Problem about subsequence of uncountable sequence.

Asked Apr 01 '19 at 11:37

Active Apr 01 '19 at 11:54

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Let $(a_n )_{n\in\mathbb{N}}$ be a real number sequence with $a_n\ne a_m$ if $n\ne m$.

(a) Show that : $\{(a_{k_n })_{n\in\mathbb{N}} : k_1 < k_2< \cdots\}$ is uncountable. $(a_{k_n })$ is subsequence of $(a_n)$

(b)Show that $\mathbb{N}\times \mathbb{N}\times\cdots=\{(i_1, i_2,\ldots) ∶ i_n\in\mathbb{N}\}$ is uncountable.

$\mathbb{N}$ is natural number.

Hello, I study Analysis alone in another country not use English.

So, I'm limited to ask question around me.

Anyway, I have no clue what I should start to solve this problem (a)

In (b), I know that $\mathbb{N}\times\mathbb{N}$ is countable. And I think that $\mathbb{N}\times(\mathbb{N}\times\mathbb{N})$ also be countable.

I have a question : Don't be that countable set $\times$ countable set is countable?

I'm so DISSY. Please hele me! T.T

edited Apr 01 '19 at 11:54

asked Apr 01 '19 at 11:37

Prater Darrius

Even a countable disjoint union of countable sets is still countable. – Pink Panther Apr 01 '19 at 11:49
Can I ask why that is true? – Prater Darrius Apr 01 '19 at 11:50
The proof is a little complicated. Also, the axiom of choice is needed. I will try to find a reference. https://math.stackexchange.com/questions/603456/prove-that-the-union-of-countably-many-countable-sets-is-countable it is basically analogous to showing that the rationals are countable. – Pink Panther Apr 01 '19 at 11:52
Oh thank you for your help! so isn't problem (b) true ? – Prater Darrius Apr 01 '19 at 11:58
@PinkPanther What does your comment have to do with the question? OP is talking about products not unions. Even a countable product of finite sets can very well be uncountable. – Klaus Apr 01 '19 at 12:10
Oh my gosh, I'm gonna be hazy – Prater Darrius Apr 01 '19 at 12:18
Can I ask why that can be uncountable?? – Prater Darrius Apr 01 '19 at 12:19
@PraterDarrius See e.g. https://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument – Klaus Apr 01 '19 at 12:29
@PraterDarrius $2^{\aleph_0}$ is uncountable. Think of representing the numbers in $[0,1)$ in binary. – saulspatz Apr 01 '19 at 12:49
@Klaus you are right, I read that wrongly. However, one can think of products as disjoint unions by for example $$\mathbb N\times\mathbb N=\prod_{n\in\mathbb N}{n}\times\mathbb N$$ although I agree that this is probably not helpful here... My apologies. – Pink Panther Apr 01 '19 at 13:01
Everyone thanks so much... I'm so late since I study that link... I almost understand it!! – Prater Darrius Apr 01 '19 at 15:16

Problem about subsequence of uncountable sequence.

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