How to show every half-open interval is a $G_\delta$ and $F_\sigma$

Question

Every half-open interval $[a,b)$ is a $G_\alpha$ and an $F_\delta$ in $R^1$.

My attempt: What I know is every $(a,b)$ is a $G_\alpha$ and an $F_\delta$ in $R^1$. Since

$(a,b)= \cap (a+\frac{1}{n}, b-\frac{1}{n})$ and obviously, (a,b) is a $G_\alpha$. But how to show $[a,b)$ is a $G_\alpha$ and an $F_\delta$ ?

My attempt:

$[a,b) = \cap (a-\frac{1}{n}, b)$ So [a,b) is a $G_\alpha$.

$[a,b) = [a+\frac{1}{n}, b-\frac{1}{n}] \cup \{a\}$

Am I correct?

Answer 1

You seem to be mostly correct about all the facts, but there are several issues with what you write, mostly in the notation:

1. It's $G_\delta$ and $F_\sigma$, not $G_\alpha$ and $F_\delta$.
2. $(a,b)$ is trivially $G_\delta$, because it is open itself ($(a,b)=\bigcap_n (a,b)$, if you want). It is not equal to what you wrote. If you replace $\cap$ with a $\bigcup$, though, then it is true: $(a,b)=\bigcup_n (a+\frac1n,b-\frac1n)$, which shows that the open interval is $F_\sigma$
3. Indeed, $[a,b)=\bigcap_n (a-\frac1n,b)$ (note the $\bigcap$: the small $\cap$ is a binary operation symbol, and similarly for $\bigcup$ vs $\cup$).
4. In the last equation, you missed the union symbol: $[a,b)=\bigcup_n [a+\frac1n,b-\frac1n]\cup \{a\}$. It can be made slightly simpler than that, though: $[a,b)=\bigcup_n[a,b-\frac1n]$.