0
$\begingroup$

Question : Let $A_0=\emptyset $ (the empty set).For each $i=1,2,3,...,$ define the set -

$A_i=A_{i-1}$ $\cup$ {$A_{i-1}$}. The set $A_3$ is :

  • $\emptyset $

  • $\{\emptyset\}$

  • $\{\emptyset, \{\emptyset\}\}$

  • $\{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}$

My Work : I Googled this, but didn't get anything useful.

|cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ Yes they are different. $\{ \emptyset \}$ is a singleton set and $\emptyset$ is null set. $\endgroup$ – Error 404 Mar 12 '17 at 13:28
  • $\begingroup$ Set $\{\varnothing\}$ has an element: $\varnothing$. Set $\varnothing$ has no elements (another notation for this set is $\{\}$). So the sets are different. $\endgroup$ – drhab Mar 12 '17 at 13:29
  • 2
    $\begingroup$ I am curious, how have you googled that? As for your question, think about sets as boxes. Is an empty box different from a box containing an empty box? $\endgroup$ – user251257 Mar 12 '17 at 13:30
  • 2
    $\begingroup$ One of infinitely many: math.stackexchange.com/questions/1951267/…; another would be math.stackexchange.com/questions/1845389/… $\endgroup$ – Asaf Karagila♦ Mar 12 '17 at 13:38
  • 1
    $\begingroup$ Yes, the set $A_n$ has $n$ elements; it's the standard representative of the ordinal number $n$ $\endgroup$ – Henno Brandsma Mar 12 '17 at 14:13
2
$\begingroup$

$\emptyset \subset \emptyset$ but $\emptyset \in \{\emptyset\}$ (and also $\emptyset \subset \{\emptyset \}$ since $\emptyset $ is included in every set.)

Now $$\{\emptyset \}\in \{\emptyset ,\{\emptyset \}\}\quad \text{and}\quad \{\emptyset \}\subset \{\emptyset ,\{\emptyset \}\}.$$

|cite|improve this answer
$\endgroup$
  • $\begingroup$ That last example is amusing. A set is an element as well as a subset of another set. +1 $\endgroup$ – Paramanand Singh May 21 '17 at 17:19
3
$\begingroup$

Yes, they're different.

$\emptyset$ denotes a set with no elements (null set) whereas $\{\emptyset\}$ denotes a set with 1 element, namely, the empty set being the element of this set.

Therefore, $A_i \neq A_j$ $\forall i \neq j$

Also, observe that, $|A_i| = i$ $\forall i \in \mathbb{Z}$

|cite|improve this answer
$\endgroup$

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.