For a set $A$, the power set of $A$ is denoted by $2^A$. If $A = \{5, \{6\}, \{7\}\}$, which of the following options are TRUE? I. $\emptyset \in 2^A$ II. $\emptyset \subseteq 2^A$ III. $\{5, \{6\}\} \in 2^A$ IV. $\{5, \{6\}\} \subseteq 2^A$

Given $A = \{5, \{6\}, \{7\}\}$. The power set $2^A$ is the set of all subsets of $A$.

1. Statement I ($\emptyset \in 2^A$): Since the empty set $\emptyset$ is a subset of every set, $\emptyset \subseteq A \implies \emptyset \in 2^A$. Hence, statement I is TRUE.

2. Statement II ($\emptyset \subseteq 2^A$): The empty set $\emptyset$ is a subset of any set, including $2^A$. Hence, statement II is TRUE.

3. Statement III ($\{5, \{6\}\} \in 2^A$): The elements $5$ and $\{6\}$ both belong to $A$, so $\{5, \{6\}\} \subseteq A$, which means $\{5, \{6\}\} \in 2^A$. Hence, statement III is TRUE.

4. Statement IV ($\{5, \{6\}\} \subseteq 2^A$): For $\{5, \{6\}\}$ to be a subset of $2^A$, every element of $\{5, \{6\}\}$ must be an element of $2^A$. But $5 \in \{5, \{6\}\}$ is not a subset of $A$ (it is a number, not a set), so $5 \notin 2^A$. Thus, $\{5, \{6\}\}

ot\subseteq 2^A$. Hence, statement IV is FALSE.

Thus, statements I, II, and III only are TRUE.