretation is that $A \cap D$ is a set with one member and that member is itself a set with one element, the integer 3. So $\{\{3\}\}$ is not a "set of integers", it is a "set of sets of integers". So we can't say that that $A \cap D \in A$.

 

[gopher_p]

In the set-theoretic construction of the natural numbers, where $0=\{\}$ and $n+1=n\cup\{n\}$, this sort of "weird" stuff happens all of the time; e.g. if $a<b$, then $a\in b$ and $a\subset b$.

 

[Stephen Tashi]

Speaking of wierd, It's counterintuitive that $A \cap B \in A$ does not imply $A \cap B \in B$.

For arbitrary sets $A$ and $B$, does $A \cap B \in A$ imply $A \cap B = \emptyset$ ?

 

[gopher_p]

Speaking of wierd, It's counterintuitive that $A \cap B \in A$ does not imply $A \cap B \in B$.

Well, in naive/intuitive set theory you have two classes* of objects, elements and sets, that are different types* of things (*though these words do have a technical meaning in foundations of math, my use of them here is completely informal). A such the relation $\in$ between an element and a set is viewed as a different kind of relation than the relation $\subset$ between two sets.

Dig a little deeper into naive set theory, and you start to see some examples of sets that also act like elements of other sets. But in an intuitive sense, we want to view the "bigger" set and its subsets as living on some higher level of "sethood" than the set that is an element. So we still see $\in$ and $\subset$ as different kinds of relations; $\subset$ is a relation between sets of the same level of sethood, and $\in$ is a relation between a lower-level set and on that has more sethood.

When you move to a more formal set theory, all of the objects are sets and elements. The relation $\in$ is a relation between sets that is not significantly different than the order relation $<$ on $\mathbb{N}$. In the same way that $1$ and $2$ are on the same level of "numberhood" to most people despite the fact that $1<2$, there are no differing degrees of sethood between $a$ and $b$ when $a\in b$. And so you start to throw away your intuition regarding the similarities and differences between the relations $\in$ and $\subset$ and begin to see them as being more similar. But you swing too far the other way and build a new false intuition in which they are too similar.

I guess this is the long way of saying that the naive approach to set theory leads us to think intuitively that $a\subset b$ excludes the possibility of $a\in b$, and vice versa. Once we come to the realization that $a\subset b$ and $a\in b$ aren't mutually exclusive, it's easy to begin to conflate $\in$ with $\subset$.

For arbitrary sets $A$ and $B$, does $A \cap B \in A$ imply $A \cap B = \emptyset$ ?

No. Similar to the example I gave in post #2, $b<a$ implies that $a\cap b=b$, which in general is not the empty set. Note that both $b\in a$ here and $b\subset a$, so $a\cap b\subset a$ (which is totally expected) and $a\cap b\in a$ (unexpected and relevant to this question).

For a more immediate counterexample, take $A=B\cup \{B\}$ where $B\neq\emptyset$.

 

[mathman]

What are some examples where the intersection of two sets is a member of one of the sets?

Let $A,B,C,D$ be sets whose elements are sets of integers.


$$A = \{\emptyset, \{1,2\},\{3\} \}$$

$$B = \{\{4,5,6\},\{7,8\} \}$$

$$C = \{ \emptyset, \{7,8\} \}$$

$$D = \{ \{3\}, \{4,5,6\} \}$$


Then $A \cap B = \emptyset$ and taking the empty set to be unique (as mentioned in the recent thread https://www.physicsforums.com/threa...ms-specify-that-the-empty-set-is-open.773047/ ) this is the same empty set that is an element of $A$ so have that $A \cap B \in A$


On the other hand $A \cap C = \{\emptyset \}$, which is to say that $A \cap C$ is a set with one element and that element is the empty set. This set is not the empty set, so we can't say that $A \cap C \in A$.


$A \cap D = \{\{3\}\}$. My interpretation is that $A \cap D$ is a set with one member and that member is itself a set with one element, the integer 3. So $\{\{3\}\}$ is not a "set of integers", it is a "set of sets of integers". So we can't say that that $A \cap D \in A$.

$\emptyset$ is not the same as $\{\emptyset \}$. The former means there is nothing in the intersection. The latter means there is a set, which is empty, in the intersection.