compact metric spaces and infimum

Question

I am currently revising metric spaces and have come across a question which I am unable to answer and have no idea how to begin with.

Let $(M,d)$ be a compact metric space. Suppose $T \colon M \to M$ is a map such that for any $x,y \in M$ with $x \neq y$, $d(Tx,Ty) < d(x,y)$.

1. By considering the function $e(x) = d(x,Tx)$, $x \in M$ prove that $\inf\{d(x,Tx) : x \in M\}$ is attained
2. Prove that $\inf\{d(x,Tx) : x \in M\} = 0$.

I tried using contraction mappings but I'm not entirely sure if this can be applied here. Any hints would be greatly appreciated.

Answer 1

So let’s do as the question suggests and consider the function $$ e \colon M \to \mathbb{R}, \quad x \mapsto d(x,Tx). $$ Notice that $e$ is continuous (even Lipschitz-continuous) because \begin{align*} |e(x) - e(y)| &= |d(x,Tx) - d(y,Ty)| \\ &\leq |d(x,Tx) - d(y,Tx)| + |d(y,Tx) - d(y,Ty)| \\ &\leq d(x,y) + d(Tx,Ty) \leq d(x,y) + d(x,y) = 2d(x,y). \end{align*} Because $M$ is compact it follows that $e$ attains its minimum on $M$, i.e. there exists some $x \in M$ with $$ d(x,T(x)) = e(x) = \min_{y \in M} e(y) = \min \{d(y,Ty) \mid y \in M\}. $$ That shows the first part. For the second part notice that if $x \neq Tx$ then $$ d(Tx,T(Tx)) < d(x,Tx) = \min\{d(y,T(y)) \mid y \in M\}, $$ a contradiction. So we must have $x = Tx$ and thus $$ \min \{d(y,T(y)) \mid y \in M\} = d(x,Tx) = d(x,x) = 0. $$