Null Sets $\{\{\emptyset\}\} \subset\{\emptyset, \{\emptyset\}\}$ Regarding null sets, I'm wondering if anyone can explain this $\{\{\emptyset\}\} \subset \{\emptyset, \{\emptyset\}\}$ I don't understand how the left set is a proper set of the right set.
In particular, I'm wondering what the extra brace on the left means and how it is different from say just plain $\{\emptyset\}$. It seems if I disregard the outer brace in the left hand set, my answer matches that of the answer key (namely, true).
I'm posting this again because it was in the wrong stackexchange.
Thanks!
 A: The only element of $\{\{\varnothing\}\}$ is $\{\varnothing\}$ which is in $\{\varnothing, \{\varnothing\}\}$.
Thus it is a subset...
A: a general rule:
$$
a \in S \Rightarrow \{a\} \subset  S
$$
since, in your example
$$
\{\emptyset\} \in \{\emptyset,\{\emptyset\}\}
$$
the result follows
A: $\{\{\emptyset\}\}$ is a set that has a set containing the empty set as a member.
$\{\emptyset\}$ is a set that has the empty set as a member.
$\{\emptyset,\{\emptyset\}\}$ is a set that has the empty set and a set containing the empty set as members.
Now $\{\{\emptyset\}\}\subset\{\emptyset,\{\emptyset\}\}$ because all the members of $\{\{\emptyset\}\}$, namely $\{\emptyset\}$, are in $\{\emptyset,\{\emptyset\}\}$.
But it is also true that $\{\emptyset\}\subset\{\emptyset,\{\emptyset\}\}$ since $\emptyset$ is also a member of $\{\emptyset,\{\emptyset\}\}$.
A: $\{\emptyset\}$ is a set containing the null set.
$\{\{\emptyset\}\}$ is a set containing a set which in turn contains the null set.
So we can write, for example, $\{\emptyset\} \in \{\{\emptyset\}\}$.
$\{\emptyset, \{\emptyset\}\}$ is a set with two elements: one element is the null set, and the other element is a set containing the null set.
So it's true that $\{\{\emptyset\}\} \subset \{\emptyset,\{\emptyset\}\}$: the only element of the left side ($\{\emptyset\}$) is also an element of the right side, but the right side also contains a different second element ($\emptyset$).
A: More generally, try proving that $\lbrace A \rbrace \subset \lbrace B,A \rbrace$.
A: When you see curly braces around something, that means you have a set with that something as an element. For example, $\{ S \}$ is a set, and its only element is $S$.  Similary, $\{1 \}$ is a set, and its only element is $1$.  $\{Hi, Bye \}$ is a set, and it has two elements: Hi and Bye.
Similarly, when you have $\{ \{ \emptyset \} \}$, this is a set, and its only element is $\{ \emptyset \}$.  It just so happens that the element of our set is a set itself, but there is nothing wrong with that.
Now, why is $\{ \{ \emptyset \} \} \subset \{ \emptyset, \{ \emptyset \} \}$?  Well, to show a set $A$ is a subset of a set $B$, you need to show every element of $A$ is an element of $B$... But $\{ \{ \emptyset \} \}$ has only one element: $\{ \emptyset \}$.  And is this element in the set $\{ \emptyset, \{ \emptyset \} \}$?  Of course, it is the second element of that set.
So, every element of $\{ \{ \emptyset \} \}$ is an element of $\{ \emptyset, \{ \emptyset \} \}$.
Now the last question is: why is this containment proper?  Well, to show a set $A$ is properly contained in a set $B$, you need to show that every element of $A$ is in $B$ (this shows $A$ is contained in $B$), and also that there is some element in $B$ that is not in $A$ (this shows $A$ and $B$ are not equal, i.e., $A$ is properly contained in $B$).  Well, our $A$ is $\{ \{ \emptyset \} \}$ and our $B$ is $\{ \emptyset, \{ \emptyset \} \}$.  Is there an element of our $B$ that is not in our $A$?  Our $B$ has two elements and our $A$ only has one element.  So there has to be an element in $B$ that is not in $A$, and actually, it is exactly the element $\emptyset$.
So, this explains why $\{ \{ \emptyset \} \} \subset \{ \emptyset, \{ \emptyset \} \}$.
A: Here, {$a,b,c,d,e,....z$} means a set with elements $a,b,c...z$. Now whenever a set is written , everything inside {} are called elements( they can be numbers, alphabets, sets, elephants....) so in your case, on Left hand side, {{$\phi$}} means the set {$\phi$} is an element in the set {{$\phi$}} and as {$\phi$} is an element on R.H.S, {{$\phi$}} $\subset${$\phi,${$\phi$}}
A: maybe, you can think of a set as a bag containing somethings. while $\emptyset $ is a empty bag, $\{\emptyset\}$ is a bag containing a empty one,and so on.
{{ϕ}} ⊂ {ϕ, {ϕ}} is right,because every element in $\{\{\emptyset\}\}$ ,namely $\{\emptyset\}$, is a element in {ϕ, {ϕ}}. it is proper because $\emptyset \in \{\emptyset,\{\emptyset\}\}$
