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Given the set $A = \{\{$∅$\},\{2\},2\}$, determine if the following statements are false. If false, then correct the statement to be true

Determine the validity of: $\{$∅$,\{$∅$\}\} ⊆ A$

Knowing that the null set is a subset of all other sets, I don't understand why this expression is false. Clearly ∅ $⊆ A$ by this definition and it also seems like $\{$∅$\} ⊆ A$ since A contains an element which is a set.

For any set A, the empty set is a subset of A:

$\forall A:$ ∅ $\subseteq A$

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To say that $\{\varnothing,\{\varnothing\}\}\subseteq A$ is the same as saying $$\varnothing\in A\text{ and }\{\varnothing\}\in A$$

While it is true that $\varnothing\subseteq A$, it is certainly not necessarily true that $\varnothing\in A$. And in this case, it is indeed false.

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  • $\begingroup$ Whichever you like. Like $\Bbb N$ and $\bf N$. $\endgroup$ – Asaf Karagila Apr 13 '15 at 20:17
  • $\begingroup$ Ah that makes sense. I didn't realize $\{$∅$,\{$∅$\}\}⊆A$ was the same thing as saying ∅ $∈A$ and $\{$∅$\}∈A$. Thank you! I'm still slightly unclear about why ∅ $∉A$ since if there was a set with no other subsets (just a list of elements), ∅ $⊆A$ would still be a valid statement. Where's the invisible ∅? $\endgroup$ – James Taylor Apr 13 '15 at 20:25
  • $\begingroup$ $A$ has three elements. $\{\varnothing\}$ and $\{2\}$ and $2$. Is any of them the empty set? $\endgroup$ – Asaf Karagila Apr 13 '15 at 20:27
  • $\begingroup$ Well no. But since ∅ $⊆A$ always holds true no matter what, I assume it would have to be an element of the set (or subset of the set) somewhere. $\endgroup$ – James Taylor Apr 13 '15 at 20:29
  • $\begingroup$ But being a subset and being an element are two different things. $\varnothing\subseteq\varnothing$, does that mean that $\varnothing\in\varnothing$? $\endgroup$ – Asaf Karagila Apr 13 '15 at 20:39
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It's true that the empty set is a subset of every set: $$\forall A\ \ \ \ \ \emptyset \subseteq A$$

But the question $\{\emptyset,\{\emptyset\}\} ⊆ A$ signifies that $\emptyset\in A$ and $\{\emptyset\}\in A$ which's different from the inclusion, in particular the statement is false because $\emptyset \not \in A$

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