Theorem: Every neighborhood is an open set.   Proof: Consider a neighborhood $E=N_r(p)$, and let $q$ be any point of $E$. $$\text{There is a positive real number $h$ such that $d(p,q)=r-h$} \tag 1$$ $$\text{For all points s such that $d(q,s) \lt h$, we have $d(p,s) \leq d(p,q)+d(q,s) \lt r-h+h=r$, so that $s \in E$.} \tag 2 $$ Thus $q$ is an interior point of $E$.  Definition: A neighborhood of $p$ is a set $N_r(p)$ consisting of all q such that $d(p,q) \lt r$, for some $r \gt 0$. The number $r$ is called the radius of $N_r(p)$.  I know from definition, to prove a set $E$ is open, we need to show every point of $E$ is an interior point of $E$. I still don't get those tagged equations as to why.
Previously, someone posted his/her proof of "every neighborhood is open" . I am focusing on the same statement, but it seems that my approach is different from his/hers. Could someone comment on my own proof? Here we go:  Let $X$ be a metric space and $p\in X$. Fix $r&gt;0$. Define a neighborhood around $p$ as $N_r (p)=\{q\in X| d(p, q)&lt;r\}$. Now, take $w = \frac{r}{2}$. Then, $N_w (p)\subset N_r(p)$. In other words, every point of $N_r(p)$ is interior. Therefore, $N_r (p)$ is open. Since $r$ is arbitrary, this proves every neighborhood is open.  Update:  Under my definition of "neighborhood," can I prove in this way instead: For all $m\in N_r(p)$, take $w=\frac{r−d(m,p)}{2}$. Then, $N_w(m)\subset N_r(p)$.   Update2  From the hints given in comments, I solve it:  Assume $m\in N_r(p)$ and $q\in N_w(m)$. Take $w=r-d(m, p)$. Then, $d(p, q)\leq d(p, m) + d(m, q)&lt; d(p, m) + r-d(m,p) =r$.   Thanks to @JoséCarlosSantos and @fleablood !
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