# Why $\{2,3\}$ isn’t a subset of B?

I have this set: $$B=\{1, \{1\}, \{2,3\}\}$$ and its elements are: $$1, \{1\}, \{2,3\}$$. This means that $$\{1\}$$ is a subset of $$B$$, but also its element. $$1$$ is a element of $$B$$. For example $$2$$ isn’t a element of $$B$$, but why $$\{2,3\}$$ is a element of $$B$$, but not it’s subset like $$\{1\}$$?

• Subset means that for all $x\in\{2,3\}$ one must have $x\in B$. In particular $2\in \{2,3\}$ should imply that $2\in B$. – conditionalMethod Nov 21 '19 at 16:10
• set $a:=1, b:=\{1\}, c:=\{2,3\}$ then $a,b,c,$ are elements of $B$ and $\{a\}$ is a subset of $B$ – Jno Nov 21 '19 at 16:11
• Oh, thank you, I understand now. – Mia09 Nov 21 '19 at 16:12
• Note that $\{\{2, 3\}\}$ is a subset of $B$; every element of this subset (namely, the element that is the set $\{2,3\}$) is an element of $B$. – Steven Stadnicki Nov 21 '19 at 16:14

For a set $$C$$ to be a subset of another set $$B$$, every element of $$C$$ must be an element of $$B$$. In your case, the set $$\{1\}$$ has the element $$1$$ which is also an element of $$B$$. But the set $$\{2, 3\}$$ has elements $$2$$ and $$3$$, neither of which are elements of the original set $$B$$.

$$\{\{2,3\}\}$$ is a subset of $$B$$ but not $$\{2,3\}$$

$$1\in B$$. Therefore the set containing $$1$$, that is $$\{1\}$$ is a subset of $$B$$.

If $$gipple \in M$$ then $$\{gipple\}\subset M$$. That is a basic definition of subset. A set $$\{gipple\}$$ is a subset of $$M$$ if every element of $$\{gipple\}$$ is an element of $$M$$. And as the only element of $$\{gipple\}$$ is $$gipple$$ and $$gipple \in M$$ then $$\{gipple\}\subset M$$.

So $$1\in B$$ so $$\{1\}\subset B$$.

Now by COMPLETE COINCIDENCE, and nothing more than complete coincidence, we ALSO have that $$\{1\}$$ is an element of $$B$$, so $$\{1\}\in B$$. There is utterly NO reason this had to happen and no reason we should have expected it to happen.

It could just as easily have happened that $$1 \in B$$ and $$\{1\}\not \in B$$. Or that $$\{1\}\in B$$ but $$1 \not \in B$$. Or that neither were in $$B$$.

TOTAL COINCIDENCE... well, whim of the author, but that's pretty much the same thing.

I should point out that because $$\{1\}$$ is an element of $$B$$ then the set containing $$\{1\}$$, that is to say, the set $$\{\{1\}\}$$ is a subset of $$B$$. That is $$\{\{1\}\}\subset B$$ BUT $$\{\{1\}\} \not \in B$$.

The elements of $$B$$ are $$1, \{1\},$$ and $$\{2,3\}$$ and none of them are $$\{\{1\}\}$$.

====

Now $$2 \not \in B$$ and $$3\not \in B$$ to so $$\{2,3\} \not \subset B$$. That's basic definition. It is not the case that all the elements of $$\{2,3\}$$ are elements of $$B$$ so $$\{2,3\}$$ is not a subset of $$B$$.

But one of the elements of $$B$$ just happens to be $$\{2,3\}$$. $$\{2,3\}$$ is an element of the set $$B$$. So $$\{2,3\}\in B$$. Because .... it is.

But saying a set $$A$$, itself, is an element of $$B$$ doesn't mean that the elements of $$A$$ are elements of $$B$$. If $$A \in B$$ it doesn't follow that $$A \subset B$$.

By COINCIDENCE we had $$\{1\} \in B$$ and by coincidence $$1\in B$$ as well so we did have $$\{1\} \subset B$$ but that was a coincidence.

$$\{2,3\} \in B$$... because it is... but $$2,3 \not \in B$$ ... because then aren't. So $$\{2,3\} \not \subset B$$.

But if it makes thing feel better, the set containing just the element $$\{2,3\}$$, that is to say, the set $$\{\{2,3\}\}\subset B$$. (Because the only element of $$\{\{2,3\}\}$$ is the element $$\{2,3\}$$ and that element is in $$B$$.