Properties of an enumeration of $\mathbb{Q}$ Multiple select correct options.
Let $(a_n)$ be a sequence where all rational numbers are terms(and all terms are rational).Then
a)-no subsequence of $(a_n)$ converges.
b)-there are uncountably many convergent subsequences of $(a_n)$.
c)-every limit point of $(a_n)$ is a rational number.
d)-no limit point of $(a_n)$ is a rational number.
I think that B is correct because for any irrational number $j$ there is a sequence of rationals converging $j$.And since there are uncountably many irrational numbers there are uncountably many convergent subsequences.
c) is incorrect because there exists the sequence $(1+\frac{1}{n})^n$ of rational numbers(A subsequence of $(a_n)$ which converges to $e$-An Irrational number.
Can someone help me out.
Please provide reasons for each for the 4 Options as to conclude-why are they CORRECT or INCORRECT.
Is the explanation for b) correct.
Help me out with a,d..
 A: $\Bbb Q$ is dense in $\Bbb R$ . This is enough to solve your problem.
(a) incorrect . Consider any $b \in \Bbb R$ then due to the above property of $\Bbb Q$, there exists a sequence of distinct rational numbers say, $\{b_n\}_{n=0}^\infty$ such that $b_n \to b$ as $n \to \infty$. 
 $\{b_n\}_{n=0}^\infty$ is a converging subsequence $\{a_n\}_{n=0}^\infty$
(b)correct. Just by the previous argument, since for each $b \in \Bbb R$ , $\exists$ a sequence of distinct rational numbers say, $\{b_n\}_{n=0}^\infty$ such that $b_n \to b$ as $n \to \infty$.Combining it with the fact that $\Bbb R$ is uncountable.
(c)incorrect. Our choice of $b \in \Bbb R$ can be an irrational!
(d)incorrect. Our choice of $b \in \Bbb R$ can be an rational!
EDIT :
I write in details the construction of $\{b_n\}_{n=0}^\infty$ that I had in mind while posting my answer. 
Note that $ \overline{\Bbb Q} = \Bbb R$ $\ldots (*)$ 
Let $b \in \Bbb R$ . Consider a monotonically decreasing sequence of positive real numbers (say $\{\epsilon_n\}_{n=0}^\infty$) converging to 0 i.e. $ \epsilon_i >0$ $ \forall i \geq 0$ $$\epsilon_0 > \epsilon_1 > \epsilon_2 > \ldots >\epsilon_r>\ldots \text{ and } \epsilon_n \to 0 \text{ as } n \to \infty$$ Choose $b_0$ to be a rational number in $[b-{\epsilon_0},b-{\epsilon_1}]$ . It follows from the definition of $\{a_n\}_{n=0}^\infty$ that $\exists$ unique $k_0 \geq 0$ so that $b_0 = a_{k_0}$ . 
Next, consider $b_1$ to be a rational number in $[b-{\epsilon_1},b-{\epsilon_2}]$ so that it is an element of $\{a_n\}_{n=0}^\infty$ say, $a_{k_1}$ so that $k_1 > k_0$ .
We argue that such a choice of $k_1$ is valid. Since, otherwise it would imply that all rational numbers of $[b-{\epsilon_1},b-{\epsilon_2}]$ appear left to $a_{k_0}$ i.e. $[b-{\epsilon_1},b-{\epsilon_2}] \cap \Bbb Q$ (an infinite set) $\subseteq \{a_0,a_1,\ldots,a_{k_0}\}$ (a finite set) , a contradiction arises! 
Similarly, consider $b_j$ to be a rational number in $[b-{\epsilon_j},b-{\epsilon_{j+1}}]$ so that it is an element of the form $a_{k_j}$ so that $k_j > k_{j-1}$ , $\forall j \geq 1$ . Continuing in this way, we construct $\{b_n\}_{n=0}^\infty$ and it follows directly from definition that $\{b_n\}_{n=0}^\infty$ is a subsequence of $\{a_n\}_{n=0}^\infty$ and $b_n \to b \text{ as } n \to \infty$ . Note that this construction is valid due to $(*)$ .
