# Countability of the set $(0,1)$

I am trying to prove that the set $$(0,1)$$ is uncountable from "A First Course in Analysis by Yau". I have a question about a particular step.

In the text, the result is proved by contradiction. It is supposed that the set $$(0,1)$$ is countable, which it is then written that there must exist a bijection $$f:\mathbb{N}\rightarrow (0,1)$$ (which is ultimately contradicted).

My question is, why does the bijective map have to exist? If we suppose that $$(0,1)$$ is countable, shouldn't there exist an injective map $$g:(0,1)\rightarrow\mathbb{N}$$?

• Well, if $f$ is a bijection, what is $f^{-1}$? – Asaf Karagila Jun 5 '20 at 13:12
• @AsafKaragila meaning what is the map from $(0,1)\rightarrow\mathbb{N}$? – Steven Jun 5 '20 at 13:16
• Meaning what kind of object is that? – Asaf Karagila Jun 5 '20 at 13:16
• @AsafKaragila The preimage of $f$? – Steven Jun 5 '20 at 13:17
• Be careful as to whether "Countable" means "countably infinite" or "countable but possibly finite". If you mean "countably infinite", then yes, a bijection must exist between $\mathbb N$ and $X$. – Prime Mover Jun 5 '20 at 13:33

There is a 1-1 correspondence between $$A$$ and $$B$$, if and only if a) there is an injection from $$A$$ to $$B$$ and b) at the same time there is an injection from $$B$$ to $$A$$.
So, if you can demonstrate that injection from $$(0,1)$$ to $$\mathbb N$$, then yes, you have demonstrated that $$(0,1)$$ is countable -- but more, you have proved it to be '''countably infinite'''.
• If there is an injection from $A$ to $\mathbb N$, then that proves that the cardinality of $A$ is at most countably infinite. If there is an injection from $\mathbb N$ to $A$, that proves that $A$ is at least countably infinite. So if there is an injection both ways, that is, that there is a bijection, that proves that $A$ is exactly countable infinite. – Prime Mover Jun 5 '20 at 13:26