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Q1, Does a metrizable space $X$ with $e(X)=\omega$ (i.e., it has countable extent) which is not lindelof exist?

Q2, Let $X$ be the one point lindefication of a discret space of cardinality $\omega_1$ and $Y$ is any Lindelof space. Is $X \times Y$ always Lindelof?

Thanks for any help.

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What do you mean by $e(X)$ or the extent of a space? –  Mark Grant Feb 21 '13 at 12:22
It denote the extent of the space $X$. –  Paul Feb 21 '13 at 12:41
Let me rephrase that: what is the "extent" of a space? –  Mark Grant Feb 21 '13 at 12:45
The maximumm of the caerdinality of closed discrete subspace of $X$. –  Paul Feb 21 '13 at 12:48
Q1. For metrizable spaces, lindelof and separable are equivalent. So the question is: Must a nonseparable metric space contain an uncountable closed discrete subset? –  Gerald Edgar Feb 21 '13 at 15:00
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4 Answers

I shall answer question 1 here by proving by contrapositive that every metric space with countable extent is Lindelof. Assume that $X$ is a metric space that is not Lindelof. Then $X$ is not separable (for metric spaces, the properties of second countability, Lindelof, and separability are all equivalent as commented above. See Dugundji p. 187). We shall now construct a sequence $(x_{\alpha})_{\alpha<\omega_{1}}$ by transfinite induction. For each $\alpha<\omega_{1}$, let $U_{\alpha}=X\setminus\overline{\{x_{\beta}|\beta<\alpha\}}$. Clearly $U_{\alpha}$ is a non-empty open set for all $\alpha$ since $X$ is not separable. Let $\epsilon_{\alpha}=\sup\{\epsilon|B_{\epsilon}(x)\subseteq U_{\alpha}\,\textrm{for some}\,x\in X\}$. Let $x_{\alpha}$ be a point where $B_{\epsilon_{\alpha}/2}(x_{\alpha})\subseteq U_{\alpha}$ for all $\alpha$. We observe that $\epsilon_{\alpha}$ is a decreasing sequence of positive real numbers of length $\omega_{1}$. Therefore, the sequence $\epsilon_{\alpha}$ is eventually some constant $\epsilon>0$. Therefore $B_{\epsilon/2}(x_{\alpha})\subseteq U_{\alpha}=X\setminus\overline{\{x_{\beta}|\beta<\alpha\}}$, so if $\beta<\alpha$, then $d(x_{\beta},x_{\alpha})\geq\epsilon/2$. Therefore the set $\{x_{\alpha}|\alpha<\omega_{1}\}$ is a closed discrete set(in fact uniformly discrete), so $X$ does not have countable extent.

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The mathod very good! –  Paul Feb 22 '13 at 0:23
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The answer to the first question is no. In fact, every metric space of countable extent even has a countable base. That's because every metric space has a base which is a countable union of discrete collections. (See, for example, Engelking's General Topology book. A collection is called discrete if every point in the space has a neighbourhood which meets at most one of its members).

The answer to the second question is yes. Let $X$ be a Lindelof space and $Y= \omega_1 \cup \{ p \}$ be the one point-Lindelofication of the discrete space $\omega_1$, where $p$ is the unique non-isolated point of $Y$. Let $\mathcal{U}$ be an open cover of $X \times Y$. Then we can assume that every element of $\mathcal{U}$ is of the form $U \times V$, where $U$ is open in $X$, $V$ is open in $Y$, and if $p \in V$ then there is an ordinal $\gamma$ such that $V=[\gamma, \omega_1)$. Let $\mathcal{V}=\{U: (\exists V)(U \times V \in \mathcal{U} \wedge p \in V)\}$. Then $\mathcal{V}$ covers $X$ and, since $X$ is Lindelof, we can find a countable subcover $\mathcal{C} \subset \mathcal{V}$. For every $U \in \mathcal{C}$ choose $V(U)$ such that $U \times V(U) \in \mathcal{U}$ and $p \in V(U)$, and define $\mathcal{G}=\{ U \times V(U): U \in \mathcal{C}\}$. Let $\alpha= \sup \{\min (V(U)): U \in \mathcal{C} \}$, and for every $\beta \leq \alpha$ let $\mathcal{V}_\beta =\{U: (\exists V)(U \times V \in \mathcal{U} \wedge \beta \in V) \}$. Let $\mathcal{C}_\beta$ be a countable subfamily of $\mathcal{V}_\beta$ covering $X$ and for every $U \in \mathcal{C}_\beta$ choose $V_\beta(U)$ such that $U \times V_\beta(U) \in \mathcal{U}$ and $\beta \in V_\beta(U)$. Finally, let $\mathcal{U}_\beta=\{U \times V_\beta(U): U \in \mathcal{C}_\beta \}$. Then $\mathcal{G} \cup \bigcup_{\beta \leq \alpha} \mathcal{U}_\beta$ is a countable subcover of $\mathcal{U}$.

Edit: the positive answer to the second question can be generalized as follows: "The product of a Lindelof space X and a Lindelof P-space Y is Lindelof" (a P-space is a space where $G_\delta$ sets are open). Here is a sketch of the proof, which is quite similar to the proof of the special case above. Let $\mathcal{U}$ be an open cover for $X \times Y$ consisting of basic open sets. Use the fact that $Y$ is a $P$-space to construct, for every $y \in Y$, a countable subfamily $\mathcal{U}_y \subset \mathcal{U}$ and an open neighbourhood $U_y$ of $y$ such that $\mathcal{U}_y$ covers $X \times U_y$. Use the Lindelof property of $Y$ to find a sequence $\{y_n: n \in \mathbb{N}\}$ such that $\bigcup \{U_{y_n}: n \in \mathbb{N} \}=Y$. Then $\bigcup \{\mathcal{U}_{y_n}: n \in \mathbb{N} \}$ is a countable subfamily of $\mathcal{U}$ covering $X \times Y$.

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For a metric space $(X,d)$ and an infinite cardinal number $\kappa$, the following are equivalent:

  1. $X$ has a base of cardinality $\le \kappa$.
  2. X has a network of cardinality $\le \kappa$. (A network is a collection $\mathcal{N}$ of subsets of $X$ such that every open set is a union of elements from $\mathcal{N}$; a base is just a network that consists of open sets.)
  3. Every open cover of $X$ has a subcover of cardinality $\le \kappa$.
  4. Every closed discrete subspace $A$ of $X$ has cardinality $\le \kappa$.
  5. Every discrete subspace $A$ of $X$ has cardinality $\le \kappa$.
  6. Every pairwise disjoint family of non-empty open sets of $X$ has cardinality $\le \kappa$.
  7. $X$ has a dense subspace of cardinality $\le \kappa$.

$1)\rightarrow 2)$ is obvious, and true for all topological spaces $X$.

$2)\rightarrow 3)$ is true in general as well: Let $\mathcal{N}$ be a network with $\left|\mathcal{N}\right| \le \kappa$. If $\mathcal{U} = \left\{ U_i : i \in I \right\}$ is an open cover of $X$, then for each $x \in X$ we pick $i(x) \in I$ and $N_x \in \mathcal{N}$, such that $x \in N_x \subset U_{i(x)}$. Then $\left\{N_x : x \in X\right\} = \mathcal{N}'$ has cardinality $\le \kappa$, and for each distinct element $A$ from $\mathcal{N}'$ we pick $U(A)$ from $\mathcal{U}$ with $A \subset U(A)$ ($A = N_x$ for some $x$, and we pick $U(A) = U_{i(x)}$). Then $\left\{U(A) : A \in \mathcal{N}'\right\}$ is the required subcover.

$3)\rightarrow 4)$ is always true as well: Let $A$ be closed and discrete. Each $x \in A$ has an open neighbourhood $U_x$ that intersects $A$ in $\{x\}$ only. The open cover $\mathcal{U} = \left\{U_x : x \in A\right\} \cup \{X \setminus A\}$ cannot spare any $U_x$ (or $x$ will not be covered), so the cover $\mathcal{U}$ has cardinality $|A|$ and no subcover of cardinality strictly less than $|A|$. So $|A| \le \kappa$, or we'd have a contradiction with 3).

$4)\rightarrow 5)$ Here we need only perfect normality of $X$, in the sense only that each open set is a countable union of closed sets, or equivalently that each closed set is a $G_\delta$. Let $A$ be discrete, then I claim that $A$ is open in $\overline{A}$.

Proof of claim (needs only that singletons are closed): let $x$ be in $A$ and let $U_x$ be an open neighbourhood of $x$ that intersects $A$ only in $\{x\}$. This $U_x$ has the property that $\overline{A} \cap U_x = \{x\}$ as well: $y \neq x$ and $y \in \overline{A} \cap U_x$, then $U_x\setminus\{x\}$ is an open neighbourhood of $y$, $y \in \overline{A}$ so $U_x\setminus\{x\}$ must intersect $A$, but this can only happen in $\{x\}$, contradiction, so that $\{x\}$ is open in $\overline{A}$.

But then, as $A$ is perfectly normal (being metrisable), $A = \cup_{i \in \mathbb{N}} A_i$ where the $A_i$ are closed in $\overline{A}$ (and thus closed in $X)$. So the $A_i$ are closed and discrete, and by 4) we have $|A_i| \le \kappa$. So $|A| \le \aleph_0 \cdot \kappa = \kappa$, as well.

$5)\rightarrow 6)$ is true for all topological spaces: pick $x_i \in U_i$ for any pairwise disjoint family $\left\{U_i : i \in I\right\}$ of non-empty open sets. By definition we have that $\left\{x_i: i \in I\right\}$ is discrete (as witnessed by the $U_i$), and so $\left|I\right| \le \kappa$, and 6) has been proved.

$6)\rightarrow 7)$ Here we need the metric in a more essential way. For each $n \in \mathbb{N}$, let $D_n$ be a family of points with the property that $x,y \in D_n$ with $x \neq y$ implies $d(x,y) \ge \frac{1}{n}$, and $D_n$ is maximal with that property. Here we use Zorn's lemma, or some equivalent principle. Note that the balls with radius $\frac{1}{2n}$ around the points of $D_n$ are disjoint so that $|D_n| \le \kappa$ by 6).

Let $D = \cup_n D_n$, we claim that $D$ is dense in $X$. We already see that $D$ is of the right size, as $|D| \le \aleph_0 \cdot \kappa = \kappa$. For if $x$ is not in $\overline{D}$, we have that $d(x,\overline{D}) > 0$ and so for some $m \in \mathbb{N}$ we know that $d(x,\overline{D}) > \frac{1}{m}$. But then, for this $m$, $d(x,\overline{D_m}) \ge d(x,\overline{D}) > \frac{1}{m}$ and in particular: $d(x,y) > \frac{1}{m}$ for all $y \in D_m$. But then we could have added $x$ to $D_m$ and would have obtained a strictly larger $D_m$, and this cannot be. So $D$ is dense.

$7)\rightarrow 1)$ This needs the metric "most". Let $D$ be the dense subset of cardinality at most $\kappa$. Let $\mathcal{B} = \left\{B(x,r): x \in D; r \in \mathbb{Q}\right\}$, then $\left|\mathcal{B}\right| \le \aleph_0 \cdot \kappa = \kappa$. I claim that $\mathcal{B}$ is a base for $X$: let $U$ be open and $x \in U$. Some $\epsilon>0 $ exists such that $B(x,e) \subset U$, and as $D$ is dense there is some $y \in D$ in $B(x,\frac{\epsilon}{3})$. Now pick $r \in \mathbb{Q}$ such that $\frac{\epsilon}{3} < r < \frac{\epsilon}{2}$, then $x \in B(y,r)$ (which is from $\mathcal{B}$) and $B(y,r) \subset B(x,\epsilon)$: if for some $z$, $d(z,y) < r$ then $d(z,x) \le d(z,y) + d(y,x) < r + r < \epsilon$, and so there is a $B_x = B(y,r)$ from $\mathcal{B}$ such that $x \in B_x \subset U$, as required for a base.

This concludes the proof of the equivalence, which shows that weight, network weight, Lindelöf number, extent, cellularity and other cardinal invariants are all the same for metrisable spaces.

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"Network"? Is it a proper terminology choice? Archangielski's term was "net". I like to call it "Archangielski net". I like proper names to be attached to notions; it makes mathematics more humane, it gives me historical awareness. –  Wlodzimierz Holsztynski Feb 24 '13 at 4:15
The term was probably in Russian first; it was indeed Arhangel'skij that first introduced it. It's called a network in Engelking, e.g., and in many papers as well. For me, and also in Engelking, a net is a convergence notion (a generalisation of a sequence), so it's good to have 2 words for them. –  Henno Brandsma Feb 25 '13 at 19:02
2 words? Like "Archangielski's net"? I am just joking. @Henno, you're right, it's good to have "net" and "network" to tell apart the generalized sequences from "Archangielski's nets"; single words are more convenient than 2-word or longer phrases. –  Wlodzimierz Holsztynski Apr 30 '13 at 4:40
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Answer to Q1:

Let   $(X\ d)$   be a metric space. I call  $A\subseteq X$  $\epsilon$-dispersed $\quad\Leftarrow:\Rightarrow\quad\forall_{x\ y\in A}\ \left(\left(x\ne y\right)\Rightarrow d(x\ y)\ge \epsilon\right)$.

Let   $A_\epsilon$   be a maximal $\epsilon$-dispersed set in   $(X\ d)$   for every   $\epsilon > 0$   (apply Kuratowski-Zorn theorem). Then   $\bigcup_{n=1}^\infty\ A_{\frac 1n}$   is dense in   $(X\ d)$. (The rest is obvious).

Answer to Q2:

(I don't see any use for $\omega_1$--am I wrong?)

I call a topological space singular $\quad\Leftarrow:\Rightarrow\quad$ it has exactly one limit point (i.e. non-isolated).

THEOREM   The topological product of an arbitrary Lindelöf space by an arbitrary singular Lindelöf space is Lindelöf.

PROOF   In arbitrary singular Lindelöf space the complement of any open set, which contains the limit point, is countable. The rest is obvious.

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