- Thread starter
- #1

Let $G_n$ be a countable collection of open sets.

Then we would have 2 cases either $x\in\bigcap G_n$ which is a point which is closed.

Or we could $(a,b)$ in all $G_n$ but how to show that would be closed?

- Thread starter dwsmith
- Start date

- Thread starter
- #1

Let $G_n$ be a countable collection of open sets.

Then we would have 2 cases either $x\in\bigcap G_n$ which is a point which is closed.

Or we could $(a,b)$ in all $G_n$ but how to show that would be closed?

Note that this property is true for all metric spaces.

- Thread starter
- #3

I don't see how that union works.

Note that this property is true for all metric spaces.

- Thread starter
- #5

Why does each $G_n$ contain $F$? How is that guaranteed?

Because $x\in B(x,n^{-1})$.Why does each $G_n$ contain $F$? How is that guaranteed?

- Thread starter
- #7

Why is $F$ in there? Is $F=\{x\}$?Because $x\in B(x,n^{-1})$.

- Thread starter
- #9

I still don't understand how we can ensure $F$ is a closed set just by taking unions of open sets.No, just an element of $F$. We have $\{x\}\subset B(x,n^{-1})$, then take the union over $x\in F$.

- Thread starter
- #10

What you have to show is that each closed subset of $\Bbb R$ can be written as a countable intersection of open sets.