We are given that the set $A = \{1, \{\emptyset\}, \{1, \emptyset\}\}$. Now, let's evaluate each of the statements:

1. Statement (a): $\{\emptyset\}, \{1, \emptyset\} \in A$

   • $A$ contains the element $\{\emptyset\}$ and $\{1, \emptyset\}$.
   • Hence, both $\{\emptyset\}$ and $\{1, \emptyset\}$ are elements of $A$, making statement (a) true.

2. Statement (b): $\{\emptyset\}, \{1, \emptyset\} \subseteq A$

   • For this to be true, each element of the set $\{\emptyset\}, \{1, \emptyset\}$ must be an element of $A$.
   • Both $\{\emptyset\}$ and $\{1, \emptyset\}$ are elements of $A$, so this is true.

3. Statement (c): $\{\{\emptyset\}, \{1, \emptyset\}\} \subseteq A$

   • This statement claims that the set $\{\{\emptyset\}, \{1, \emptyset\}\}$ is a subset of $A$.
   • For it to be true, each element of $\{\{\emptyset\}, \{1, \emptyset\}\}$ must be an element of $A$.
   • Both $\{\emptyset\}$ and $\{1, \emptyset\}$ are elements of $A$, so this is true.

4. Statement (d): $\{1, \emptyset\} \subseteq A$

   • The set $\{1, \emptyset\}$ is a subset of $A$ if both 1 and $\emptyset$ are elements of $A$.
   • The element 1 is in $A$, but $\emptyset$ is not an element of $A$; hence, this statement is false.

5. Statement (e): $\{\{1\}, \{\emptyset\}\} \subseteq A$

   • This statement claims that the set $\{\{1\}, \{\emptyset\}\}$ is a subset of $A$.
   • The set $\{1\}$ is not an element of $A$, but $\{\emptyset\}$ is. Therefore, this statement is false.

Conclusion:

The correct statements are (a), (b), and (c), so the correct answer is:

D) a and c

Would you like more details on this solution or any further clarification?

Here are 5 related questions for further practice:

1. What is the difference between an element of a set and a subset?
2. How would the answer change if $\emptyset$ were an element of $A$?
3. What does it mean for a set to be a subset of another set?
4. How would you represent the power set of $A$?
5. Can a set contain both elements and subsets simultaneously?

Tip: When evaluating membership ($\in$) and subset ($\subseteq$) relationships, always carefully distinguish between sets and their elements.