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Let $A = \{1,3, \{1,2\}, \{\{1,3\}\}\}$

Clearly, $\{1,2\}∈A$ is true because $\{1,2\}$ is a element in set $A$, but I'm confused as to why $\{1,2\}⊂A$ is false. Similarly, in this example, $\{\{1,3\}\}∈A$ is true but $\{\{1,3\}\}⊂A$ is false

If an element is in a given set, is that element not a subset of the parent set?

I could make these subsets using only the members of $A$, so it seems like it should work. I would think that the subsets I could make of $A$ would be denoted by $P(A)$ (all the power sets of $A$) and including the null set $\{$∅$\}$.

Any clarification you could provide on this would be greatly appreciated!

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    $\begingroup$ $\{ 1,2 \}$ is not a subset of $A$ because $2$ is not an element of $A$, i.e. $2 \notin A$ ($2$ is not "listed" inside the brackets that "include" all elements of $A$). $\endgroup$ – Mauro ALLEGRANZA Mar 23 '15 at 9:06
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The elements of $A$ are:

$$1, 3, \{1,2\}, \{\{1,3\}\}$$

in which there's no $1$ and $2$. So $\{1,2\}$ is not a subset of $A$.

If an element is in $A$, then the set containing that element is a subset of $A$. So $\{\{\{1,3\}\}\}\subset A$.

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