working through some notes and, in essence, the following is said:
"$A=\{1,2\}$ and $B=\{3,4\}$
$A\cap B = \{\emptyset\}$"
Is this a mistake? Should it not be $\emptyset$?
$\begingroup$
$\endgroup$
3
-
14$\begingroup$ You are correct. It is a mistake. The intersection is $\emptyset$, not $\{\emptyset\}$. $\endgroup$– MJDDec 28, 2019 at 17:22
-
$\begingroup$ "why is the intersection of two sets with no common element equal to the subset containing the empty set and not simply the empty set?" For the same reason when you go to a zoo and look at the elephant you see a dictionary describing what an elephant is rather than an actual animal.... (Which is my snarky way of answering "why is the intersection of two sets with no common element equal to the subset containing the empty set and not simply the empty set?" with "It isn't. The intersection is the empty set and most certainly is NOT $\{\emptyset\}$". $\endgroup$– fleabloodDec 28, 2019 at 17:26
-
3$\begingroup$ This is a mistake, but I think it's a fairly common one - sometimes authors (very reasonably) want to avoid writing $A\cap B = \{\}$ because it looks weird and doesn't emphasize that the lack of terms inside the bracket was intentional. Of course, the appropriate solution is to write $A\cap B = \emptyset$, but it's easy for the gears in an author or editor's brain to slip and incorrectly write $\{\emptyset\}$ instead. $\endgroup$– Milo BrandtDec 29, 2019 at 1:57
Add a comment
|
4 Answers
$\begingroup$
$\endgroup$
Yes, this is a mistake, it should be $A\cap B=\emptyset$.
answered Dec 28, 2019 at 17:21
$\begingroup$
$\endgroup$
Definitely a written/typed error.
If we consider the "$\{$" "$\}$" as notational placeholders to indicate "the list/description between them is the list/description of elements in a set then:
$\emptyset = \{\}$ is the set that when you list its elements.... doesn't have any.
$\{\emptyset\}=\{\{\}\}$ is a set that when you list its elements contains as its single element: the emptyset.
$A\cap B = \{x| x\in A; x \in B\}$. And there is nothing that is in both $A$ and $B$ so $A\cap B=$ a set who if you listed its elements would contain.... nothing $= \{\}=\emptyset$.
If we had a set $C= \{\emptyset, Barbara, \text{a mushy banana}\}$ then $\emptyset$ is an an elements of $C$ so $\emptyset \in C$ but as $A=\{1,2\}$ and neither $1$ nor $2$ is the same thing as the empty set, so $\emptyset$ is NOT an elements of $A$ and $\emptyset \not \in A$. Likewise if $B = \{3,4\}$ then $\emptyset \not \in B$.
And if $\emptyset$ is in neither $A$ nor $B$ it certainly can't be in both $A$ and $B$.
But, you knew all that reasoned it out yourself. I just want to pound home that this is utterly consistent and logical and you should trust that if you look down you will see that both your feet actually are firmly on the ground.
$\begingroup$
$\endgroup$
Indeed it is a mistake. $\emptyset$ doesn't belong to $A$ (nor to $B$) and therefore can't be an element of $A \cap B$ which is equal to $\emptyset$.
answered Dec 28, 2019 at 17:23
$\begingroup$
$\endgroup$
Hint, The set $\{\emptyset \}$ has one element, but the set $\emptyset$ has no element.
