prove the set $x: d(p,x) < d(q,x)$ is open

Question

In trying to prove this, I had R^2 in mind. But the statement to be proved is true for any metric space. I claim for any $x$ in this set, $B(x,r)$ is inside the set too for $$r= min (d(x,p), |(d(x,p) - d(p,q)/2)| $$

I am trying to use triangle inequality to establish that for all $y$ inside the ball, $$d(y,p) < d(y,q)$$ holds, but somehow I can't extract this information. Is this assertion wrong?

Answer 1

Your claim is correct, and has to be proven. Note that your definition of $r$ is too complicated. It is enough to define $$r:={1\over2}\bigl(d(q,x)-d(p,x)\bigr)>0\ .$$ Let $X$ be your metric base space and $\Omega:=\bigl\{x\in X\bigm|d(p,x)<d(q,x)\bigr\}$ the set you are interested in. Consider a point $x_0\in\Omega$. I claim that the open ball $B:=B_r(x_0)$ is contained in $\Omega$.

Proof. Consider an arbitrary point $y\in B$. Then by the triangle inequality we have $$\eqalign{d(p,y)&\leq d(p,x_0)+d(x_0,y)\cr &<d(p,x_0)+r=d(q,x_0)-r \cr &<d(q,x_0)-d(x_0,y)\cr &\leq d(q,y)\ .\cr}$$

Answer 2

If $X$ is your metric space, then the map $f\colon X\longrightarrow\mathbb R~$ defined by $f(x)=d(q,x)-d(p,x)$ is continuous. Since your set is $f^{-1}\bigl((0,\infty)\bigr)$ and $(0,\infty)$ is an open subset of $\mathbb R$, your set is open.