0

how can we prove countable function like:

If $f : X \to Y$ is a function and $A$ is a countable subset of $X$, then $f(A)$ is countable.

Asaf Karagila
  • 405,794
qwer tyui
  • 419
  • The function must be one-to-one and onto. If it is then f(A) should be countable by definition. – Ziad Fakhoury Mar 10 '17 at 16:59
  • here A is a countable subset of X – qwer tyui Mar 10 '17 at 17:02
  • @ZiadFakhoury: It is true no matter which properties $f$ has, as long as "countable" means "finite or countably infinite". – hmakholm left over Monica Mar 10 '17 at 17:02
  • 2
    @qwertyui: What is the precise definition of "countable" you're working from? There are several possible ones, but proving it from one you're not already familiar with will lead to a significant detour for a conclusion as simple as this. – hmakholm left over Monica Mar 10 '17 at 17:05
  • I am more than a 100% sure that sometime, not long ago, this exact question was asked. And in any question, this is an almost duplicate. – Asaf Karagila Mar 10 '17 at 17:06
  • Note that a set $E$ is countable if there is a map $g:\mathbb{N}\rightarrow E$ such that $g$ is onto, that is; the cardinal number of $E$ is at most the same as $\mathbb{N}$. So, our goal is find a onto map from $\mathbb{N}$ to $f(A)$.

    Since $A$ is countable, there is a map $g:\mathbb{N}\rightarrow A$ such that $g$ is onto. Also, $f:A\rightarrow f(A)$ is onto since $f(A)$ is the collection of all $f(x)$ for $x\in A.$ Thus, we consider the composition map $f\circ g: \mathbb{N}\rightarrow f(A)$. Since both $f$ and $g$ are onto, $f\circ g$ is also onto, which implies that $f(A)$ is countable.

     – Pei-Lun Tseng Mar 10 '17 at 17:24

2 Answers2

1

Since $A$ is countable we have an injective function $h:A\to\mathbb N$.

For every $y\in f(A)$ choose some $x_y\in A$ such that $f(x_y)=y$.

Prescribe $g:f(A)\to A$ by $y\mapsto x_y$ and note that $g$ is injective.

Now observe that $h\circ g:f(A)\to\mathbb N$ is also an injective function, and we conclude that $f(A)$ must be countable.

edit:

More directly (and without an appeal on choice) function $k:f(A)\to\mathbb N$ prescribed by: $$y\mapsto\min\{h(x)\mid f(x)=y\}$$ can be shown to be injective.

drhab
  • 153,781
0

We can refer to $f\restriction_A : A \to f(A)$. This is a surjective map, so that we may find an injection $g:f(A)\to A$ by mapping any element to something in its fiber. Then, $A$ injects to $\mathbb{N}$ by countability, so their composition is an injection from $f(A)$ to $\mathbb{N}$. Now there are two possibilities:

An injection exists from $\mathbb{N}$ to $f(A)$, in which case we can create a bijection by Schroeder-Bernstein.

No injection exists, which means that $f(A)$ is finite (every infinite set has a countable subset, so that there exists an injection from $\mathbb{N}$ into it).

user326159
  • 3,277
Amin Idelhaj
  • 1,793