# A question on metrizable space

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

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

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. – Włodzimierz Holsztyński 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. – Włodzimierz Holsztyński Apr 30 '13 at 4:40

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).

(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).