Why does an injection from a set to a countable set imply that set is countable? I'm reading a proof, and it concludes that a set $A$ is countable after finding an injection from $A$ to a countable set. Why is this true? I thought that we need to find a bijection from $A$ to a countable set to prove $A$ is countable.
Shouldn't $A$ be at most countable?
 A: If $B$ is countable denote it $B = \{b_n\}_{n \in \mathbb{N}}$.
If $f : A \to B$ is injective for all $a \in A$ there is a $b_n = a$. Since the map is injective two different elements in $A$ map to different points in $B$, so can you see how to enumerate $A$ now?
A: Unfortunately, there is no uniform agreement to the meaning of "countable". Specifically, does it mean only countably infinite, or do we include also finite sets?
Well. The answer depends on context, convenience, and author. Sometimes it's easier to separate the finite and infinite, and sometimes it's clearer if we lump them together.
A: Use induction! Well, more conveniently, in well ordering-principle form.
Suppose that $f:A\to N'$ is a bijection (basically $N'$ is the range of $A$) where $N'\subseteq \mathbb{N}$. Now we consider elements in $N'$. Take the smallest element in $N'$ (which exists by the well-ordering principle), say $x_1$. Then consider the second smallest element (which exists because $N'\backslash\{x_1\}$ is a set), and call this $x_2$. Repeat with $x_3$, etc. (if we ever run out of elements in $N'$ then we know $A$ is finite which is fine).
Now we know that $f:A\to \{x_i: i\in\mathbb{N}\}$ is a bijection. This is good news, because this is a bijection from $A$ to $\mathbb{N}$ if you think about it carefully. In other words, ordering our set $N'$ from smallest to largest makes it a bijection to $N$.
A: Consider this:
Let $A$ be an arbitrary set, $M$ be a countable set and $f:A \to M$ injective.
It holds that the preimage $f^{-1}(m_i) \subset A$ of each $m_i \in M$ ($i=1,2,...$) is a single pointed set.
In other words: each individual element $m_i \in M$ has an individual preimage $f^{-1}(m_i) = \{a_i\} \subset A$
Since $M = \bigcup_i^n\{m_i\}$ was countable, $$f^{-1}(M) = f^{-1}\left(\bigcup_i^n\{m_i\}\right) = \left(\bigcup_if^{-1}\{m_i\}\right)= \bigcup_{i=1}^n\{a_i\} = A$$ is countable itself.
A: To me, countable means capable of being put in a list.  Thus, any set in bijection with $\omega$, or any finite set, is countable. 
I think it's more about comprehension than definition.   Any one is free to define anything they like (alá Smale), but are there good reasons for doing that?
The other thing to comprehend here, and which the OP indicates he does understand (but then why ask the question) is that an injection from set $A$ to set $B$ indicates that the cardinality of $A$ is less than or equal to that of $B$.
This is just personal preference, I guess,  but I don't find the definition that excludes finiteness from countability as very interesting or exciting.  That is, to reiterate,  I really don't know who in their right mind would exclude finite sets from being considered countable.   After all, countable is a descriptive and suggestive word.  And it suggests that finite should be considered countable. 
If you want to distinguish the two, there is always countably infinite.
But, this is all just my two cents.  If you have some reason why you prefer it the other way, be my guest.   To each his own. 
