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In Rudin's Principle of Mathematical Analysis, in claims that in Theorem 2.8: Every infinite subset of a countable set $A$ is countable. Could someone explain why the function $f: \mathbb{N} \rightarrow E$ is surjective?

Proof Suppose $E \subset A$, and $E$ is infinite. Arrange the elements of $x$ of $A$ in a sequence $\{x_{n} \}$ of distinct elements. Construct a sequence $\{n_{k} \}$ as follows: Let $n_{1}$ be the smallest positive integer such that $x_{n_{1}} \in E$. Having chosen $n_{1}, ... n_{k-1} (k = 2, 3, 4, ...)$, let $n_{k}$ be the smallest integer greater than $n_{k-1}$ such that $x_{n_{k}} \in E$.

Putting $f(k) = x_{n_{k}} (k = 1, 2, 3, ...)$, we obtain a bijection between $\mathbb{N}$ and $E$.

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Please insert the text in question so we can see it. – ncmathsadist Feb 27 '12 at 2:12
up vote 3 down vote accepted

If $A$ is a countable set, there is a bijective map $f:\mathbb{N}\to A$, which is the same as an order on the set: $f(1)$ is the first element, then $f(2)$, etc. If you have an infinite subset $E$ of $A$, we can order $E$ by letting $f(1)$ be the first element of $E$ (first with respect to the ordering on $A$, that is). Since every element of $E$ appears somewhere in the ordering on $A$, it will appear either in that spot or closer to the front in the ordering on $E$. (This last sentence tells you that $f:\mathbb{N}\to E$ is surjective.)

NB: Many books, and mathematicians, use the word countable to include both finite sents and countably infinite sets. It's clear from context here that we're only dealing with countably infinite sets, but you should be careful when looking at other references.

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Very helpful, thanks! – Student Feb 27 '12 at 2:23

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