Prove that $\mathbb R/\mathbb Z$ is sequential and not first countable

Question

Let's say I have a a space X, the quotient space of R (the reals) obtained by identifying all points of Z (the integers).

How do I prove that X is sequential but not first countable?

(sequential means every sequentially open set is open. I was thinking I might be able to show that R is sequential and then that any quotient space of a sequential space is sequential. Not sure how to show it's not first countable.)

I need help not just with the idea, but with actually writing out the rigorous proof. Thank you.

Answer 1

As I said in comments you can imagine the space $X$ as an infinity loops passing through a point (this point is $\mathbb{Z}$). Remember that a set $U$ in $X$ is open if and only iff $\pi^{-1}(U)$ is open. To prove that $X$ is not a first countable space you should find an element in $X$ such that for every countable collection of neighborhoods is not enough to be a basis, i.e, there exist another open not in the basis which contains the point but does not exist any open in the countable collection contained in the open.

You should have the following things in mind:

1. The open set of $\mathbb{Z}$ in $X$ are those sets image by $\pi$ of a union of neighborhoods of each integer. I mean for example a union like $\bigcup_{n\in\mathbb{N}}(-\frac{1}{2}+n,n+\frac{1}{2})$.

2. For $n\in \mathbb{N}$. We define the function $f(n)=-\frac{n-1}{2}$ if $n$ odd and $f(n)=\frac{n}{2}$ if $n$ is even. We use this later.

I claim that the point $\mathbb{Z}$ which has not a countable basis. To prove this take a countable collection $\{U_{n}\}_{n\in\mathbb{N}}$ of neighborhoods of $\mathbb{Z}$ (remember that this is a point in the quotient space).

Now we are going to construct a new open $U$ in $X$ which does not contain any $U_n$. For 1. the open $\pi^{-1}(U_1)$ contains a open interval $(-\epsilon, \epsilon)$ about $0$. Choose an interval $(-\delta_1,\delta_1)$ contains in $(-\epsilon, \epsilon) $ (and then in $U_1$ ). Now $\pi^{-1}(U_2)$ contains an interval $(-\epsilon+1,\epsilon+1)$ and choose an interval $(-\delta_2+1,1+\delta_2)$ contained in $(-\epsilon+1,\epsilon+1)$ (and then in $U_2$ ).

Following this process we get a family $\{(-\delta_n+f(n),\delta_n+f(n))\}_{n\in \mathbb{N}}$ such that $(-\delta_n+f(n),f(n)+\delta_n)\subset \pi^{-1} (U_n )$ for each $n\in \mathbb{N}$. Then for 1. the set $U=\pi\left[\bigcup_{n\in \mathbb{n}}(-\delta_n+f(n),f(n)+\delta_n)\right] $ is open in $X$. Prove that $U$ does not contain any $U_n$ and then $X$ is not first countable.

The second problem is rather easy. Because of the definition of quotient space if you take a set $U$ in $X$ and prove that $\pi^{-1}(U)$ is open then you can assure that $U$ is open. Furthermore, you have that $\mathbb{R}$ is sequentially open, then if you prove that $\pi^{-1}(U)$ is sequentially open then $\pi^{-1}(U)$ is indeed open.

My principal suggestion to you is: Draw a picture!