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I've got some difficult for understanding Theorem 2.14 in baby rudin.

Theorem 2.14. Let $A$ be the set of all sequences whose elements are the digits 0 and 1. This set $A$ is uncountable.

The elements of $A$ are sequences like 1, 0, 0, 1, 0, 1, 1, 1, ...

Rudin gave a proof looks correct.

But with Theorem 2.12, I figure out a proof that gets a opposite conclusion.

Theorem 2.12. Let $\{E_n\}, n=1, 2, 3,...,$ be a sequence of countable sets, and put $S=\bigcup_{n=1}^{\infty}E_n$. Then S is countable.

Collary. Suppose $A$ is at most countable, and, for every $\alpha\in{A}$, $B_\alpha$ is at most countable. Put $T=\bigcup_{\alpha\in{A}}B_\alpha$. Then $T$ is at most countable.

And my proof for the set $A$ in Theorem 2.14 is countable:

My proof: Let $A_n$ be the set of length $n$'s sequences whose elements are digits 0 and 1. $A_n$ is at most countable. So, with Theorem 2.12, $A=\bigcup_{n=1}^\infty{A_n}$ is at most countable. That means the set of all sequences whose elements are the digits 0 and 1 is countable.

I know I definitely made a mistake, but where is it?

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    $\begingroup$ You mistake is that you are only counting sequences which are eventually zero. $\endgroup$ Jun 26, 2017 at 10:45
  • $\begingroup$ could you explain more? @uniquesolution $\endgroup$
    – Johnny Ji
    Jun 26, 2017 at 10:56

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In $\bigcup_{n=1}^\infty{A_n}$ are only sequences of finite length ! For example the sequence $(1,1,1,....)$ is not in $\bigcup_{n=1}^\infty{A_n}$

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  • $\begingroup$ I think the notation $\infty$ means infinite length... $\endgroup$
    – Johnny Ji
    Jun 26, 2017 at 10:50
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    $\begingroup$ No. It means the union of all sequences of finite length $\endgroup$
    – Fred
    Jun 26, 2017 at 10:52
  • $\begingroup$ In Definition 2.9 in baby rudin, it says "The symbol $\infty$ merely indicates that the union of a countable col­lection of sets is taken", why not infinite length? $\endgroup$
    – Johnny Ji
    Jun 26, 2017 at 10:56
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    $\begingroup$ To say the sequence $(1, 1, 1, \ldots)$ is in the union, is to say that there exists an $n \in \mathbb{N}$ with $(1, 1, 1, \ldots) \in A_n$. It therefore means that $(1, 1, 1, \ldots)$ is of (finite) length $n$. This is absurd! Therefore it cannot belong to any $A_n$, so it's not in the union. $\endgroup$ Jun 26, 2017 at 12:16
  • $\begingroup$ That make sense. Thanks very much. $\endgroup$
    – Johnny Ji
    Jun 26, 2017 at 17:58
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The union of all $A_n$'s is the set of all finite seqeunces of $0$'s and $1$'s.

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  • $\begingroup$ The notation $\infty$ means all the integers, doesn't it? $\endgroup$
    – Johnny Ji
    Jun 26, 2017 at 10:53
  • $\begingroup$ @JohnnyJi The notation $\bigcup_{n=1}^\infty A_n$ means $A_1\cup A_2\cup A_3\cup\ldots$ Does this answer your question? $\endgroup$ Jun 26, 2017 at 11:04
  • $\begingroup$ $\bigcup_{n=1}^\infty{A_n}$ means $A_1\bigcup{A_2}\bigcup{A_3}\bigcup...\bigcup{A_\infty}$, is this right? $\endgroup$
    – Johnny Ji
    Jun 26, 2017 at 13:44
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    $\begingroup$ @JohnnyJi No, you're wrong. It means what I wrote: $A_1\cup A_2\cup A_3\cup\ldots$ Besides, what is $A_\infty$? Where did you define it? What makes you think that it is countable? $\endgroup$ Jun 26, 2017 at 14:11
  • $\begingroup$ Yeah, it makes sense. I've checked the definition of $\bigcup_{n=1}^\infty{A_n}$, you're right, there is no $A_\infty$, it only means that the union of a countable collections of sets. But still, I don't understand why the set of all positive integers doesn't include the infinite posivite number. $\endgroup$
    – Johnny Ji
    Jun 26, 2017 at 16:59
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You are counting only sequences of finite length. Rudin considers sequences of infinite length.

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Your set $A_7$ consists of sequences of length $7$. Rudin's set $A$ had no sequences of length $7$. So your union is not a subset of $A$.

Does your union involve a set $A_\infty$, sequences of infinite length? if so, then you are wrong when you say $A_n$ is at most countable, since you have not proved that $A_\infty$ is at most countable.

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