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Given is the Baire space $\mathscr{N}$. The elements are functions (or sequences) $f : \mathbb{N} \to \mathbb{N}$ and the metric $d$ is given by $d(f, g) = \frac{1}{k}$ if $f(i) = g(i)$ for all $1 \leq i \leq k-1$ and $f(k) \neq g(k)$.

Problem: show that it is separable.

So we have to find a countable dense subset. I suppose the set of sequences $\{ (x_1, x_2, \ldots, x_k) : k \in \mathbb{N}, x_i \in \mathbb{N} \}$ is dense in $\mathscr{N}$, but I don't think it's countable, since $k \to \infty$ (or is it?).

Math Student 020
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    Your notation is off, an $(x_1,x_2,\dotsc,x_k)$ is not an element of $\mathscr{N}$. I suppose you meant to extend that initial segment of a sequence with - for example - all following values $1$. That gives you an embedding $j_k \colon \mathbb{N}^k \hookrightarrow \mathscr{N}$. Then $$\bigcup_{k=1}^\infty j_k(\mathbb{N}^k)$$ is the set you're talking about? (It is dense, and it is countable.) – Daniel Fischer Feb 11 '14 at 21:17
  • Indeed, my notation is off. By $(x_1, \ldots, x_k)$ I meant a function/sequence with a fixed start and a random ending (starting at the $k+1$-th entry. I think that is equal to the embedding $j_k$. – Math Student 020 Feb 11 '14 at 21:33
  • Yes. It's easier if you don't let the ending be random, though. – Daniel Fischer Feb 11 '14 at 21:36

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The set of all finite tuples from $\mathbb{N}$ is countable. Here is an injection from your set to $\mathbb{N}$:

Let $x=(x_1,...,x_k)$ be an element and define $f(x)=\prod p_k^{x_k+1}$ where $\{p_k\}_{k=0}^\infty$ is an enumeration of the prime numbers (we need the $x_k+1$ in case $x_k=0$).

Now there is a slight matter that this set is not in $\mathcal{N}$, but it does clearly inject into it. Send $x=(x_1,...,x_k)$ to $(x_1,...,x_k,m_x,m_x,m_x,...)$ where $m_x=\max(x_1,...,x_k)+1$.

  • Why should $m_x$ equal $\operatorname{max}(x_1, \ldots, x_k) + 1$? I don't think the functions (the elements of $\mathscr{N}$) have to be increasing. Is there another reason? – Math Student 020 Feb 11 '14 at 21:40
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    No real important reason, but it allows us to see where the tail 'begins'; thus you can easily explicitly describe this set as a subset of $\mathcal{N}$. –  Feb 11 '14 at 21:50