Originally Posted by

**aliceinwonderland** It seems like you don't need to find a "c" to show the existence of a fixed point of x such that $\displaystyle \Phi (x) = x $.

First, choose a point $\displaystyle x_{1}$ in M and define

$\displaystyle x_{2} = \Phi (x_{1}), x_{3} = \Phi (x_{2}),...,x_{n}= \Phi (x_{n-1})$ for $\displaystyle n \ge 2$.

You might need to mark some points in M (to figure out the points indeed converge) and make sure each $\displaystyle d(x_{k-1}, x_{k}) > d(\Phi (x_{k-1}), \Phi (x_{k})) $ where $\displaystyle k=2,3,...,n$.

Since $\displaystyle \color{blue}\Phi$ is a contractive function, $\displaystyle \color{blue}\{x_{n}\}_{n=1}^{\infty} $ is a Cauchy sequence. Since M is a complete metric space, the sequence has a limit in M and we call it x. A contractive function in a metric space is a continuous, so $\displaystyle \{\Phi (x_{n})\}_{n=1}^{\infty} $ converges. We know that $\displaystyle \{\Phi (x_{n})\}_{n=1}^{\infty} $ is simply $\displaystyle \{x_{n}\}_{n=2}^{\infty} $ whose limit is x. Thus $\displaystyle \Phi (x) = x$.

To show the uniqueness, suppose on the contrary that you have another point $\displaystyle \Phi (y) = y$. Now you can draw a contradiction if you check your distance function formula $\displaystyle d(\Phi (x), \Phi (y)) < d(x, y) $