Why is the equation between these intervals true?

Question

                       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?

$$ \left]a;b\right] =\bigcap_{n=1}^\infty \left]a-\frac 1n ;b\right]$$

Answer 1

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 $I = \left]a;b\right]$.

Answer 2

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:

1. $I \subseteq [a,b]$
2. $ [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 $\mathbb{R}$.