1
$\begingroup$
  • Show that the numbers of the form $\sum_{k=1}^{\infty} \frac{a_j}{3^k}$ , where $a_j = 0$ or $a_j = 1$ is countable .

  • If $A = \cap_i^n A_1$ is countably infinite, then atleat one $A_i$ is counntable.

I know that the series $\sum_{k=1}^{\infty} \frac{a_j}{3^k}$ is convergent , because $\sum_{k=1}^{\infty} \frac{a_j}{3^k} \leq \sum_{k=1}^{\infty} \frac{1}{3^k}$, I want to prove that every series is convergent to rational number, but i am unable.

For second statement , I think it is false, please give me any counter example

any help would be appreciated. Thank you

$\endgroup$
2
$\begingroup$

Both statements are false, actually.

For (1): You've already shown that every such series converges; it's not hard to show that any two series which differ at some point (e.g. $a_n\not=b_n$ for some $n$) converge to different reals. Thus, you're really just counting the number of sequences of possible $a_i$s.

Using this, can you construct a bijection between the set of reals which are representable by such a series, and some set you already know is uncountable? (Or, apply a diagonal argument directly.)

For (2): Can you think of two uncountable sets whose intersection is "very small"?

$\endgroup$
4
  • $\begingroup$ For (1) : I think let two series $\sum_{k=1}^{\infty} \frac{a_k}{3^k}$ and $\sum _{k=1}^{\infty} \frac{a_k}{3^k}$ differ finite many points says $a_{n_1}, \dots a_{n_k}$ differ by $b_{n_1} \dots b_{n_k}$, remove these points from the sequences and converges same points, but if two sequence differ by infinite many points, how to prove that they converges at different pints. $\endgroup$ – user120386 Nov 29 '15 at 4:33
  • $\begingroup$ For (2) $A = \{ a+ \iota b : a \in \mathbb N , b\in \mathbb R \}$ and $B = \{ a + \iota b : a \in \mathbb R, b \in \mathbb N \}$ $\endgroup$ – user120386 Nov 29 '15 at 4:35
  • $\begingroup$ Your idea for (2) works (I assume "$\iota$" is "$i$", the (a :P) square root of $-1$) - those are two uncountable sets whose intersection is countable - but it's more complicated than necessary: how about just $A=(0, 1)$ and $B=(2, 3)$? For (1), suppose I have two such sequences $a_i$ and $b_i$, and for the first $i$ where they disagree we have $a_i<b_i$; what can you say about the corresponding sums? $\endgroup$ – Noah Schweber Nov 29 '15 at 5:54
  • $\begingroup$ @ Noah : I have question in my mind : Numbers of the form contains an open inerval of $\mathbb R$ $\endgroup$ – user120386 Dec 14 '15 at 13:56

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.