# Showing that the set of natural number, $\omega$, is Dedekind infinite

Showing that the set of natural number, $$\omega$$, is Dedekind infinite. It is an easy task to show this directly by sending $$n$$ to $$2n$$, then we produces a injective map that is not surjective.

But suppose I want to make use of this fact that there is a bijection between $$\omega$$ and $$\omega+1 = \omega\ \cup \{\omega\}$$, how might one produce an injective map from $$\omega$$ to $$\omega$$ that is not surjective ?

Cheers and thanks

## 1 Answer

Simply compose a bijection $$f\colon\omega+1\to\omega$$ with the inclusion map $$\omega\subseteq\omega+1$$. In other words, simply restrict $$f$$ to $$\omega$$.

Since $$\omega$$ is a proper subset of $$\omega+1$$, the result will be an injection from $$\omega$$ to itself whose range is not $$\omega$$.

• Ah! Because I need the '+1' to hit every $x \in X$ using $f$, so if I just restrict $f$, then I am bound to miss some guy. It remains injective because restriction does not affect injectivity. – some1fromhell Nov 27 '19 at 15:44