TITLE: Generalization of Mathematical Induction?
QUESTION [6 upvotes]: Yesterday one of my friend and I was discussing about the Principle of Mathematical Induction. After some time he made the following,

Theorem. Let $\emptyset \subset X\subseteq Y$ and $f:X\to Y$ be a function such that,

for all $y\in Y\setminus X$ there exists $x\in X$ such that $f(x)=y$.

$f(X)\subseteq X$.


Then $X=Y$.
Sketch of the Proof of the Claim. If $X=Y$ then we have nothing to prove. Otherwise suppose that $X\subset Y$. Then there exists $y\in Y\setminus X$. But by the first property we can conclude that there exists $x\in X$ such that $f(x)=y$. By the second property we can conclude that $y(=f(x))\in X$, a contradiction. So we are done.

Now the interesting thing is that from the above theorem we can prove the so called Weak Principle of Mathematical Induction by taking $Y=\mathbb{N}$ and $X\subseteq \mathbb{N}$ such that $1\in X$ and by taking $f:X\to \mathbb{N}$ defined by $f(n)=n+1$ for all $n\in X$.
The above observation suggests that the above theorem is more general than the Principle of Mathematical Induction. But I couldn't find any literature regarding this type of 'induction'. The question is,

Is this type of 'induction' well known? If so, can some literature regarding this be mentioned?

REPLY [6 votes]: Like Hayden pointed out in the comments, your theorem doesn't really have much to do with mathematical induction. Notice that the first hypothesis of your theorem says that $Y\backslash X \subseteq f(X)$. So the two hypotheses of your theorem combined says that 
$$Y\backslash X \subseteq f(X) \subseteq X.$$
Your theorem effectively says that if $X\subseteq Y$ and $Y\backslash X\subseteq X$, then $Y=X$. It has nothing to do with $f$ in particular. This is not a statement about induction or even mappings, but rather a statement about set inclusions.