Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
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
2  
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
1  
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
1  
Hint: it is ccc; it should follow using the fact that the whole product is. –  Henno Brandsma Mar 10 '12 at 19:09

1 Answer 1

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 $p_\alpha=0$.

share|improve this answer
    
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. –  Arthur Fischer 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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.