(Lower case) sigma-products of separable metric spaces are Lindelof

Consider the product space X=\prod_{\alpha \in A} X_\alpha. Fix a point b \in \prod_{\alpha \in A} X_\alpha, called the base point. The \Sigma-product of the spaces \left\{X_\alpha: \alpha \in A \right\} is the following subspace of the product space X:

    \Sigma_{\alpha \in A} X_\alpha=\left\{ x \in X: x_\alpha \ne b_\alpha \text{ for at most countably many } \alpha \in A \right\}

In other words, the space \Sigma_{\alpha \in A} X_\alpha is the subspace of the product space X=\prod_{\alpha \in A} X_\alpha consisting of all points that deviate from the base point on at most countably many coordinates \alpha \in A. We also consider the following subspace of \Sigma_{\alpha \in A} X_\alpha.

    \sigma=\left\{ x \in \Sigma_{\alpha \in A} X_\alpha: x_\alpha \ne b_\alpha \text{ for at most finitely many } \alpha \in A \right\}

For convenience , we call \Sigma_{\alpha \in A} X_\alpha the (upper case) Sigma-product (or \Sigma-product) of the spaces X_\alpha and we call the space \sigma the (lower case) sigma-product (or \sigma-product). Clearly, the space \sigma is a dense subspace of \Sigma_{\alpha \in A} X_\alpha. In a previous post, we show that the upper case Sigma-product of separable metric spaces is collectionwise normal. In this post, we show that the (lower case) sigma-product of separable metric spaces is Lindelof. Thus when each factor X_\alpha is a separable metric space with at least two points, the \Sigma-product, though not Lindelof, has a dense Lindelof subspace. The (upper case) \Sigma-product of separable metric spaces is a handy example of a non-Lindelof space that contains a dense Lindelof subspace.

Naturally, the lower case sigma-product can be further broken down into countably many subspaces. For each integer n=0,1,2,3,\cdots, we define \sigma_n as follows:

    \sigma_n=\left\{ x \in \sigma: x_\alpha \ne b_\alpha \text{ for at most } n \text{ many } \alpha \in A \right\}

Clearly, \sigma=\bigcup_{n=0}^\infty \sigma_n. We prove the following theorem. The fact that \sigma is Lindelof will follow as a corollary. Understanding the following proof for Theorem 1 is a matter of keeping straight the notations involving standard basic open sets in the product space X=\prod_{\alpha \in A} X_\alpha. We say V is a standard basic open subset of the product space X if V is of the form V=\prod_{\alpha \in A} V_\alpha such that each V_\alpha is an open subset of the factor space X_\alpha and V_\alpha=X_\alpha for all but finitely many \alpha \in A. The finite set F of all \alpha \in A such that V_\alpha \ne X_\alpha is called the support of the open set V.

Theorem 1
Let \sigma be the \sigma-product of the separable metrizable spaces \left\{X_\alpha: \alpha \in A \right\}. For each n, let \sigma_n be defined as above. The product space \sigma_n \times Y is Lindelof for each non-negative integer n and for all separable metric space Y.

Proof of Theorem 1
We prove by induction on n. Note that \sigma_0=\left\{b \right\}, the base point. Clearly \sigma_0 \times Y is Lindelof for all separable metric space Y. Suppose the theorem hold for the integer n. We show that \sigma_{n+1} \times Y for all separable metric space Y. To this end, let \mathcal{U} be an open cover of \sigma_{n+1} \times Y where Y is a separable metric space. Without loss of generality, we assume that each element of \mathcal{U} is of the form V \times W where V=\prod_{\alpha \in A} V_\alpha is a standard basic open subset of the product space X=\prod_{\alpha \in A} X_\alpha and W is an open subset of Y.

Let \mathcal{U}_0=\left\{U_1,U_2,U_3,\cdots \right\} be a countable subcollection of \mathcal{U} such that \mathcal{U}_0 covers \left\{b \right\} \times Y. For each j, let U_j=V_j \times W_j where V_j=\prod_{\alpha \in A} V_{j,\alpha} is a standard basic open subset of the product space X with b \in V_j and W_j is an open subset of Y. For each j, let F_j be the support of V_j. Note that \alpha \in F_j if and only if V_{j,\alpha} \ne X_\alpha. Also for each \alpha \in F_j, b_\alpha \in V_{j,\alpha}. Furthermore, for each \alpha \in F_j, let V^c_{j,\alpha}=X_\alpha- V_{j,\alpha}. With all these notations in mind, we define the following open set for each \beta \in F_j:

    H_{j,\beta}= \biggl( V^c_{j,\beta} \times \prod_{\alpha \in A, \alpha \ne \beta} X_\alpha \biggr) \times W_j=\biggl( V^c_{j,\beta} \times T_\beta \biggr) \times W_j

Observe that for each point y \in \sigma_{n+1} such that y \in V^c_{j,\beta} \times T_\beta, the point y already deviates from the base point b on one coordinate, namely \beta. Thus on the coordinates other than \beta, the point y can only deviates from b on at most n many coordinates. Thus \sigma_{n+1} \cap (V^c_{j,\beta} \times T_\beta) is homeomorphic to V^c_{j,\beta} \times \sigma_n. Note that V^c_{j,\beta} \times W_j is a separable metric space. By inductive hypothesis, V^c_{j,\beta} \times \sigma_n \times W_j is Lindelof. Thus there are countably many open sets in the open cover \mathcal{U} that covers points of H_{j,\beta} \cap (\sigma_{n+1} \times W_j).

Note that

    \sigma_{n+1} \times Y=\biggl( \bigcup_{j=1}^\infty U_j \cap \sigma_{n+1} \biggr) \cup \biggl( \bigcup \left\{H_{j,\beta} \cap (\sigma_{n+1} \times W_j): j=1,2,3,\cdots, \beta \in F_j \right\} \biggr)

To see that the left-side is a subset of the right-side, let t=(x,y) \in \sigma_{n+1} \times Y. If t \in U_j for some j, we are done. Suppose t \notin U_j for all j. Observe that y \in W_j for some j. Since t=(x,y) \notin U_j, x_\beta \notin V_{j,\beta} for some \beta \in F_j. Then t=(x,y) \in H_{j,\beta}. It is now clear that t=(x,y) \in H_{j,\beta} \cap (\sigma_{n+1} \times W_j). Thus the above set equality is established. Thus one part of \sigma_{n+1} \times Y is covered by countably many open sets in \mathcal{U} while the other part is the union of countably many Lindelof subspaces. It follows that a countable subcollection of \mathcal{U} covers \sigma_{n+1} \times Y. \blacksquare

Corollary 2
It follows from Theorem 1 that

  • If each factor space X_\alpha is a separable metric space, then each \sigma_n is a Lindelof space and that \sigma=\bigcup_{n=0}^\infty \sigma_n is a Lindelof space.
  • If each factor space X_\alpha is a compact separable metric space, then each \sigma_n is a compact space and that \sigma=\bigcup_{n=0}^\infty \sigma_n is a \sigma-compact space.

Proof of Corollary 2
The first bullet point is a clear corollary of Theorem 1. A previous post shows that \Sigma-product of compact spaces is countably compact. Thus \Sigma_{\alpha \in A} X_\alpha is a countably compact space if each X_\alpha is compact. Note that each \sigma_n is a closed subset of \Sigma_{\alpha \in A} X_\alpha and is thus countably compact. Being a Lindelof space, each \sigma_n is compact. It follows that \sigma=\bigcup_{n=0}^\infty \sigma_n is a \sigma-compact space. \blacksquare

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A non-Lindelof space with a dense Lindelof subspace

Now we put everything together to obtain the example described at the beginning. For each \alpha \in A, let X_\alpha be a separable metric space with at least two points. Then the \Sigma-product \Sigma_{\alpha \in A} X_\alpha is collectionwise normal (see this previous post). According to the lemma in this previous post, the \Sigma-product \Sigma_{\alpha \in A} X_\alpha contains a closed copy of \omega_1. Thus the \Sigma-product \Sigma_{\alpha \in A} X_\alpha is not Lindelof. It is clear that the \sigma-product is a dense subspace of \Sigma_{\alpha \in A} X_\alpha. By Corollary 2, the \sigma-product is a Lindelof subspace of \Sigma_{\alpha \in A} X_\alpha.

Using specific factor spaces, if each X_\alpha=\mathbb{R} with the usual topology, then \Sigma_{\alpha<\omega_1} X_\alpha is a non-Lindelof space with a dense Lindelof subspace. On the other hand, if each X_\alpha=[0,1] with the usual topology, then \Sigma_{\alpha<\omega_1} X_\alpha is a non-Lindelof space with a dense \sigma-compact subspace. Another example of a non-Lindelof space with a dense Lindelof subspace is given In this previous post (see Example 1).

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\copyright \ 2014 \text{ by Dan Ma}

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