Why is point $b$ an adherent point to the interval $(a,b)$?

Question

Explain: Why point $b$ is an adherent point to the interval $(a,b)$.

My answer: Consider the point y expressed as $y=b - \frac{b-a}{n}, n\in \mathbb{N}$. Then $y$ belongs to the set $(a,b)$. The distance between $y$ and $b$ is given as: $$|b-y|=\frac{b-a}{n}$$ Hence, to get $\epsilon$-closeness, choose $n=\frac{b-a}{\epsilon}$.

Hence, $b$ is an adherent point of $(a,b)$.

Is my proof correct?

Definition ($\epsilon$-adherent points) Let $X$ be a subset of $R$, let $\epsilon > 0$, and let $x\in R$. We say that $x$ is $\epsilon$-adherent to $X$ iff there exists a $y\in X$ which is $\epsilon$-close to $x$ (i.e., $|x- y|\leq\epsilon$).

Definition (Adherent points)Let $X$ be a subset of $R$,let $x\in R$. We say $x$ is an adherent point of $X$ iff it is $\epsilon$ adherent to X for all $\epsilon>0$

Answer 1

Notice that $I_{\epsilon} = (b - \epsilon, b + \epsilon)$ centered in $b$ you have that $(a,b) \cap I_{\epsilon} \neq \emptyset$. Because $b = \sup (a,b) $ and there exists $b - \epsilon < x \in (a,b)$ and $x \in I_{\epsilon}$.

Answer 2

Here is my asnwer. I think it is correct, but I am happy to discuss any of its details.

Assuming $b > a$ to avoid the interval being empty one has

Proof: Suppose towards contradiction that $b$ is not adherent to $(a,b)$. Then, there exists an $\epsilon > 0$ such that for all $y \in (a,b)$ one has $|b-y| > \epsilon$. By trichotomy either $\epsilon < b-a$ or $\epsilon \geq b-a$. If $\epsilon < b-a$, then let $y = b-\epsilon$. Thus, $|b-y| = \epsilon$. Contradiction. If $\epsilon \geq b-a$ let $y = (b+a)/2$. Then, $|b-y| = |(b-a)/2| < \epsilon$. Contradiction. Therefore, $b$ must be an adherent point of $(a,b)$.