# 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?

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A hint that may help you think about the problem more simply: Elements of $C(X,\mathbb{R})$ are simply sequences of real numbers. –  Ragib Zaman Jun 9 '12 at 14:47
Your logic is flawed. Finding a sequence of discrete elements means nothing, after all $\mathbb N$ is such example in $\mathbb R$ which is separable. You need to show that for every countable set there is a point not in the closure of that set. Furthermore if you wish to use "distance" between elements then you need to have a metric as well. –  Asaf Karagila Jun 9 '12 at 14:49
What topology/metric on $C(X,\mathbb R)$ are you supposed to use? (It only makes sense to speak about dense sets if you know some metric or topology on the set you're working with.) –  Martin Sleziak Jun 9 '12 at 15:07
On $$C(X,\mathbb{R})$$ I have supnorm metric –  La Belle Noiseuse Jun 9 '12 at 15:09
What is $d(0,id)$, where $0$ is the constant function $0$ and $id$ sends $n$ to $n$? –  Asaf Karagila Jun 9 '12 at 15:13

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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can you help me to understand what it mean to say that metric space with discrete metric whose points are only integers ? Is it a metric defined on a set of integers ?? I didn't understand :( . –  Theorem Jun 9 '12 at 15:08
As far as I understand that question $X=(\mathbb{N},\rho)$, where $\rho$ is the discrete metric on $\mathbb{N}$. –  Norbert Jun 9 '12 at 15:10
@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. A more serious problem is that the OP did not specify which metric is he using on $C(X,\mathbb R)$. –  Martin Sleziak Jun 9 '12 at 15:30

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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What would be the norm of the function $f(n)=n$ if you take the $\sup$-norm? –  Asaf Karagila Jun 9 '12 at 15:08
@AsafKaragila I didn't even think of that :( To be honest, I'm struggling to think of the "supnorm metric" Mex says he meant in the comments below his post. For topological spaces $X$ and $Y$, is there a canonical topology on $C(X,Y)$ ? –  Ragib Zaman Jun 9 '12 at 15:12
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 $C(X,Y)$. –  Asaf Karagila Jun 9 '12 at 15:15