How to show [a,b) is not open with respect to Euclidean topology Show that, if $a,b\in\mathbb{R}$ and $a < b$, then $[a,b)$ is not open set.
I am using Topology Without Tears by Sydney Morris
My thoughts on it
Observe  $a\in [a,b)$.
Suppose that there exists $c,d\in\mathbb{R}$ such that $c\leq d$ $a\in(c,d)\subseteq[a,b)$
$a\in(c,d)$ implies $c\leq a < d$ and $c\leq 0.5(a+c) < c < b$.
So $0.5(a+c)\notin [a,b)$ and $0.5(a+c)\in (c,d)$.
So $[a,b)\not\subseteq (c,d)$. Contradiction.
So $[a,b)$ is not open.
Basically I am tweaking Morris’s argument.
It is obviously not open since it does not contain the point $b$
Let $I_{n} = [a,b-1/n) \bigcup_{n\in N} I_n = [a,b)$.
 A: You have the exact right idea!
$[a,b)$ is open if and only if each $x \in [a,b)$ has an open interval $(x-\epsilon, x+\epsilon) \subseteq [a,b)$. So to show that $[a,b)$ isn't open, we must find a point $x$ for which that condition fails.
You're exactly right to think $a$ is a good choice of point. We want to show that no matter how small an $\epsilon$ we choose, $(a-\epsilon, a+\epsilon) \not \subseteq [a,b)$.
Of course, this is easy! Since $a - \frac{\epsilon}{2} \in (a-\epsilon, a+\epsilon)$, and yet $a-\frac{\epsilon}{2} \not \in [a,b)$.
So $(a-\epsilon, a+\epsilon) \not \subseteq [a,b)$. As needed.

As a remark, I'm not sure what you mean by "It is obviously no open since it does not contain the point b". Perhaps this was a typo and you meant it is obviously not closed since it doesn't contain $b$, and this would be true. After all, $b$ is in the closure of $[a,b)$ (do you see why?).
As you saw from the argument above (and the argument you outlined), the relevant point for showing $[a,b)$ is not open is $a$, not $b$.

I hope this helps ^_^
A: I shall prove it for the case of $(0,1]$ and you shall prove this for $(\alpha, \beta]$ using a similar argument.
Consider the interval $(0,1]$, we claim that such set is not open in $\mathbb{R}$ since
$1$ is not an interior point of $(0,1]$. To see this, notice that $\forall \varepsilon>0$, the neighbourhood $(1-\varepsilon, 1+\varepsilon) \nsubseteq(0,1]$ . For instance :
$$1+\frac{\varepsilon}{2} \in(1-\varepsilon, 1+\varepsilon)$$
However :
$$1+\frac{\varepsilon}{2} \notin(0,1]$$
