# Subsets of Sets- confused with the bracketing!

I'm a bit confused with the following sets!

$$S_1= \{\{B\}, B\}$$

$$S_2= \{\{B\}\}$$

In terms of $S_1$, I'm not sure whether both members are also subsets and how to write that in terms of the brackets, and in $S_2$ will it be $\{\{B\}\}$ or $\{B\}$? The brackets are really confusing me.

Jess

• Some relations that might help: (One) $\{B\}\in S_2$, (Two) $\{\{B\}\}\not\in S_2$, (Three), $\{\{B\}\}\subseteq S_2$, (Four) $\{\{B\}\}\not\subsetneq S_2$ (it's hard to see but there are two slashes there which say that $\{\{B\}\}$ is not a proper subset of $S_2$), and (Five) $S_2 \cup \{B\} = S_1$. – user137731 Oct 16 '16 at 20:18
• $B \in S_1$, $\{B\} \in S_1$, $\{B\} \subseteq S_1$, and $\{\{B\}\} \subseteq S_1$ – Alexis Olson Oct 16 '16 at 20:21
• I don't understand what you mean with {{B}} is not a proper subset though? I thought a set can be a subset of itself and therefore it will be? – Jessica Poyanzadeh Oct 16 '16 at 20:31
• @JessicaPoyanzadeh There's a difference between a (regular) subset and a proper subset. – Bobbie D Oct 17 '16 at 3:19

The elements of $S_1$ are $\{B\}$ and $B$. So:
• Is $\{B\}$ a subset of $S_1$? Yes. Why? By definition, all the elements of $\{B\}$ are in $S_1$, because "all the elements of $\{B\}$" is actually only $B$, which is in $S_1$.
• Is $B$ a subset of $S_1$? For this to be true, we would need to check that all elements of $B$ are in $S_1$. If $B$ is the empty set, this is trivially true and the answer is yes. If not, the situation becomes tricky: if $B$ is any simple known set like $\{1\}$ or the set of integers, the answer is no. But in the general case or we would be dealing with things like $B \in B$ or $\{B\} \in B$. In abstract axiomatic set theory, this is called an improper class, and it is not exactly a set. Russell's paradox is enough to illustrate something that is not a set. In any case I think it's enough to consider that without more information about $B$, there is more than one answer.
On the other hand, $S_2$ has one member, $\{B\}$. For this to be a subset of $S_2$, all of its members should be in $S_2$, but $B$ is a member of $\{B\}$ which is not. So $\{B\}$ is only a member of $S_2$, not a subset of it.