According to the [following lecture](https://www.youtube.com/watch?v=lIA__6thuo4) (or [Wikipedia](https://en.wikipedia.org/wiki/Fr%C3%A9chet_space#Constructing_Fr%C3%A9chet_spaces)), 

> 
> Let $F(\omega)$ be the set of all real sequences and let $x=(x\_1,x\_2,...,x\_k...) ,y=(y\_1,y\_2,...,y\_k,...) \in F(\omega)$. Define the function $d : F(\omega) \times F(\omega) \to \mathbb{R} $, $~$ $d(x,y)= \sum\_{i=1}^{\infty} \frac{1}{2^i}\frac{|x\_i-y\_i|}{1+|x\_i-y\_i|} $. $~~$(Clearly, the function is well-defined) Then, $(F(\omega),d)$ is a metric space which is well known as **Fréchet space**.
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And according to the Wikipedia again on the Fréchet space, they are special [topological vector space](https://en.wikipedia.org/wiki/Topological_vector_space) (TVS). I wonder that they really satisfy the TVS condition. More precisely, if $(F(\omega),d)$ is a TVS, following conditions

$~~$ (i) The vector addition $+: F(\omega) \times F(\omega) \to F(\omega) $ is continuous. 

$~~$ (ii) The scalar multiplication $\cdot : \mathbb{R} \times F(\omega) \to F(\omega) $ is continuous.

would be satisfied. 

Since $(F(\omega),d)$ is a metric space, I thought that it would be OK by using [the continuous functions between metric spaces](https://en.wikipedia.org/wiki/Continuous_function#Continuous_functions_between_metric_spaces), **but is it right idea?** In order to check the condition, $F(\omega) \times F(\omega)$ must have the metric space, and I thought that the natural metric space on the (product) space is [product metric space](https://en.wikipedia.org/wiki/Metric_space#Product_metric_spaces). 

And how to approach in case (ii)? $\mathbb{R} \times F(\omega)$ also have a metric space. However, since the component of product is not coincidence, I don't imagine why scalar multiplication is continuous.