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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?

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    But if $X$ is infinite discrete space, then $\|f\|=\sup_{x\in X}|f(x)|$ can be infinite. Namely if $X=\mathbb N$ you can take $f(x)=x$.2012-06-09

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Hint:

  1. 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.

  2. Take a look at binary sequences.

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    @Ananda Discrete metric is usually understood as the metric given by $d(x,y)=1$ for $x\ne y$ (and, of course, $d(x,x)=0$), see [Wikipedia](http://en.wikipedia.org/wiki/Discrete_space). A more serious problem is that the OP did not specify which metric is he using on $C(X,\mathbb R)$.2012-06-09
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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?

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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.

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    You can think of this as a subspace of the product $Y^X$. I'm not sure if it is the canonical way to get a nice topology. I do recall that under certain conditions (which are fulfilled, if I recall, in this case) there is a canonical way of topologizing $C(X,Y)$.2012-06-09