Finding a subset for which the following logical statements hold I don't know much about sets and their operations, yet I have attempted a couple of these questions. But I'd like to have some direction on how to tackle this question. The question is as follows:
Let $S$ be the set of all real sequences. For each of the following statements, find the subset $A$ of $S$ such that the statement is true for all the members of $A$ and the statement is false for all the members of $S\setminus A$.


*

*$\langle a_n\rangle$ converges to $0$ and $\langle a_n\rangle$ converges.

*$\langle a_n\rangle$ converges to $0$ or $\langle a_n\rangle$ converges.

*If $\langle a_n\rangle$ converges to $0$ then $\langle a_n\rangle$ converges.

*$\langle a_n\rangle$ converges to $0$ iff $\langle a_n\rangle$ converges.


My attempt:
Let $C_k = \{\langle a_n\rangle\vert \lim_{x\to\infty}=k\in\mathbb{R}\}$ and $C_0 = \{\langle a_n\rangle\vert \lim_{x\to\infty}=0\}$ 


*

*$A=C_k\cup C_0$

*$A=C_k\cap C_0 = \emptyset$

*$A= C_0^\complement \cap C_k$

*$-$
 A: Your first two answers are incorrect. Think about what the logical statements are asking for. Question 1 wants convergent sequences that converge to 0. So the resulting subset should be smaller than the set $C_k$. In that case, does a union make sense? 
Similarly, for question 2, you got the empty set as a result. But as I'm sure you know, there are lots of sequences that converge to 0, like $(\frac{1}{n})$ and lots of convergent sequences. So you have defined a set that is too restrictive. What set operation would be more appropriate in the place of the intersection? 
A: Maybe something like this? 
Short Answers
Let $A = \{\langle a_n \rangle \in S \, | \lim a_n = 0\}$ and $B = \{\langle a_n \rangle \in S \, | \,\exists \, k \in \mathbb{R} : \lim a_n = k\}$. Note that the way $B$ is formally written simply means that it contains all convergent real sequences.
Furthermore, the set complement $A^c$ of $A$ contains all sequences that do not converge to $0$, and the set complement $B^c$ of $B$ contains all divergent sequences.


*

*$A$ and $S\setminus A = A^c.$

*$A \cup B$ and $S\setminus(A \cup B) = A^c \cap B^c = B^c$

*$A^c \cup B$ and $S\setminus(A^c \cup B) = \emptyset$.

*$(A^c \cup B) \cap (A \cup B^c) = (A \cap B) \cup (A^c \cap B^c)$ and $S\setminus((A\cap B) \cup (A^c \cap B^c)) = A^c \cap B$. 





Long Answers
Problem 1
The first set is $A \cap B$, which contains all convergent sequences that converge to $0$. However, notice that all sequences which converge to $0$ are convergent sequences, and all sequences which converge to $0$ and converge are sequences that converge to $0$. Therefore, $A \cap B = A$. Furthermore, the set $S\setminus(A \cap B)$ contains all sequences that do not converge to $0$ nor converge.
But this means that $S\setminus(A \cap B)$ contains all sequences that either do not converge to $0$ or diverge, so $S \setminus (A \cap B) = A^c \cup B^c.$ However, sequences that do not converge to $0$ or diverge do not converge to $0$ in either case, and sequences that do not converge to $0$ either do not converge to $0$ or diverge. Hence, $A^c \cup B^c = A^c,$ so
$$S\setminus (A\cap B) = A^c.$$


Problem 2
By definition, the set $A \cup B$ satisfies the given condition, which says that this set contains all real sequences $\langle a_n \rangle$ such that $\lim a_n = 0$ or $\lim a_n$ converges. What about $S\setminus (A \cup B)$? Well, this is saying that this set contains all real sequences $\langle a_n \rangle$ such that $\lim a_n \neq 0$ and $\lim a_n$ diverges; or, in short, $S\setminus(A \cup B)$ contains all divergent real sequences. Therefore,
$$S\setminus(A\cup B) = B^c.$$


Problem 3
This one is a bit tricky. One way to tackle this is to recall from logic that $p \implies q \equiv \neg p \lor q$; for this problem, the statement "If $\langle a_n\rangle$ converges to $0$, then $\langle a_n \rangle$ converges" is logically equivalent to the statement "Either $\langle a_n \rangle$ does not converge to $0$ or $\langle a_n \rangle$ converges." We can now formalize this notion using sets. 
Using the set complement $A^c = \{\langle a_n \rangle \in S \, | \langle a_n \rangle \notin A\} = \{\langle a_n \rangle \in S \, | \lim a_n \neq 0\}$, we can form the union
$$A^c \cup B,$$
denoting the set of all sequences $\langle a_n \rangle$ such that either $\langle a_n \rangle$ does not converge to $0$ or $\langle a_n \rangle$ converges.
Further, the set $S\setminus (A^c \cup B)$ contains all real sequences that converge to $0$, but diverge. However, this is ridiculous, so it follows that $S \setminus (A^c \cup B) = \emptyset$. 
Don't be alarmed that the empty set has arisen; you should be able to convince yourself that this was an inevitable situation. To say that the statement $p \implies q$ is false means that $p \text{ and } \neg q$ is true. But this translates to $\langle a_n \rangle$ converges to $0$ and diverges, which is a contradiction. Hence, no sequence in $S$ can make the statement "If $\langle a_n \rangle$ converges to $0$, then $\langle a_n \rangle$ converges" false, and thus the set containing all such sequences is empty. 


Problem 4
Recall that $p \iff q$ means $p \implies q \text{ and } q \implies p$. So the statement, "$\langle a_n \rangle$ converges iff $\langle a_n \rangle$ converges to $0$" is equivalent to the statement, 
$$\text{If } \langle a_n \rangle \text{ converges, then } \langle a_n \rangle \text{ converges to } 0 \text{ and if } \langle a_n \rangle \text{ converges to } 0, \text{ then } \langle a_n \rangle \text{ converges.}$$ 
We already saw what the set looks like for the statement "If $\langle a_n \rangle$ converges to $0$, then $\langle a_n \rangle$ converges," which was $A^c \cup B$. 
A similar idea is employed on the statement "If $\langle a_n \rangle$ converges, then $\langle a_n \rangle$ converges to $0$," which         translates to the statement "Either $\langle a_n \rangle$ diverges or $\langle a_n \rangle$ converges to $0$." 
If we let the set complement $B^c = \{\langle a_n \rangle \in S \, | \langle a_n \rangle \notin B\} = \{\langle a_n \rangle \in S\, | \langle a_n \rangle \text{ diverges}\}$, then the above statement is encapsulated with the set $B^c \cup A$.
Therefore, using the set inclusion for "and", we have the final set 
$$(A^c \cup B) \cap (B^c \cup A),$$
which is actually equal to the set
$$(A \cap B) \cup (A^c \cap B^c),$$
which says that this set contains all sequences $\langle a_n \rangle$ that either converge to $0$ or diverge.
Therefore, the set $S\setminus((A \cap B) \cup (A^c \cap B^c))$ contains all convergent sequences that do not converge to $0$. Consequently, 
$$S\setminus((A \cap B) \cup (A^c \cap B^c)) = A^c \cap B.$$
