3
$\begingroup$

Let $(X_1,d_1), (X_2,d_2)$ be metric spaces and $X_1\cap X_2=\emptyset$. Define $d_3$ on $X_1\cup X_2$ by $d_3(x_1,x_2)=1$ when $x_1\in X_1 ,x_2\in X_2,$ and $d_3(x,y)=d_1(x,y)$ when $x,y\in X_1,$ and $d_3(x,y)=d_2(x,y)$ when $x,y\in X_2.$ Prove that $d_3$ is a metric on $X_1 \cup X_2$.

We must show $d_3(x,z) \leq d_3(x,y) + d_3(y,z)$ $\forall x,y,z \in X_1 \cup X_2$. Any help would be much appreciated.

EDIT: I should clarify that we should assume $d_1$ and $d_2$ are metrics that produce real numbers less than 1.

$\endgroup$
3
  • $\begingroup$ What happens in the triangle inequality if $x, z\in X_1$ and $y\in X_2$? $\endgroup$
    – anomaly
    Sep 29 '17 at 18:11
  • $\begingroup$ Well, have you tried just writing out all the cases based on which of $X_1$ and $X_2$ your points are in? $\endgroup$ Sep 29 '17 at 20:04
  • $\begingroup$ I suppose that's the best way to go about this. Thank you, Eric. $\endgroup$
    – Darkdub
    Sep 29 '17 at 20:15
1
$\begingroup$

The inequality is valid when either $x,y,z\in X_1$ or $x,y,z\in X_2$. So we need to prove it when one point is in $X_1$ and two are in $X_2$ (the same applies when two points are in $X_1$ and one in $X_2$).

Suppose $x,z\in X_1$ and $y\in X_2$. Then $$ d_3(x,z)\le 1<d_3(x,y)+d_3(y,z)=2 $$ Suppose $x,y\in X_1$ and $z\in X_2$. Then $$ d_3(x,z)=1\le d_3(x,y)+d_3(y,z)=d_1(x,y)+1 $$ Suppose $y,z\in X_1$ and $x\in X_2$. Then $$ d_3(x,z)=1\le d_3(x,y)+d_3(y,z)=1+d_1(y,z) $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.