• 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


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"?

  • $\begingroup$ 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

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