9
$\begingroup$

If $ \{\emptyset\} ∈ \{\emptyset,\{\emptyset\}\} $ is true, does it mean this $ \emptyset \in \{\{\emptyset\}\} $ true ? If it is not, why it is false?

Also, does $ \{\{\emptyset\}\}$ mean $\{\emptyset,\{\emptyset,\{\emptyset\}\}\}$ ?

$\endgroup$
31
$\begingroup$

The notation $a \in A$ says that among the elements of $A$ there is one element that is exactly equal to $a.$

The notation $\{\emptyset, \{\emptyset\}\}$ describes a set with exactly two elements.

The first element is $\emptyset.$ The second element is $\{\emptyset\}.$ Is one of those two elements exactly equal to $\{\emptyset\}$?

The notation $\{ \{\emptyset\}\}$ describes a set with one element. That element is $\{\emptyset\}.$

Which element of $\{ \{\emptyset\}\}$ do you think is exactly equal to $\emptyset$? Hint: there's only one element you have to check.

The notation $\{\emptyset, \{\emptyset, \{\emptyset\}\}\}$ again describes a set with two elements. One element is $\emptyset$ and the other is $\{\emptyset, \{\emptyset\}\}.$ So this is definitely not the same thing as any set that has only one element.

$\endgroup$
  • $\begingroup$ This is not an answer but a counter-question. $\endgroup$ – rexkogitans Jan 21 at 9:36
  • 2
    $\begingroup$ @rexkogitans Also known as a "hint" $\endgroup$ – TreFox Jan 21 at 19:36
2
$\begingroup$

The set $\{\{\emptyset\}\}$ is a set with one element: $\{\emptyset\}$ (which in turn is a set with one element, the empty set, which in turn is a set with no elements).

Therefore the empty set is not an element of the set you described. The way of thinking about it is: the more external (outer) brackets determine which elements compose the set.

This also explains why $\{\{\emptyset\}\} \not\equiv \{\emptyset,\{\emptyset,\{\emptyset\}\}\}$.

$\endgroup$
0
$\begingroup$

Let us begin with naming the sets in the question: The set $B=\{a\}$ has a single element, which is $a$. Now, let $a=\emptyset$, then we have $B=\{\emptyset\}$, which is of course also a set with a single elementy, namely $\emptyset$.

If we put $A$ into a set $C=\{B\}$, then we have $C=\{\{\emptyset\}\}$. It can be easily seen that $a \not\in C$. Hence, $\emptyset \not\in \{\{\emptyset\}\}$.

The set $A=\{\emptyset, b\}$ has two elements. Now, let $b=\{\emptyset\}$. Then we have the set $A=\{\emptyset, \{\emptyset\}\}$. If we now take a look at $b$, then, since $b\in A$, also $\{\emptyset\} \in A$, which is $\{\emptyset\} \in \{\emptyset, \{\emptyset\}\}$.

To answer if $\{\{\emptyset\}\}$ means $\{\emptyset,\{\emptyset,\{\emptyset\}\}\}$: Look at $C$, it has only one element, which is $B$. It follows that $\emptyset \not\in C$, so these sets are different (the second set has two elements, which are the empty set and set set containing the empty set).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.