Let $B$ be a countable set and let $f$ be a surjective function ($f:A\to B$), then $A$ is also countable

Question

Question: Given a countable set $B$ and a function $f:A\to B$ which is surjective, then $A$ is also countable.

I think it's a false affirmation, but I have no idea of a counter example I can use here.

I basically know that: every subset $X \subseteq N$ is a countable set. And the corollary: given a countable set $A$ and a function $f:A\to B$ which is surjective on $B$, then $B$ is also countable.

Answer 1

Take $f:\Bbb R \to \Bbb N$ defined as

$$f(x) = \begin{cases} x \text{ if } x \in \Bbb N\\ 1 \text{ if } x \not\in \Bbb N \end{cases}$$

Edit: I remembered that there's no concensus on the use of "countable". I've assumed it meant "countable infinite", that is, the same size of $\Bbb N$. But it's also used to denote "finite or of the size of $\Bbb N$", in which case a simpler example would be $f:\Bbb R \to \{0\}$ given by $f(x)=0$ for all $x$.

Answer 2

Consider the function $f: \mathbb{R} \longrightarrow \mathbb{N} \cup \{0\}$ given by $f(x) = \lceil |x| \rceil$. This is clearly a well defined function since if $x_1=x_2$, then $|x_1|=|x_2|$ and thus $\lceil |x_1| \rceil = \lceil |x_2| \rceil$.

Further, given any $y \in \mathbb{N}$, we can surely find $x \in \mathbb{R}$ such that $\lceil |x| \rceil = y$. Namely, $x=y-0.5$. If $y=0$, then $f(0)=0$. Hence $f$ is surjective.

However, $\mathbb{R}$ isn't countable.

Answer 3

Simple counter:

f: R--->{x} be a function such that f(y) = x for all y in R

Co domain is a singleton set ,so it is countable.

But domain is set of all real numbers which is uncountable.