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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 $\{1\}$?

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

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

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$1\in B$. 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 $\{\{1\}\}$.

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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 $B$.

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