Singletons are included in Borel $\sigma$-algebra on $[0,1]$

Question

Let $F$ be the Borel $\sigma$-algebra on $[0,1]$ relatively to the metric topology. We want to show that singletons in $[0,1]$ are in $F$.

We know that $F$ contains the metric topology so it contains open sets of the form $(a,b)$ with $0<a,b<1$, it also contains $[0,1]$, etc.

$F$ contains open sets of the form $(x-\frac1n, x+\frac1n)$ $\forall n \geq N$ from a certain $N \in \mathbb N$. Since $F$ is a $\sigma$-algebra, it is closed under uncountable intersections so $$\bigcap_{n \geq N} (x-\frac1n, x+\frac1n) = \{x\} \in F$$ This works for $x \in (0,1)$. How can I prove that the singletons $\{0\}$ and $\{1\}$ are also in $F$ ? Maybe I can use the fact that $(0,1) \in F \implies (0,1)^C=\{0,1\} \in F$ but it isn't very useful.

Answer 1

All sets $[0,\frac1n), n \in \Bbb N^+$ are relatively open in $[0,1]$ (as the can be written as $(-\frac1n, \frac1n) \cap [0,1]$ e.g.) and their intersection is $\{0\}$. For $1$ a similar argument holds.

Answer 2

$[0, 1)$ is open in $[0, 1]$, so $[0, 1) \in F$ and $\{ 1 \} = [0, 1)^c \in F$.

The same argument works for $\{ 0 \}$.