Proof that $b$ is point of closure of $[a,b)$

Question

Let $E=[a,b)$. I want to exhibit that $\{b\}$ is a point of closure of $E$. Attempt at a proof:

My goal is to show that for all $\epsilon>0$ there exists a $y\in E$ such that $|b-y|<\epsilon$. Let, therefore, $\epsilon>0$ be arbitrary. I need to construct a $y=y(\epsilon;a,b)\in E$ such that $|b-y|<\epsilon$. Are there any hints as to the construction of $y$?

Answer 1

As a first step, you may take $y=b-\frac\varepsilon2$. Of course, this will not work if, say, $a=0$, $b=1$, and $\varepsilon=10$. Now, see what you will have to do then.

Answer 2

You need $|b - y| < \epsilon$ and $a \le y < b$. So

$-\epsilon < b-y < \epsilon$ and $a \le y < b$. So

$\epsilon > y-b > -\epsilon$ and $a\le y < b$. So

$b-\epsilon < y < b+\epsilon$ and $a\le y < b$. Or in other words

$b-\epsilon < y < b$ and $a \le y$.

Does such a $y$ exist? Sure just let $y$ be any point $y \in (\min(a, b-\epsilon), b)$.

If you want to be specific take $y = b -\frac {\epsilon}2$. That assumes that $\frac \epsilon 2 < b-a$. Usually we only worry about small epsilons so that isn't really an issue but we can just let $y =\max (b-\frac \epsilon 2, b-\frac {b-a}2)$ to cover that possibility.