Sachchidanand Prasad

01-08-2024

Problem: Let $(X,d)$ be a metric space and let $f: X \to X$ be a function such that \[ d(f(x), f(y)) \leq d(x,y)\quad \forall\ x,y \in X. \] Which of the following statements is necessarily true?
  • $f$ is continuous
  • $f$ is injective
  • $f$ is surjective
  • $f$ is injective if and only if $f$ is surjective
topology metric-space CSIR-NET-FEB-2022
Solution: We will verify each options.
  • $f$ is continuous.

    This is TRUE. We recall that $f$ is continuous if given any $\varepsilon >0$ there exists $\delta >0$ such that \[ d(x,y) < \delta \implies d(f(x),f(y) < \varepsilon ). \] Let $\varepsilon >0$ be given. Take $\delta = \varepsilon $. Then we have \begin{align*} d(f(x), f(y)) \leq d(x,y) < \delta = \epsilon . \end{align*} Thus, $f$ is continuous function.

  • $f$ is injective.

    This is FALSE. For example, take $X = \mathbb{R} $ with the Euclidean metric. Consider the constant function $f(x) = 0$ for any $x \in \mathbb{R} $. Then it satisfies the given hypothesis as \[ d(f(x), f(y)) = d(0,0) = 0 \leq d(x,y). \] But clearly $f$ is not injective.

  • $f$ is surjective.

    This is FALSE. The same example given in the previous option will work.

  • $f$ is injective if and only if $f$ is surjective.

    This is FALSE. The same example given in the previous option will work.