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This is a question from Gemignani's Elementary Topology. Here is the question:

Let $X,D$ be a metric space. For each $x,y \in X$, define $H_1(x,y)$ to be $\{ w\in X \, | \, D(x,w) >D(y,w)\}$ and $H_2(x,y)$ to be $\{ w\in X \, | \, D(x,w) <D(y,w)\}$. Prove that $H_1(x,y)$ and $H_2(x,y)$ are open with respect to $D$.

I tried proving it but could not succeed. I noticed that when $X=\mathbb{R}^2$ and $D$ is the Euclidean metric then $H_1(x,y)$ and $H_2(x,y)$ are half planes. In order to show that $H_1(x,y)$ is open, I took an arbitrary $w \in H_1(x,y)$ tried to show that it is an interior point. I conjectured that $N(w,r) \subset H_1(x,y)$ where $r=\frac{(D(x,w))^2-(D(y,w))^2}{2D(x,y)}$ but couldn't prove it.

Hints will appreciated.

ashK
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2 Answers2

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The map $D:(X,D)\times (X,D)\to\Bbb R$ is continuous. Therefore, the map $G(w)= D(x,w)-D(y,w)$ is continuous on $(X,D)$. Those two sets are $G^{-1}[(0,\infty)]$ and $G^{-1}[(-\infty,0)]$.

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This looks like a job for the triangle inequality!

Choose $w \in H_1(x, y)$ and define $k=D(x, w)-D(y, w).$ Then $k \gt 0$ by the definition of $H_1$.

Now consider $B(w, \frac k4)$. You can easily use the triangle inequality and the definition of $H_1$ to prove that $B(w, \frac k4) \subseteq H_1(x, y)$.

The proof for $H_2$ is identical, because $H_2(x, y)=H_1(y, x)$.

Robert Shore
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