Is this function a bijection? $f: \Bbb N \to P(\Bbb N)$ be given by $f(n) = \{n+1,n+2,n+3,\ldots\}.$
From general intuition and reasoning I think the function is not injective here is my work.
If $n = 1$
$f(1) = \{2,3,4,5,\ldots\}$


If $n = 2$ 
$f(2) = \{3,4,5,\ldots\}$
Since, $f(2)$ is a subset of $f(1)$ the function is not injective.
 A: The fact that $f(2) \subseteq f(1)$ does not show the function is not injective.
A function is injective if for $n,m \in \Bbb N$ you have $n\neq m \implies f(n) \neq f(m)$, so to show a function is not injective you would have to find two different numbers that have the same image under $f$ (you have one containment, not equality).
The function in this exercise is actually injective. It's easier to show this using the contrapositive of the implication in the definition:  $f(n)=f(m) \implies n=m$.
Try writing out $f(n)$ and $f(m)$ as sets and proving the implication above.

EDIT: I see in the title you're wondering if it's a bijection. To show the function is not surjective, find a set in $P(\Bbb N)$, a subset of $\Bbb N$, that is not of the form $\{n+1, n+2, \dots\}$.
A: I would have put this as a comment but I don't have the reputation. If I understand you correctly, you have a function mapping a set N into its power set. From Cantor's theorem the cardinality of a power set is strictly greater than the original set therefore no bijection exists. 
