a non separable metric space Let $X$ be a metric space  with discrete metric whose points are the positive integers. We have to show $C(X,\mathbb{R})$ is non separable. Well, what I have to do is to show $C(X,\mathbb{R})$ has no countable dense subset. I have no idea how to show that It has no countable as well as dense subset of $C(X,\mathbb{R})$, so far I guess to show it has  non dense subset we need to find a sequence of functions $f_n\in C(X,\mathbb{R})$ which has some constant distance to the element of that set. Please, will any one help me to solve the problem?
 A: Hint:


*

*Prove the follwing lemma. If $\{x_i:i\in I\}$ - is an uncountable family in metric space $(M,d)$ such that 
$$
\exists \delta>0\quad\forall i\in I\quad\forall j\in I\quad (i\neq j\Longrightarrow d(x_i,x_j)>\delta)
$$
then $(M,d)$ is not separable.

*Take a look at binary sequences.
A: Hint: any function from a discrete space to any space is continuous, so sharpening Ragib's comment we can say that $\,\mathcal{C}(X,\mathbb{R}) \,$ contains all the functions $\,X\to\mathbb{R}\,$ , i.e. all the sequences of real numbers (indexed by the naturals, of course).
Now, since $\,X\,$ is not compact I am not sure what topology are you taking for $\,\mathcal{C}(X,\mathbb{R})\,$...The supremum wrt the usual metric in the reals?
A: Assuming something like the sup-norm, we can prove the result with a diagonal argument. 
Suppose we have any countable collection of sequences in $\mathbb{R}.$ We can list this collection into a sequence, say $f_1, f_2, \cdots$ where each $f_n$ is a sequence of reals (denote the i-th term of $f_n$ by $f_n^{(i)}.$)
Define a sequence of real numbers by $g_n = f_n^{(n)}+1.$ Then $g$ has distance at least $1$ from any $f_n$ so $\{ f_n \}$ is not dense in the sequences of real numbers. 
