Is $\emptyset \subseteq P(A)$ or $\{\emptyset\}\subseteq P(A)$, where $A$ is a set and $P(A)$ is the power set of $A$ I think my professor made an error on his answer key and I'm trying to confirm it before I bring it to his attention. He asserts only 1.) is false. I believe both 1 and 4 are false. This class is only using naïve set theory.

$A = \{1,2,3,4\}$
Select the statement that is false
1.) $\{2,3\} \subseteq P(A)$
2.) $\{2,3\} \in P(A)$
3.) $ \emptyset \in P(A)$
4.) $\emptyset \subseteq P(A)$


*

*$P(A)$ is the set of all the subsets of $A$

*

*$P(A) = P(A) = \{\emptyset,\{1\},\{2\},\{3\},\{4\},\{1,2\},\{1,3\},\{1,4\},\{2,3\},\{2,4\},\{3,4\},\{1,2,3\},\{1,2,4\},\{1,3,4\},\{2,3,4\},\{1,2,3,4\}\}$




*

*FALSE

*

*$2$ is not an element of $P(A)$

*$3$ is not an element of $P(A)$

*∴ $\{2,3\}$ cannot be a subset of $P(A)$

*$\{{2,3}\}$ would be a subset of $P(A)$



*TRUE

*

*The element $\{2,3\}$ can be found in the set $P(A)$



*TRUE

*

*The element $\emptyset$ can be found in the set $P(A)$



*FALSE

*

*Both operands of the subset operator requires a set. $\emptyset$ is the empty set were as $\{\emptyset\}$ is an element that is the empty set. Therefore $\emptyset$ is not a subset of $P(A)$.




You are wrong: both $\varnothing$ and $\{\varnothing\}$ are subsets of $\mathcal P(A)$. $\varnothing$ is a subset of any set $X$, because the condition of all its elements being elements of $X$ is vacuously true; $\{\varnothing\}$ is a subset of $\mathcal P(A)$ because $\varnothing\in\mathcal P(A)$, seeing as $\varnothing\subseteq A$.
$\{\varnothing\}$ is a set, specifically the set containing exactly the element $\varnothing$. Its existence is implied by the axiom of pair and the axiom of empty set together.
