Let be $X$ a finite set and $Y$ an infinite set, show that exists a surjective function $g: Y \longrightarrow X$

Question

I was trying solve this question, but I didn't solve. I'm trying construct a surjective function $h: Y \longrightarrow I_n$ at the moment, but I don't know if I'm in the right way. Someone can give me a hint to solve this problem? Thank you in advance!

Answer 1

Actually it's not quite true: there is no surjective function from $Y$ to the empty set.

EDIT: Assuming $X$ is nonempty, we proceed by induction on the cardinality $n$ of $X$.

For $n = 1$, let $X = \{x\}$ and take $f(y) = x$ for all $y \in Y$.

Induction step: Suppose the statement is true for $n-1$. If $x$ is one member of $X$, then $X = \{x\} \cup X'$ where $X'$ is a finite set of cardinality $n-1$, not containing $x$. Let $y$ be one element of $Y$. Then $Y = \{y\} \cup Y'$ where $Y'$ is infinite and does not contain $y$. Let $g$ be a surjective function from $Y'$ to $X'$, and define $$ f(t) = \cases{ y & if $t=x$\cr g(t) & otherwise\cr}$$

Answer 2

A construction of such a surjection (if $X$ is not empty !) :

We can assume without loss of generality that $X = \{ 1,2, \cdots, n\}$

Now, as $Y_1 = Y$ is infinite, it's not empty, so there exists an $y_1 \in Y_1$.

As $Y_1$ is infinite, $Y_2 = Y \backslash \{ y_1 \}$ is infinite (if $Y_2$ was finite, it would have $k$ elements ($k \in \Bbb N$), and $Y_1$ would then have $k+1$ elements and would be finite)

But as $Y_2$ is infinite, it's not empty, so there exists an $y_2 \in Y_2$. And $Y_3 = Y_2 \backslash \{y_2\}$ is infinite.

Iterate this up to have constructed $y_{n-1}$ and $Y_n$. Remark that $Y_n = Y \backslash \{ y_1, \cdots, y_{n-1} \}$

Now define $f$ this way :

If $y \in Y_{n}$, then $f(y) = n$. Else, there exists $i\in \Bbb N , 0 < i < n$ such that $y = y_i$ and then define $f(y) = f(y_i) = i$

$f$ is well defined and is surjective by construction

Answer 3

Let $X$ be a finite set and $Y$ be an infinite set. Now there exists a function $f: Y \to \mathbb{N}$ such that $f$ is a bijection. There also exists $g: \mathbb{N} \to X$ such that $g$ is a surjection. Now, $g \circ f: Y \to X$ and is surjective. This works as long as $Y$ is countable. If $Y$ is uncountable, there exists $h: Y \to \mathbb{N}$ such that $h$ is a surjection. Now, $ g\circ h: Y \to X$ is the desired surjection.

Answer 4

First define some arbitrary order on X.

Let $Y$ be your infinite set construct a contiguous set of ordinal numbers $I$ such that $ |Y| = |I| $.

There exists a function $ h : Y \to I $ due to the definition of bijection. Then create a function $ g : I \to X $ that assigns the first $ |X| $ elements of $I$ to elements of $X$ in the order given and the rest to the last element of X. $ g \circ h : Y \to X $ is a surjective function.