# Sets with all or none of the elements also being subsets - Cohn - Classic Algebra Page 11

Give examples of sets such that $(i)$ all and $(ii)$ none of their members are also subsets

Firstly I should make sure I understand this correctly: The subsets of the set $S=\{\emptyset,1,2,3\}$ are $\emptyset,\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},S$

So $i)$ wants me to find a set, with all elements being subsets as well. With my above understanding of subsets, I can't see how this is possible, unless the answer is $S_2 = \emptyset \subseteq \emptyset$(is it?)

$ii)$ With my understanding on subsets being the way it is, the set $S$ above should meet this criteria, since $\{1\},\{2\}$ etc are all sets, and $S$ only has elements within it. But this would defeat my answer to $i)$ since it states that $\emptyset$ is not a subset, and furthermore, if $i)$ is correct, than no set meets this criteria, as all sets will have the empty set as a member and a subset.

Cohn - Classic Algebra Page 11

• The set $S=\{\emptyset, 1, 2, 3\}$ also contains the subset $\{\emptyset\}$, which is distinct from the subset $\emptyset$. Commented Oct 7, 2014 at 10:07
• So, the answer to part (ii) is let $T=\{1, 2, 3\}$. Indeed, none of the members of $T$ are subsets of $T$. The members are numbers; they have no chance of being "subset" in any sense. Commented Oct 7, 2014 at 10:09
• @Prism In set theory everything is a set. According to one standard way of defining numbers as sets, $0=\emptyset$, $1=\{0\}$, $2=\{0,1\}$, $3=\{0,1,2\}$, etc. Thus none of the elements of your set $T=\{1,2,3\}$ is a subset of $T$, not because $1,2,3$ are not sets, but because as sets they all have the element $0$ which is not an element of $T$.
– bof
Commented Oct 7, 2014 at 10:17
• @bof: Thanks for clarification! I apologize if I confused OP. Commented Oct 7, 2014 at 10:28
• @Prism You are fine, his clarification was caused by your comment, and together they helped build my understanding.
– user142198
Commented Oct 7, 2014 at 11:36

$\{\emptyset\}$ has just one element, $\emptyset$, which is a subset.

$\{\emptyset,\{\emptyset\}\}$ has two elements, both of which are subsets.

No element of $\{\{\emptyset\}\}$ is a subset.

• Is'nt $\{\{\emptyset\}\}=\{\emptyset,\{\emptyset\}\}$? (and hence an element of $\{\{\emptyset\}\}$ is a subset?)
– user142198
Commented Oct 7, 2014 at 10:15
• No, $\{\{\emptyset\}\}\ne\{\emptyset,\{\emptyset\}\}$, because $\emptyset\in\{\emptyset,\{\emptyset\}\}$ but $\emptyset\notin\{\{\emptyset\}\}$. A set with only one element cannot be equal to a set with two elements.
– bof
Commented Oct 7, 2014 at 10:21
• I thought $\emptyset$ was an element of all sets, but contributed cardinality of $0$?
– user142198
Commented Oct 7, 2014 at 10:22
• $\emptyset$ is a SUBSET of every set.
– bof
Commented Oct 7, 2014 at 10:22
• Okay so $\emptyset$ isn't necessarily an element of a set, but it is necessarily a subset of a set. It can be an element, such as in $\{\emptyset\}$, and when it is a subset, it is an element or not?
– user142198
Commented Oct 7, 2014 at 10:26