Why is the equation between these intervals true?                        Can you show me why is the following true?

$$ \left[a;b\right] =\bigcap_{n=1}^\infty \left]a-\frac 1n ;b\right]$$
                              And why is this wrong?

$$ \left]a;b\right] =\bigcap_{n=1}^\infty \left]a-\frac 1n ;b\right]$$
 A: Let $\displaystyle I =\bigcap_{n=1}^\infty \left]a-\frac 1n ;b\right]$, we want to show that $a \in I$.
To prove that $a \in I$, we have to show that $a \in \left]a - \frac{1}{n}; b \right]$ for every $n \geq 1$. In fact, if $n \geq 1$, then $\frac{1}{n} > 0$ and consequently $a - \frac{1}{n} < a$. Since we also have $a < b$, we get $a - \frac{1}{n} < a < b$, i.e. $a \in \left]a - \frac{1}{n}; b \right]$.
Now, since $a \in I$ and $a \notin \left]a;b\right]$, we cannot have $I = \left]a;b\right]$.
A: So, this will be some slow-paced hand-crafted answer. I'm sure there is a shortcut, but anyways it will state a proof and I do hope it will give you security.
Let $I := \bigcap_{n=1}^\infty \left]a-\frac{1}{n},b\right]$. There are two directions here to show:


*

*$I \subseteq [a,b]$

*$ [a,b]\subseteq I$


ad 2:
Since the upper bound is the same it remains to show that for any $x \in\mathbb{R}$ if $a \leq x \leq b$ then  $\forall n \in \mathbb{N}: a -\frac{1}{n} < x \leq b$. Obviously this holds, since 
$$
a-\frac{1}{n} < a\Leftrightarrow -\frac{1}{n} < 0
$$ 
where $n\in\mathbb{N}$ and therefore: $a - \frac{1}{n} < a \leq x$.
ad 1:
This seems to me the interesting part. Again, the upper bound is the same, it remains to show that for any $x \in \mathbb{R}$:
$$
\left(\forall n \in \mathbb{N}: a - \frac{1}{n} < x\right) \Rightarrow a \leq x
$$
which is equivalent to:
$$
a > x \Rightarrow \left(\exists n \in \mathbb{N} : a - \frac{1}{n} \geq x \right)
$$
So let's dive into $a - \frac{1}{n} \geq x$ which is equivalent to $a - x \geq \frac{1}{n}$. The aim is to prove the existence of such an $n$. Let $c := a - x$.


*

*By assumption $ a > x$ therefore $c > 0$. 

*If $c\geq 1$ then $n=1$ would be a solution, so let $c < 1$, therefore $1 > c > 0$ hence $c = \frac{1}{r}$ where $r \in\mathbb{R}^+$.


Putting it all together, under the assumption that $a > x$ (and $a-x = \frac{1}{r}$) it has to hold that
$$
\exists n \in\mathbb{N}: \frac{1}{r} \geq \frac{1}{n}
$$
which is equivalent to the Archimedean property which holds within $\mathbb{R}$.
