The question I'm stuck on is the following:

Let $(X,d)$ be a metric space and let $Y$ be a subset of $X$. If $x\in X$, define the distance $d(x,Y)$ as $\inf\{(d(x,y):y\in Y\}$. Show that the mapping from $X$ to $\mathbb{R}:x\rightarrow d(x,Y)$ is Lipschitz, i.e. that there exists a constant $C>0$ such that $|d(x,Y)-d(x',Y)|\le Cd(x,x'), x,x'\in X$.

I'm quite lost as to how to approach it because there is no upper bound for $d(x,x')$ and thus the left side can easily go off to infinity. How can I relate distance between two points and the difference in their distances to $Y$ in a way that one constant works for the entire set?

  • 2
    $\begingroup$ $C=1$ works. A picture may help. $\endgroup$ – Jochen Sep 22 at 16:34

Let $x,y \in X$

then $\forall z \in Y$ we have $d(x,Y) \leq d(x,z) \leq d(x,y)+d(y,z)$

Thus $$d(x,Y) \leq d(x,y)+d(y,Y) \Longrightarrow d(x,Y)-d(y,Y) \leq d(x,y)$$

Now similarly

$\forall z \in Y$ we have $$d(y,Y) \leq d(y,z) \leq d(x,y)+d(x,z)\Longrightarrow d(x,Y) \leq d(x,y) +d(x,Y)$$

So $d(y,Y)-d(x,Y) \leq d(x,y)$

Combining the above we have that $|d(y,Y)-d(x,Y) |\leq d(x,y)$

  • $\begingroup$ Thank you so much! $\endgroup$ – cruijf Sep 22 at 16:50
  • $\begingroup$ You are welcome. $\endgroup$ – Marios Gretsas Sep 22 at 16:51

We have $d(x, x')+d(x', y) \ge d(x, y)\ge d(x, Y)$ for any $y\in Y$.
So, taking infimum for $y\in Y$, we get $$d(x, x') +d(x', Y) \ge d(x, Y)$$ So, $d(x, Y)-d(x', Y) \le d(x, x')$.
By symmetry, we also get $d(x',Y) - d(x,Y)\le d(x, x')$.


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