Subspace of Real-valued Functions I've been looking at the product topology, and came across this question.
Let X be the set of all real-valued functions which are zero outside of a countable subset of $\mathbb{R}$. Consider X as a subspace of $\mathbb{R}^{\mathbb{R}}$ with the product topology. 
1) Is X separable? 
2) Does X has the Souslin property? That is, every collection of pairwise disjoint non-empty open subsets of X is countable.
3) Show Y = $\{f \in X: |f(x)| \leq 1 \mbox{ for every } x \in \mathbb{R} \} $ $\thinspace$ is countably compact.
4) Is X Lindelof?
I believe that X is separable, but I am having difficulty formulating a proof. Also, I know that if X is separable, then it has the Souslin property.
 A: (1) $X$ is not separable. (That would make life too easy!) Let $D$ be a countable subset of $X$. For each $x\in D$ let $S(x)=\{\alpha\in\Bbb R:x(\alpha)\ne 0\}$, the support of $x$. Let $$S=\bigcup_{d\in D}S(x)\;;$$ each $S(x)$ is countable, so $S$ is countable. Now let $\alpha_0\in\Bbb R\setminus S$, and define $p\in X$ by $$p(\alpha)=\begin{cases}1,&\text{if }\alpha=\alpha_0\\0,&\text{otherwise}\;.\end{cases}$$ Let $B=\{x\in X:x(\alpha_0)\ne 0\}$; then $B$ is an open nbhd of $p$ disjoint from $D$, and $D$ is therefore not dense in $X$.
(2) $X$ is ccc (i.e., does have the Suslin property). Let $\mathscr{I}$ be the set of open intervals with rational endpoints. For each finite $F=\{\alpha_1,\dots,\alpha_n\}\subseteq\Bbb R$ and function $\varphi:F\to\mathscr{I}$ let $$B(F,\varphi)=\{x\in X:x(\alpha_k)\in\varphi(\alpha_k)\text{ for }k=1,\dots,n\}\;;\tag{1}$$ $X$ has a base of such open sets, so show that $X$ is ccc, it suffices to show that it has no uncountable pairwise disjoint family of open sets of the form $(1)$. Suppose, then, that $I$ is an uncountable index set, and $\mathscr{B}=\{B(F_i,\varphi_i):i\in I\}$ is a family of these basic open sets. By the $\Delta$-system lemma there are a finite $F\subseteq\Bbb R$ and an uncountable $I_0\subseteq I$ such that $F_i\cap F_j=F$ for every pair of distinct $i,j\in I_0$. There are only countably many finite collections of open intervals with rational endpoints, so there are a $\varphi:F\to\mathscr{I}$ and an uncountable $I_1\subseteq I_0$ such that for each $i\in I_1$ and each $\alpha\in F$, $\varphi_i(\alpha)=\varphi(\alpha)$. (In other words, the basic open sets $B(F_i,\varphi_i)$ for $i\in I_1$ all restrict the coordinates in $F$ in exactly the same way.) But then for any $i,j\in I_1$ we have $B(F_i,\varphi_i)\cap B(F_j\varphi_j)\ne\varnothing$, and $\mathscr{B}$ is not pairwise disjoint.

Added: Alternatively, if you know that $\Bbb R^{\Bbb R}$ is separable, which follows from the Hewitt-Marczewski-Pondiczery Theorem, then you know that $\Bbb R^{\Bbb R}$ is ccc, and you can easily show that any dense subspace must also be ccc. (But the $\Delta$-system lemma is a handy tool to have anyway.)

(3) Adapt the idea that I used in (1) to show that $Y$ does not contain an infinite closed discrete subset.
(4) Show that $Y$ is a closed subset of $X$ that is not compact. Conclude that $Y$ cannot be Lindelöf. What does this tell you about the Lindelöfness of $X$?

$X$ is an example of what is known as a $\Sigma$-product. More generally, let $\{X_\alpha:\alpha\in A\}$ be a family of spaces, and for each $\alpha\in A$ fix a point $p_\alpha\in X_\alpha$. Let $$p=\langle p_\alpha:\alpha\in A\rangle\in \prod_{\alpha\in A}X_\alpha\;,$$ and let $$X=\left\{x\in\prod_{\alpha\in A}X_\alpha:\{\alpha\in A:x_\alpha\ne p_\alpha\}\text{ is countable}\right\}\;;$$ $X$ is the $\Sigma$-product of the $X_\alpha$ with base point $p$. In your case each $X_\alpha$ is $\Bbb R$, and each $p_\alpha=0$.
