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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.

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It is not separable. Let $\{f_n\}$ be a countable set of such functions. We show it is not dense. For each $n$, let $C_n=\{x:f(x)\neq 0\}$. Each $C_n$ is countable and so is $C=\bigcup_n C_n$. Let $x\neq C$. The product set of all admissible functions $f$ such that $f(x)\in(1,\infty)$ is open, but doesn't contain an element from $\{f_n\}$. – Michael Greinecker Mar 10 '12 at 18:12
For 3, any countable set essentially "lives" on a countable index set (as the previous commenter used to show non-separability) and thus lives on a countable metric space $[-1,1]^N$ and thus has a limit point. – Henno Brandsma Mar 10 '12 at 19:06
Once you know it's countably compact, and non-compact (it's dense in the product) it cannot be Lindelöf anymore. – Henno Brandsma Mar 10 '12 at 19:08
Hint: it is ccc; it should follow using the fact that the whole product is. – Henno Brandsma Mar 10 '12 at 19:09
up vote 6 down vote accepted

(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 $p_\alpha=0$.

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Nice work! As an addendum to the addition to (3), you can also show that $\mathbb{R}^\mathbb{R}$ is ccc by using the fact that a Cartesian product $\prod_{i \in I} X_i$ is ccc iff $\prod_{i \in J} X_i$ is ccc for all finite $J \subseteq I$, and noting that the "finite subproducts" of $\mathbb{R}^\mathbb{R}$ are just the Euclidean spaces $\mathbb{R}^n$ which are easily seen to be ccc. – arjafi Mar 10 '12 at 21:13
I believe the separability of $\mathbb{R}^\mathbb{R}$ is easier than the H-M-P theorem. Take, for instance, the polynomials with rational coefficients. It is clear that every basic open set contains such a polynomial. – Nate Eldredge Mar 10 '12 at 22:16
@Arthur: Yes, another good use of the $\Delta$-system lemma. – Brian M. Scott Mar 11 '12 at 6:30
@Nate: That works, but I don’t think of it that way: I tend to view $\Bbb R^{\Bbb R}$ as a product, not as a function space. – Brian M. Scott Mar 11 '12 at 6:34
@Brian: Thank you for answering my question! I understand 1 and 2, and I am able to follow the hint given for 4. For 3, I was able to figure out that any countable subset of $Y$ has limit point since it is a subspace of $[-1, 1]^{\omega}$ (thanks to Henno). However, I'm not sure how this implies every infinite subset of $Y$ has a limit point in $Y$. I'm having a hard time adapting the idea from 1. – Maria Mar 11 '12 at 16:03

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