On countable sets

Question

I cant quite understand why $S$ is countable. Is it countable because it is mapped to a countable set which is $T$? $T$ i know is countable since it is an infinite subset of a countable set which is the set of positive integers.

Thanks for the help.

I got his from rudin's principle of mathematical analysis.

Answer 1

It is countable for exactly the same reason that Rudin states it is. It's countable because there's a bijection $\mathbb{N} \to S$ which is described as writing these sequences out in rows and going about them diagonally. Eventually, for large enough integers, all members of the union are reached.

Answer 2

Correct me if wrong:

An attempt:

Let $(E_n)_{ n \in \mathbb{N}}$ be a sequence of countable sets.

All elements of $S := \displaystyle \cup E_n$ can be arranged as shown in the array.

The array can be arranged as a sequence:

$x_{11},x_{21},x_{12},x_{31},x_{22},...$

Means we could define $f: \mathbb{N} \rightarrow S:$

$f(1)=x_{11};f(2)=x_{21}; f(3)=x_{12},.$

Note : $f$ may not(!) be injective since any two sets $E_n$ may have the same element. Hence one distinct element may have been counted more than once.

Now:

There is $T \subset \mathbb{N}$ such that

($T$ is countable as a subset of $\mathbb{N}$)

$T \sim S$, i.e. there exists a bijection(!) $g$:

$g: T \rightarrow S.$

Hence :

1)$S$ is at most countable.

2) $E_1\subset S$ , $E_1$ infinite, hence $S$ countable.

Remark:

$f$ is not necessarily an injection.

Hence: Rudin goes back to $T$ to find an injection (bijection) $g: T \rightarrow S.$

Does this help?