Set membership and a box of bananas Suppose that I have two boxes $A$ and $B$, each of which contains some number of identical (indistinguishable)* bananas.
If I treat $A$ and $B$ as multisets whose elements are bananas, it follows that
$$a\in A\implies a=banana$$
$$b\in B\implies b=banana$$
As $A$ and $B$ are physically seperate boxes, it also follows that $A$ and $B$ are disjoint:
$$A\cap B=\emptyset$$
However, by way of transitivity:
$$a=banana=b\implies a=b$$
$$\therefore a\in A\implies a\in B$$
Which leads to the contradiction $a\in A\implies a\in\emptyset$
How do I resolve this?
*Indistinguishable means that a particular banana, while it is in a box, carries no information that would allow it to be differentiated from any other banana in either box. If you swap any two bananas, the result is the same as doing nothing at all.

Guesses (in order of descending ridiculousness)
Potential option 1:
The relationship between a banana and its box is not adequately described by set membership and equality, devise a more exclusive relation to prevent a banana in $A$ from being in $B$ at the same time.
Potential option 2:
The bananas in box $A$ exist in an undecidable superposition with those of box $B$. Thus neither $a\in A$ nor $a\in B$ is either true or false, but half-true and half-false, until I open the box.
Potential option 3:
The intersection of $A$ and $B$ is not empty. There exists a "virtual" set $C$ of "potential bananas" such that $a\in A\implies a\in B\implies a\in C$ and $A\cap B=C$
Potentail option 4:
Options $1-3$ represent different expressions of a fundamental truth regarding set membership, and are simultaneously true (somehow)
 A: The OP says that swapping bananas does not change anything, but it does not say the same about merging bananas in a given box. The mention of "multisets" suggests that was intentional, so that the counts of the bananas in each box matter. However, as pointed out in the OP, if we just treat the boxes as multisets, then we'd get a nonempty intersection for the boxes, contradicting our physical intuition.
In order to incorporate physical intuition like this, bananas can't be actually indistinguishable. The description about swapping bananas only forces them to be indistinguishable except for position. Even if we treat all positions in a given box as indistinguishable, we would then have a situation like "Box $A$ is a multiset containing $7$ copies of 'banana located in the box $A$ region'" and "Box $B$ is a multiset containing $3$ copies of 'banana located in the box $B$ region'." The act of swapping bananas in different boxes changes their location/region.
We can introduce a bit more notation around this idea. Suppose  is some mathematical object like a set or an urelement. Then a "physical banana" is an ordered pair $(,p)$ where $p$ is represents a "physical position" in some way. By how equality of ordered pairs works, we have $(,p_0)=(,p_0)$ for any particular physical position $p_0$.
Suppose $\{a_1,\ldots,a_n\}$ and $\{b_1,\ldots,b_m\}$ are disjoint sets of positions, to account for $A$ and $B$ being physically separated boxes. (Though it may be that some or all the $a$s are equal to each other, if you wish.) Then the multisets $A=[(,a_1),\ldots,(,a_n)]$ and $B=[(,b_1),\ldots,(,b_m)]$ have empty intersection, even though the first coordinates of all of their elements (what you get when you ignore position) are all equal to .
A: You're using "=" as a equivalence relation, but then using different sets so it breaks down.
A: The problem is that each banana is different so the fact that $a$ is a banana and $b$ is a banana doesn't allow us to say $a$ = $b$.
If $a$ and $b$ came from a set of fruits, then the banana would be unique and the fact that both sets contained a banana would indeed say they have non-empty intersection. In your case each box represents a set of physical objects so, although two bananas may be indistinguishable, they are unique.
