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I get that $∅$ is subset of every set thus $∅ ⊆ \{\{∅\}\}$. However, I'm not sure if $∅ ⊂ \{\{∅\}\}$. From definition of proper subset, the relation between two sets require the larger set to have at least one element not in the other one. What I'm confused is, does $\{\{∅\}\}$ have an element that $∅$ doesn't have?

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    $\begingroup$ "From definition of proper subset, the relation between two sets require the larger set to have at least one element not in the other one". You can also phrase it more directy as "$A$ is a proper subset of $B$ if and only if $A\subseteq B$ and $A\ne B$". $\endgroup$ Aug 2 '18 at 6:48
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    $\begingroup$ The empty set is subset of every set; thus it is a proper subset of every not-empty set: $\endgroup$ Aug 2 '18 at 7:08
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    $\begingroup$ $\{\varnothing\}$ is an element of $\{\{\varnothing\}\}$ which is not an element of $\varnothing$. I explained this in my answer over here. $\endgroup$
    – M. Winter
    Aug 2 '18 at 8:29
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The set $\{\{\emptyset\}\}$ contains the element $\{\emptyset\}$. The empty set contains no elements, thus the containment is proper i.e $\emptyset\subsetneq \{\{\emptyset\}\}$.

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Yes. $\emptyset\subsetneq\{\{\emptyset\}\}$ becasue $\{\emptyset\}\in\{\{\emptyset\}\}$ and $\{\emptyset\}\not\in\emptyset$. More generally, empty set is proper subset of every non-empty set.

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Let A={{∅}} be a set, then {∅}∈A. But always remember, ∅ ≠ {∅}, the former is an empty set and the latter is the element of a set.

Hence set A contains one element ({∅}). And as per your definition, From definition of proper subset, the relation between two sets require the larger set to have at least one element not in the other one.

∅ ⊂ {{∅}}, as ∅ is an empty set containing no element and {{∅}} is a set containing one element ({∅}).

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