# Let A = {a, c, 4, {4, 3}, {1, 4, 3, 3, 2}, d, e, {3}, {4}} Which of the following is true?

Let A = {a, c, 4, {4, 3}, {1, 4, 3, 3, 2}, d, e, {3}, {4}} Which of the following is true?

A: {1, 4, 3} and {4, {4}} are both subsets of A.

B: {1, 4, 3} and {4, {4}} are both elements of A.

C: {1, 4, 3} is a subset of A and {4, {4}} is an element of A.

D: None of the above.

I'm not really sure how subsets work since I couldn't find any examples pertaining to this. Would the answer be D? I'm guessing so because {1, 4, 3, 3, 2} is a subset but it's not the exact subset as {1, 4, 3} and the whole concept of {1, 4, 3} being a subset of A is that it's checking if {1, 4, 3} is a contained in the elements of A.

$\{4, \{4\}\}$ is a subset of $A$, since $4 \in A,$ and $\{4\}\in A$, but the set with which it is paired is not a subset of $A$, and none of the items listed as "elements of $A$ are, in fact, elements of $A$.
For example, $\{1, 4, 3\} \not \subset A$ because $1 \notin A$ (nor is $3 \in A$). But it also the case that $\{1, 4, 3\} \notin A$ because $A$ does not contain the set $\{1, 4, 3\}$ as an element. Hence, none of the options $A),\, B),\,$ or $\, C)$ are true.
That leaves only $(D)$ as being true.
Be careful though: $\{1, 4, 3, 3, 2\} = \{1, 2, 3, 4\} \in A$, but $\{1, 4, 3, 3, 2\} = \{1, 2, 3, 4\} \not \subset A$. That is, the set itself is an element in $A$. It is true, however that $\{\{1, 4, 3, 3, 2\}\} = \{\{1, 2, 3, 4\}\} \subset A$. That is, the subset of $A$ containing the element $\{1, 4, 3, 3, 2\}$ is a subset of $A$.