Can you do $A\cup B$ for $B \in A$? For instance, let $A=\{1, 2, B\}$ and $B=\{3, 4\}$. Is it then true that $A \cup B = \{1, 2, 3, 4\}$?
 A: You can do $A\cup B$ for any sets $A$ and $B$. However it is not true that in your case $A\cup B=\{1,2,3,4\}$.
Since we have $A=\{1,2,B\}$ and $B=\{3,4\}$, we have $A=\{1,2,\{3,4\}\}$. This is just inserting the value of $B$ in the description of $A$.
That is, $A$ contains three elements, namely the number $1$, the number $2$ and the set $B=\{3,4\}$
Now $A\cup B$ is the set that contains all elements of $A$ and all elements of $B$. The elements of $B$ are the numbers $3$ and $4$. Note that those numbers are not elements of $A$, as only the single set $\{3,4\}$ is an element.
Therefore we have
$$A\cup B = \{1,2,B\}\cup\{3,4\} = \{1,2,B,3,4\} = \{1,2,\{3,4\},3,4\}.$$
A: Let $X=\{A,B\}$. Then by definition, $A\cup B=\cup_{x\in X} x$. So we must take the elements of the elements of $X.$ The fact that $B\in A$ is no exception to this general rule. This gives $A\cup B = \{1,2,\{3,4\},3,4\}.$ The above definition shows that if $B\in A,$ then $B\in A\cup B.$
A: For two sets $A,B$, the union $A\cup B$ is the set which contains all elements of $A$ or $B$. That is, $A\cup B=\{x:x\in A\text{ or }x\in B\}$.
A full list of things in $A$ is "$1,2$, and $B$," and a full list of things in $B$ is "$3$ and $4$." Thus $A\cup B=\{1,2,3,4,B\}$.
Since $B=\{3,4\}$, we can also write $A=\{1,2,\{3,4\}\}$ and $A\cup B=\{1,2,3,4,\{3,4\}\}$.
The most important aspect of these types of problems is to distinguish between (for example)


*

*$x$ (which in math is just the symbol $x$)

*the set containing $x$ (in math, $\{x\})$

*the set containing the set containing $x$ (in math, $\{\{x\}\}$).

*etc.


These are all distinct objects. In other words, make sure to keep the number of braces straight. There's some orthodontics joke there but I can't find it.
A: If $B \in A \cup B$ then $B \in A$ or $B \in B$. Since we can't have $B \in B$. We have $B \in A$.
