Expressing half-open interval [a,b) as infinite intersection of open intervals. (Answer Verification)

Question

I need to express the half open interval $[a,b)$ where, $(a<b)$ as the infinite intersection of open intervals.

My answer/attempt is

$$[a,b)=\bigcap\limits^\infty_{n=1}(a-\frac{1}{n},b)$$ I'm not sure if this is correct however.

Answer 1

To prove that $L=[a,b) =\displaystyle \bigcap_{n=1}^{\infty}\left(a-\frac{1}{n},b\right)=R$, you need to show that one is the subset of the other. To show $L \subseteq R$, we start with some $x \in [a,b)$. Then $a-\frac{1}{n}<a\leq x$ for all $n \in \mathbb{N}$, so $x \in R$.

Now we start with $y \in R$, this means for all $n \geq 1$, we have $a-\frac{1}{n}<y<b$.

If $y \not\in L$, that would mean $y<a$. So let $y=a-\epsilon$, with $\epsilon >0$, by the Archimedean property, there exists $k \in \mathbb{N}$ such that $k\epsilon >1$. This would mean, $$y=a-\epsilon <a-\frac{1}{k}.$$ But then $y \not\in \left(a-\frac{1}{k},b\right)$. This contradicts the assumption we made at the very beginning. So $y \in L$.

Similarly you can check the other part.

Answer 2

The solution is correct if your intervals make sense. For instance $(a+1/n,b) $ makes sense if $a+1/n<b$

I would change the $1/n$ to $(b-a)/n$ to take care of that.