# Suppose that there is an injection $f: \mathbb{Z}^+ \rightarrow X$. Prove that X is an infinite set.

I am trying to prove this by contradiction. Can I somehow do it by restricting the domain and using a corollary that states if X and Y are non-empty finite sets and there exists an injection $f: X \rightarrow Y$ then $\lvert X\rvert \le \lvert Y\rvert$. I am just struggling because most of the theorems I know only apply to finite sets, and the set of integers is not finite.

## 2 Answers

Suppose that $|X|=n$ for some finite $n$. Then, $$X\supset f(\mathbb{Z^+})\supset f(\{0,1,2,\ldots,n\})\implies |X|\geq|f(\{0,1,2,\ldots,n\})|$$ which is a contradiction because by injectivity $f(\{0,1,2,\ldots,n\})$ is a set of $n+1$ (distinct) elements.

• G++d way to solve the case. :-) – mrs Nov 19 '16 at 9:04

If such that function is an injection so $$f:\mathbb{Z}^{+}\to\text{Im}(f)\subset X,~~~~|\mathbb{Z}^{+}|=\infty$$ is an isomorphism so $X$ cannot be a finite set.

• Come by and ping me from the constructive feedback chatroom! – Namaste Nov 18 '16 at 22:58