WTS in a metric space $X$, $B = \{q \in X | d(p, q)> \delta \}$ is open without using the fact that the complement of a closed set is open

Question

As the title says, I want to show in a metric space $X$, $B = \{q \in X | d(p, q)> \delta\}$ is open without using the fact that the complement of a closed set, which is $ \bar{A} = \{ q \in X | d(p, q)\leqslant \delta\}$ in this case, is open.

I will first try to explain how I did. We can say $\forall x \in B, \exists N_{d(p,x)-\delta}(x)=\{y \in X| d(x,y) < d(p,x) - \delta \}$. I wanted to show $N_{d(p,x)-\delta}(x) \subset B$ so that every $x \in B$ is an interior point. Because $d(x,y) < d(p,x) - \delta \Leftrightarrow d(x,y) + \delta < d(p,x)$, if we show that $d(x,y) + \delta < d(x,y) + d(y,p) \leqslant d(p,x)$, every $x$ is an interior point since $d(x,y) + d(y,p) \leqslant d(p,x)$ is true by triangle inequality.

However, I cannot figure out how to show $\delta < d(y,p)$. If I just say $\delta < d(y,p)$ is right beacuse $y \in B$ and thus $d(y,p) > \delta$, it seems like I am using what I have to prove to prove.

Thanks in advance!

Answer 1

Let $q \in B$. So $$d(p,q) > \delta \tag 1$$

So we can define $$r=d(p,q)-\delta > 0\tag 2$$

The claim is that $B(q,r) \subseteq B$, showing that $q$ is an interior point of $B$.

So let $q' \in B(q,r)$, so $d(q,q') < r$. If (assume for a contradiction) $q' \notin B$ we'd have that $d(q',p) \le \delta$ and then we'd have

$$d(p,q) \le d(p,q') + d(q,q') \le \delta + d(q,q') < \delta + r\tag 3$$

but applying the definition $(2)$ we see that $$r+\delta = d(p,q)$$

so from (3) we get the contradiction $d(p,q) < d(p,q)$. This shows that $q' \in B$ and $B(q,r) \subseteq B$ as claimed.

Answer 2

If $q\in X$ and $y\in B_{d(x,q)-\delta}(q)$, then $d(x,q)\leqslant d(x,y)+d(y,q)$ and therefore\begin{align}d(x,y)&\geqslant d(x,q)-d(y,q)\\&>d(x,q)-\bigl(d(x,q)-\delta\bigr)\\&=\delta.\end{align}