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Consider following statment $\{\{\varnothing\}\} \subset \{\{\varnothing\},\{\varnothing\}\}$

I think above statement is false as $\{\{\varnothing\}\}$ is subset of $\{\{\varnothing\},\{\varnothing\}\}$ but to be proper subset there must be some element in $\{\{\varnothing\},\{\varnothing\}\}$ which is not in $\{\{\varnothing\}\}$.As this is not the case here so it is false.

Is my explanation and answer right or not?

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    $\begingroup$ Yes. Equivalently, you could say that $\{\{\varnothing\},\{\varnothing\}\}=\{\{\varnothing\}\}$, since the two sets have the same members. $\endgroup$ – Brian M. Scott Nov 29 '13 at 21:21
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Yes, your answer and explanation is right. Because $A$ is not a prober subset of $A$, we have $\{\{\emptyset\}\} = \{ \{\emptyset\},\{\emptyset\} \}$ is not a proper subset of $\{\{\emptyset\}\}$.

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As you correctly noted, $\{\{\emptyset\}\}$ is not a proper subset of $\{\{\emptyset\},\{\emptyset\}\}$. Beware of the notation though, the symbol $\subset$ is often used to denote any subset, while the symbol $\subsetneq$ denotes a proper subset.

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