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A more proper statement:

Question: Let $B$ be not finite and $B \subset A$, then $A$ is also not finite.

Suppose that $A$ is finite. Then, by theorem 6.2 in Munkres, there exists a bijection $f: A \rightarrow \{1,2,...,n\}$. By the same theorem, since $B$ is a proper subset, there exists another bijection $g: B \rightarrow \{1,2,...,m\}$ where $m < n$. However, we are given that $B$ is not finite. This is a contradiction as the bijection $g$ cannot exist.

Is this correct?

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For clarity, I inserted question before the theorem that I was trying to prove. In a very solipsist move I phrased everything in such a way that it was clear to me. Sorry. –  emka Sep 5 '12 at 11:35

1 Answer 1

up vote 1 down vote accepted

It is not correct that "My first thought is that if B is not finite then there exists a bijection $f:B→\mathbb{Z}_+$".

If $B = (0,1)$ and $A = \mathbb{R}$, $B$ is infinite but it is well known that there is no bijection between $B$ and $\mathbb{N}$.

The usual definition of finite is that a set $A$ is finite if and only if the $A$ is in bijection with $\bar{n}$ for some $n \in \mathbb{N}$, where $\bar{n} = \{0, 1, ..., n - 1\}$.

A set $A$ is infinite if it is not finite.

Suppose now that $B$ is infinite and $A$ happens to be finite. By definition, there exists a $n \in \mathbb{N}$ such that $f : \bar{n} \rightarrow A$ is a bijection. Then by recursion define $a_0$ to be the least $k \in \bar{n}$ such that $f(k) \in B$. Let $a_{i + 1}$ (as long as $i + 1 < n$) be the least $k$ such that $k > a_i$ and $f(k) \in B$. Then $g$ defined by $g(i) = a_i$ (as long as $a_i$ is defined) is a bijection between $\bar{m}$ for some $m \leq n$ and $B$. Hence $B$ is finite by definition. Contradiction.

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@EMKA Since the question has been edited the first part of my answer is no longer relevant. The second part of my answer is similar to your argument. –  William Sep 5 '12 at 10:37
By second part, do you mean the part below the horizontal line? –  emka Sep 5 '12 at 11:09

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