# Deciding if subset and element statements involving sets are true or false

I'm trying to figure out if these statements are true or false:

(1) {∅} ∈ P(A)

(2) {A} ⊆ A

(3) A ⊆ {A}

This is what I think they are:

(1) false

• reasoning: ∅ is a set with no elements, but {∅} is a set with one element (∅). Since ∅ is a subset of every set, ∅ is a subset of A. By definition a power set of a set, in this case A, is a set whose elements are subsets of the set A. So since ∅ is a subset of A, $$∅ ∈ P(A)$$ is true but not {∅} ∈ P(A)

(2) false

• reasoning: A is contained in {A}, but {A} is not contained in A, so A ⊆ {A} is true, but {A} ⊆ A is false.

(3) true

• reasoning: see previous explaination

Is what I said correct (both the true/false answer and my reasoning)?

• You should precise if those statements are supposed to be true for some $A$ or for all $A$. Apr 24, 2020 at 20:33
• You should try to avoid ambiguous words like "is contained in". Stick to "is an element of" for expressing the relation $\in$, and stick to "is a subset of" for expressing $\subseteq$. As your reasonings are currently written, it's rather hard to understand them because of the ambiguity of "is contained in". Apr 24, 2020 at 21:01

$$(1)$$ is false: $$\emptyset \subseteq A \implies \emptyset \in \mathcal{P}(A) \implies \{\emptyset\} \subseteq \mathcal{P}(A) \implies \color{green}{ \{\emptyset\} \in \mathcal{P}( \mathcal{P}(A))}.$$
$$(2)$$ is false: $$A\subseteq A \implies A \in \mathcal{P}(A) \implies \color{green}{ \{A\} \subseteq \mathcal{P}(A)} \implies \{A\} \in \mathcal{P}( \mathcal{P}(A))..$$
$$(3)$$ is false: $$\color{green}{A \in \{A\}}.$$
The reasoning for 2 is incorrect, because of the statement $$A \subseteq \{A\}$$, replace it with $$A \in \{A\}$$. For understanding, let $$A = \{1, 2\}$$. It's clear that $$\{1, 2\} \not\subseteq \{\{1, 2\}\}$$. This also shows that correct answer for 3 is false. You can give this example to prove.
The third statement should be false because $$A \in \{A\}$$ and $$A \subseteq \{A\}$$ are different statements. Note that the set $$\{A\}$$ has $$2^1=2$$ subsets: $$\varnothing$$ and the set itself, $$\{A\}.$$ Clearly $$A$$ is neither of those two subsets.