4
$\begingroup$

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..

$\endgroup$
6
$\begingroup$

$\Bbb Q$ 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 $(*)$ .

$\endgroup$
  • 5
    $\begingroup$ As far as I can see, this answer is incomplete as written. The only information we're given about the sequence $(a_{n})$ is that it enumerates the rationals. In particular, with your notation, it's not clear why any $(b_{n})$ converging to $b$ must appear as a subsequence of $(a_{n})$, even if all of the entries of $(b_{n})$ appear in $(a_{n})$. However, it is true that some permutation of the sequence $(b_{n})$ appears as a subsequence of $(a_{n})$, and any permutation of $(b_{n})$ must also converge to $b$. $\endgroup$ – Alex Wertheim Dec 29 '17 at 6:34
  • $\begingroup$ Would you want to post an answer , Alex? $\endgroup$ – Peter Szilas Dec 29 '17 at 7:13
  • $\begingroup$ @Peter: I appreciate the suggestion, but this answer has the idea, even if it is missing a subtlety. If someone feels strongly enough though, they're welcome to post something along these lines. I should also mention that there's one more minor subtlety to address: $a_{n}$ could be a bijection of $\mathbb{N}$ with $\mathbb{Q}$, so one needs to additionally ensure that we can take the entries of $(b_{n})$ to be distinct. But this is easy to verify: if $b$ is rational, take $b_{n} = b - 1/n$; if $b$ is irrational, take the "decimal expansion" of $b$ as a sequence, and subtract $10^{-n}$, e.g. $\endgroup$ – Alex Wertheim Dec 29 '17 at 9:43
  • $\begingroup$ Technically, what is important is that ${\bf Q}$ is dense and it is dense in itself. Well, that and uncountability of the set of reals. $\endgroup$ – tomasz Dec 29 '17 at 10:10
  • $\begingroup$ @AlexWertheim Due to your comments, I've given a detailed argument, please see my answer for further corrections . :) $\endgroup$ – reflexive Jan 9 '18 at 16:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy