Let $A = \{\{1\},\emptyset\}$, $B=\{\{1\}\}$. Is it true that $A\subset B$?
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Note that $\emptyset \in A$ but $\emptyset \notin B$. However, note that $\{1\} \in B$ and $\{1\} \in A$. Hence, for all $x \in B$, we have that $x \in A$. Hence, in fact, $B \subset A$. |
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False. For $A \subset B$ we need $(\forall x \in A)x\in B$. But $(\exists x\in A)x \notin B$, namely $x = \emptyset$. This is the logical negation, so $A \not\subset B$. |
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