1. **State the problem:** We have the set of natural numbers $N$ and the set $A = \{1, 2, \{3\}\}$. We want to determine which of the following statements is true: - $\{3\} \subseteq A$ - $N \cup A = N$ - $A \setminus \{3\} = A$ - $A \subseteq N$ 2. **Recall definitions and rules:** - A set $X$ is a subset of $Y$ ($X \subseteq Y$) if every element of $X$ is also an element of $Y$. - The union $N \cup A$ contains all elements in $N$ or $A$. - The set difference $A \setminus \{3\}$ contains elements in $A$ that are not in $\{3\}$. - Elements of $N$ are natural numbers (1, 2, 3, ...). 3. **Check $\{3\} \subseteq A$:** - The set $\{3\}$ contains the element $3$. - Does $A$ contain $3$ as an element? No, $A$ contains $1$, $2$, and the set $\{3\}$ as an element. - Since $3 \notin A$, $\{3\} \subseteq A$ is **false**. 4. **Check $N \cup A = N$:** - $N$ contains natural numbers. - $A$ contains $1$, $2$, and $\{3\}$. - $1$ and $2$ are in $N$, but $\{3\}$ is not a natural number. - So $N \cup A$ contains $\{3\}$ which is not in $N$. - Therefore, $N \cup A \neq N$, so this is **false**. 5. **Check $A \setminus \{3\} = A$:** - $\{3\}$ contains the element $3$. - $A$ does not contain $3$ as an element, only $\{3\}$. - So removing $3$ from $A$ does nothing. - Hence, $A \setminus \{3\} = A$ is **true**. 6. **Check $A \subseteq N$:** - $A$ contains $\{3\}$ which is a set, not a natural number. - So $A$ is not a subset of $N$. - This is **false**. **Final answer:** The only true statement is $A \setminus \{3\} = A$.