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This should be an easy question but I just want to make sure I'm understanding the notation and what the question is really asking, correctly.

If we have two sets:

$A = \{x \in \mathbb Z: x\ \text{is an integer multiple of three}\}$

$E = \{3, 6, 9\}$

Is $E \in A$ true?

I know that all elements from set E are in set A. But, does the definition $Set \in Set$ actually mean that the set {3,6,9} has to appear in A, and not just the elements?

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No but you have $E\subset A$. In other word, if $x\in E$ then $x\in A$. An other thing, you have that $E\in \mathcal P(A)$

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"I know that all elements from set $E$ are in set $A$" is a nice way of saying: "I know that $E$ is a subset of $A$" (notation $E\subseteq A$, or - if you want to emphasize that the sets $E$ and $A$ are not the same - $E\subset A$ or excluding any ambiguity $E\subsetneq A$; see the useful comment of Asaf on this question).

$E\in A$ is not true because $E$ cannot be recognized as "an integer multiple of three". It is a set of such integers (but is not such an integer itself).

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    $\begingroup$ If you really want to emphasize the inclusion is proper $\subsetneq$ is unmistakable. In some places $\subset$ denotes improper inclusion. $\endgroup$ – Asaf Karagila Feb 3 '15 at 13:19

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