Contraction Mapping, why constant and weak inequality? From Wikipedia, a contraction mapping is a function $f: M \rightarrow M$ on a metric space $(M,d)$ such that there exists a nonnegative real number $k<1$ such that for all $x,y\in M$, 
$$
d \left(f(x),f(y)\right) \leq k \cdot d(x,y). \tag{1}
$$
I have a question that's probably pretty basic since I feel that I just have the wrong intuition for it right now.
Why could the above definition not just be replaced (unless it can be) with a simpler definition like this?:
A function $f: M \rightarrow M$ on a metric space $(M,d)$ is a contraction mapping if and only if 
$$
d \left(f(x),f(y)\right) < d(x,y) \tag{2}
$$
for all $x,y\in M$ (replacing the weak inequality and constant $k$ with just a strict inequality)? This is probably just an intuitive misunderstanding, but to me, it seems like that a contraction map is just a function where given two points, they always get mapped by $f$ to two points that are even closer together.
 A: The main conceptual difference is this: in a contraction, we want that $k$ to be the same, no matter what x and y are. If you just use $d(f(x),f(y))<d(x,y)$ (whenever $x<>y$), then you have a $k$ for each pair -- when this happens, we call f a WEAK contraction.
So $y(x)=x^2$ defined for $0<=x<=1/2$ is a weak contraction, but it is not a contraction!
A: In order to answer "why is this definition the way it is?" it helps to ask a more basic question: "what is this definition good for?" 
The property $d(f(x),f(y))\le kd(x,y)$ (for all $x,y $) implies the existence of a unique fixed point of $f$, provided that $f:X\to X$ is a map on a complete metric space $X$. (As Daniel Fischer said in a comment.) In other words, the equation $f(x)=x$ has a unique solution. This is fantastic, getting the existence of a  solution  of an equation (where $x$ may well be a vector or a function) from  a mild assumption. This theorem is currently high on the list of overpowered mathematics results, whatever that means.
The property $d(f(x),f(y))<d(x,y)$ (for all $x\ne y$) implies the uniqueness of a fixed point, should it exist. It does not imply existence, and uniqueness with no clue to existence is not nearly as useful. For a concrete example, 
$$f(x)=x-\arctan  x$$
is a map from $\mathbb R$ to $\mathbb R$ which satisfies this property, but has no fixed points.
A: please note that, the contraction mappings are mainly used for obtaining the fixed point of mappings, which can be further applied to obtain the existence of solutions of differential equations, integral equations etc.. Note that, the condition $$d(f(x),f(y))\leq kd(x,y) \; \forall x,y\in X$$ implies both the existence and uniqueness of fixed point, and so, the existence of solution of various problems including the differential equations, integral equations etc.. But the condition $$d(f(x),f(y))<d(x,y) \; \forall x,y\in X,x\neq y$$ can not assure the existence of solution (fixed point). The second condition can be used for existence, but with some additional condition. Although, you can use any of the conditions as your requirements. 
Best Wishes 
