Subset S is subset of a set containing s

Question

Let S be a set, determine the following is true or false

S ⊆ {S}

I know this is false, because S can be mulitple elements but {S} contain only one element. However what if I put some numbers into S, for example, S = {1,2,3} Therefore now:

{1,2,3} ⊆ {{1,2,3}},

Is it true or false?

Answer 1

The set $\{1,2,3\}$ has three elements, the numbers 1 2 and 3. On the other hand the set $\{\{1,2,3\}\}$ contains only one element which is the set $\{1,2,3\}$.

By definition, a set $A$ is a subset of $B$ $(A\subseteq B)$ if every element in $A$ is an element in $B$.

In the previous example $\{1,2,3\}$ is not a subset of $\{\{1,2,3\}\}$ as $1$ is an element in $\{1,2,3\}$ yet is not an element in $\{\{1,2,3\}\}$ (simply because $1\not = \{1,2,3\}$).

Answer 2

It is false in general, and it is false in your example. $\{1,2,3\}\not\subseteq \{\{1, 2, 3\}\}$, because $1\in \{1, 2, 3\}$ but $1\notin \{\{1, 2, 3\}\}$. However, we do have $\{1,2,3\}\in \{\{1, 2, 3\}\}$.

There are special cases where it's true, though. For instance, $\varnothing \subseteq \{\varnothing\}$. You could also, in theory, have that $S$ is an element of itself, in other words, $S\in S$. In that case we do have $S\subseteq \{S\}$. However, the existence of a set $S$ with $S\in S$ contradicts one of the axioms of ZF set theory (the axiom of foundation, also known as the axiom of regularity), so it would be somewhat non-standard.

Answer 3

You should distinguish three cases: if $S$ has more than $1$ element, then, as you say, the assertion cannot be true. If $S$ is the empty set, then it is definitely true! And if $S$ has only one element, well... $S$ cannot be an element of itself (if you allow this kind of things, you run into trouble), so it is not true again. As for the example you make: it is exactly the same as in the general case. Notice that the set $\{\{1,2,3\}\}$ does not contain any number, only the element $\{1,2,3\}$.

Answer 4

Assuming axiom of regularity, we have $S\subseteq\{S\}$ if and only if $S=\varnothing$.

For if $S=\varnothing$, we have $S=\varnothing\subseteq\{S\}$.

On the other hand, if $S\subseteq\{S\}$ and $S\neq\varnothing$, there exists $x\in S$, hence $x\in\{S\}$, thus $x=S$ from which $S\in S$. This contradicts axiom of regularity, hence $S=\varnothing$.