# If $f: A \to B$ is injective and $B$ is countable, then so is $A$. [closed]

I would appreciate any hint or idea to prove this, thank you.

• What's your definition of "countable"? Some people take this to mean "countably infinite" and some intend "finite or countably infinite". Depending on which definition you intend, this statement could be true or false. – Rick Decker Sep 13 '13 at 0:46

I guess in your source being countable also includes being finite, in some sources like Walter Rudin's principles of mathematical analysis, they use the term 'at most countable' and when they talk about countability they exclude the case when the set is finite.

HINT: Every subset of a countable set is countable. Now think about $f(A)$ and use injectivity of $f$ to create your bijection.

HINT

Create an enumeration of $A$ from your enumeration of $B$.

It really depends on exactly how you’ve defined countable. One definition is that $B$ is countable if and only if there is an injection $\varphi:B\to\Bbb N$. If that’s your definition, consider the map $f\circ\varphi$.

Any subset of a countable set has a least element.

• That's not true. $\mathbb{Z}$ is a countable set. But assuming the standard ordering on $\mathbb{Z}$ there is no least element in $\mathbb{Z}$. – user66733 Sep 12 '13 at 22:33
• But we are assuming OP is asking about a set where there IS a well-ordering ... – Don Larynx Sep 12 '13 at 22:37
• No, we are not. There is no need for a well-ordering. Your previous answer was OK. – user66733 Sep 12 '13 at 22:38