Basic Probability Theory: establish $(a, b) = \bigcup_{n=1}^{\infty}\left(a, b - \frac{1}{n}\right]$

Question

I'm having trouble with the following question, which comes from Chapter 1 of Basic Probability Theory by Robert Ash. The book assumes some calculus, so I'm hoping for a hint that is at that level.

Question

Let $\Omega$ be the reals. Establish the following $$(a, b) = \bigcup_{n=1}^{\infty}\left(a, b - \frac{1}{n}\right] = \bigcup_{n=1}^{\infty}\left[a + \frac{1}{n}, b \right)$$

The book provides a short answer in the back

If $a < x < b$ then $x \leq b - \frac{1}{n}$ for some $n$, hence $x \in \bigcup_{n=1}^{\infty}\left(a, b - \frac{1}{n}\right]$. Conversely if $x \in \bigcup_{n=1}^{\infty}\left(a, b - \frac{1}{n}\right]$, then $a < x \leq b - \frac{1}{n}$ for some $n$, hence $x \in (a, b)$. The other arguments are similar.

My attempt

I recently learned the least upper bound axiom and subsequent properties and I'm trying to prove it using those properties, the one that I think is relevant is

Any set of real numbers with supremum contain points arbitrarily close to supremum. If $S$ has supremum, then for some $x \in S$ we have $x > \sup S - h$

Intuitively I see whats happening. We are taking the union of sets of reals where the supremum of each set approaches b. The supremums (of each set) are the points that are arbitrarily close to b.

I'm stuck on how to move forward analytically. I start from $x \in (a, b)$ and conclude $b$ is the supremum. Therefore there exists an $x \in (a, b) \text{ where } x > b - \frac{1}{n}$ for some $n$. I don't see a path forward from here, especially since it completely contradicts (I think) the approach of the book solution.

Hoping for a hint/direction that I should be exploring. For context, I have not taken any analysis, but I have been working through Apostol Calculus so a helpful hint would be targeted at that level.

an aside

One direction I tried but doesn't seem promising is going from $x > b - \frac{1}{n}$ to $x + \frac{1}{n} > b > x$. This is close to another theorem, but we're missing the equality.

If three real numbers a, x and y satisfy the inequalities $a \leq x \leq a + \frac{y}{n}$ for all $n \geq 1$ then $x = a$

Answer 1

I will do the union part. You want to show that $(a,b)$ is equal to the union of $(a,b-\frac{1}{n}]$ for all $n\in \mathbb{N}\setminus\{0\}$. To prove two sets are equal, the first thing to try is to show that any element from the LHS belongs in the RHS and any element of the RHS belongs in the LHS. We proceed:

• Pick an element $x$ in the RHS. Since it is in the union, it is in at least one of the sets $(a,b-\frac{1}{k}]$ but, since $b>b-\frac{1}{k}$ for any $k$, it must be in the LHS $(a,b)$ as well.
• Now pick an element $x$ in $(a,b)$. We have $x<b$, which in turns means $x<b-\frac{1}{n}$ for $n$ sufficiently large, say $n>m$. Therefore $x\in(a,b-\frac{1}{n}]$ for $n>m$, which means $x$ is in the RHS.

The trick of the second point is that for any two real numbers $a<c$, one can always find $b$ such that $a<b<c$ (for instance, $b=\frac{a+c}{2}$).