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Let be $X$ a metric space and $a \in X$. Fix $r > 0$. Show that $\{ x \in X \ ; \ d(x,a) > r \}$ is an open set.

Probably, this question was asked here, but I didn't found it here and I would like to know how to proof this without using sequence, is it possible?

$\textbf{My attempt:}$

Define $s := d(x,a) - r$. Note that if $B \left( x, \frac{s}{2} \right) \ \cap \ B[a,r] = \emptyset$, $\forall x \in A := \{ x \in X \ ; \ d(x,a) > r \}$, then $B \left( x, \frac{s}{2} \right) \subset A$ and follows that $A$ is an open set. Now, I try to prove that $B \left( x, \frac{s}{2} \right) \ \cap \ B[a,r] = \emptyset$.

Given $y \in B \left( x, \frac{s}{2} \right)$, we have

$ r < d(x,a) \leq d(x,y) + d(y,a) < \frac{s}{2} + d(y,a)$

$\Longrightarrow d(y,a) > r - \frac{s}{2} = r - \left( \frac{d(x,a) \ - \ r}{2} \right) = \frac{3r \ - \ d(x,a)}{2}$

$\Longrightarrow d(y,a) > \frac{3r \ - \ d(x,a)}{2}$

but now I don't know what to do. Anyone can help me? Thanks in advance!

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George
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2 Answers2

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This is another aproach:

  1. $d(x,a)-d(y,a)\leq d(x,y)$. By symmetry, $|d(x,a)-d(y,a)|\leq d(x,y)$. So, $d(\cdot,a)$ is continuous.
  2. $d(\cdot,a)^{-1}\left(]r,+\infty[\right)=\{x\in X:d(x,a)>r\}$ is open.
Veridian Dynamics
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Let $S = \{x\in X\,\vert\, d(x,a) > r\}$. We want to show that for any point in $S$, we can find an open ball centered around that point contained in $S$. Thus, let $x\in S$. Then let $\rho = d(x,a)-r$. Note that $\rho > 0$ since $d(x,a)>r$. Now suppose $y\in B(x,\rho)$. Then $$d(y,a) \geq d(x,a)-d(x,y) > d(x,a)-\rho =r$$ and so $y\in S$, thus proving that $B(x,\rho)\subseteq X$, and so we're done.

florence
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