Metrizability of a topological space of all real sequences Let X denote the set of all sequences of real numbers and $A$ consists of all subsets G of X such that: for each $x=(x_n)$ in G, there are integers $n_1 <...< n_N$ and an $\epsilon >0$ such that $\{y= (y_n): |x_{n_k}- y_{n_k}| < \epsilon, \forall 1 \leq k \leq N\}$ is contained in G. Then, $(X,A)$ is a topological space. How to check that whether it is metrizable or not?
I have checked that it is Hausdorff and first countable.
 A: That's just the product topology. A countable product of metrizable spaces is metrizable.
Sketch: First, if $d$ is a metric on $X$ then $d/(1+d)$ is a bounded metric inducing the same topology. Now say $X_1,\dots$ are metrizable spaces. Let $d_j$ be a metric on $X_j$ with $d_j\le 1$. Define $$d(x,y)=\sum_{n=1}^\infty2^{-n}d_n(x_n,y_n).$$You can check that $d$ is a metric on $\prod X_j$ that induces the product topology.
A: This space is $$\prod_{n \in \Bbb{N}}\Bbb{R}$$ the product of countable many copies of $\Bbb{R}$. There is a standard theorem in topology saying that this is metrizable with the following metric:
$$d((x_n),(y_n)) = \sup \left\{ \frac{1}{n} \min(|x_n-y_n|, 1) : n \in \Bbb{N} \right\}$$
A: I see two ways to prove that $(X,A)$ is metrizable:


*

*Notice that this topology is the product topology as already mentioned.

*Or use a theorem stating that a topological vector space is metrizable if and only if it is Haussdorf and has a countable local base.


Denote by $\mathbb N^{(\mathbb N)}$, $\mathbb Q^{(\mathbb N)}$ the countable sets of finite sequences of integers and rationals. For $m \ge 1$ integer, $(a_1, \dots ,a_m) \in \mathbb N^{(\mathbb N)}$, $(q_1, \dots ,q_m) \in \mathbb Q^{(\mathbb N)}$ and an integer $n \ge 1$ denote by $$V(a_1, \dots, a_m,q_1,\dots,q_m,n)=\{(x_n) \in X : \vert x_{a_i} - q_i \vert \le \frac{1}{n}, 1 \le i\le m \}$$
It is a good exercise to  prove that those sets $V$ are countable and form a local base of your topology.
