Show that the function is a bijection.

Question

Let $A = \lbrace 1,2,6, 24, 120, .... \rbrace $. We define the following map $\alpha: \mathbb{N} \rightarrow A, \; \alpha(a) = a!.$ I'm not able to show by the definition that this map is bijective. I tried but it seems to be very simple, I could not visualize maybe a property. A little help would be very helpful

Answer 1

I'm assuming by the obvious pattern that you're defining

$$A=\{n! : n\in\mathbb{Z}^+\}.$$

So we're considering the mapping

$$\alpha : \mathbb{Z}^+\to A,\quad n\mapsto n!.$$

To show that $\alpha$ is bijective, we show that it's injective and surjective. Injectivity we can easily prove by noticing that

$$(n+1)!=n!\cdot\underbrace{(n+1)}_{>1}>n!$$

for $n\in\mathbb{Z}^+$, from which it follows easily that if $n>m$, then

$$\alpha(n)>\alpha(m),$$

which proves injectivity. Now for surjectivity, if $a\in A$, then $a=n!$ for some $n\in\mathbb{Z^+}$ by definition, and so $a=\alpha(n)$, which proves surjectivity.