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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 $A$
    • $P(A) = P(A) = \{\emptyset,\{1\},\{2\},\{3\},\{4\},\{1,2\},\{1,3\},\{1,4\},\{2,3\},\{2,4\},\{3,4\},\{1,2,3\},\{1,2,4\},\{1,3,4\},\{2,3,4\},\{1,2,3,4\}\}$
  1. 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 $P(A)$
  2. TRUE
    • The element $\{2,3\}$ can be found in the set $P(A)$
  3. TRUE
    • The element $\emptyset$ can be found in the set $P(A)$
  4. 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 $P(A)$.
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#4 is true. The empty set is a subset of every set.

It happens that the empty set us also an element of this set.

So both $\varnothing \subseteq \mathscr{P}(A)$ and $\varnothing \in \mathscr{P}(A)$ are true.

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    $\begingroup$ Okay, so the empty set is implicit. Does this mean that 1st element I listed in P(A) can express the following statement: {{}} ⊆ P(A)? $\endgroup$ – Shawn Armstrong Jan 7 at 2:27
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    $\begingroup$ You’ve made several errors. You have correctly included the empty set, but you have omitted all subsets with 3 or 4 elements, and note that $\{2\}$ and $\{2,2\}$ are the same set containing the single element $2$. $\endgroup$ – MPW Jan 7 at 2:35
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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.

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The empty set is a subset of every set, so it is also true that $\emptyset \subset P(A)$.

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    $\begingroup$ @Sumanta: the power set is always non-empty, so it is appropriate to use the symbol ‘$\subset$‘. $\endgroup$ – Pietro Paparella Jan 7 at 3:10

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