if $\displaystyle X$ is any infinite set show that $\displaystyle |X| + |\mathbb{N}| = |\mathbb{N}|$
thats it thanks
Not true, with the wording you have $\displaystyle X=\mathbb{R}$ is permitted, and
$\displaystyle |\mathbb{R}| + |\mathbb{N}| = |\mathbb{R}| \ne |\mathbb{N}|$
May be you mean:
if $\displaystyle X$ is any finite set show that $\displaystyle |X| + |\mathbb{N}| = |\mathbb{N}|$
CB
Well, that depends on what you know about infinity.
If you know there exists a $\displaystyle B \subset X$ such that $\displaystyle |B|=|\mathbb{N}|$, you can find a bijection between $\displaystyle B\cup \mathbb{N}$ and $\displaystyle \mathbb{N}$ (that just means $\displaystyle B\cup \mathbb{N}$ is countable), and you've won.
Indeed, you get $\displaystyle X \equiv (X-B)\cup B \equiv (X-B)\cup \mathbb{N} \equiv (X-B)\cup (B\cup \mathbb{N}) \equiv X\cup \mathbb{N}$
(here $\displaystyle \equiv$ means "equipotent to")