Some cardinal functions in lexicographic products of LOTS

Abstract

The Lindelöf degree and extent were studied in Buhagiar et al. (Cent Eur J Math 12(8):1249–1264, 2014) in relation to lexicographic products of linearly ordered spaces. In this paper we consider the behaviour of other important cardinal functions, such as spread, density, weight and character, in such lexicographic products. Namely, we study the relation between a particular cardinal function on a lexicographic product of linearly ordered spaces and that cardinal function on each factor of the product.

1 Introduction

The lexicographic product of a collection , $\mu > 1$, of linearly ordered sets is an unexpectedly useful and ubiquitous set theoretic structure. When each linearly ordered set is given the order topology, it is important to study the behaviour of topological properties when lexicographic products are taken. Such investigations were done, for example, by Faber [3]. In this paper, the behaviour of some cardinal functions is investigated when lexicographic products of linearly ordered spaces are taken.

A linearly ordered topological space (abbreviated LOTS) is a triple $(X,\lambda (\leqslant ),\leqslant )$, where $(X,\leqslant )$ is a linearly ordered set (abbreviated LOS) and $\lambda (\leqslant )$ is the usual interval topology defined by $\leqslant$ [7, 8].

An ordered pair (A, B) of disjoint subsets of a LOS X is said to be a jump if (i) , (ii) $x<y$ whenever $x\in A$ and $y\in B$, and (iii) A has a maximal element a, and B has a minimal element b. Thus $a<b$ and . For a LOTS X we will let

We use standard notation for the cardinal functions on a topological space X. Namely,

As usual, we require all cardinal functions to take only infinite cardinals as values. The reader is referred to [4,5,6] for a thorough study of cardinal functions. Figure 1 shows the relations between the above cardinal functions in LOTS (see [2, p. 222]).

Fig. 1

Cardinal functions in LOTS

The behaviour of the cardinal functions $\ell$ and e in lexicographic products of LOTS was studied in [1], in this paper we study the behaviour of the other cardinal functions.

If X is a LOS and for some $x\in X$ we write $x=0$, then this would mean that X has a minimal element denoted by 0. Analogously, $x=1$ means that x is the maximal element of X denoted by 1. In particular, when we consider the two-point LOS or LOTS we understand that $0 < 1$. In the rest of the paper we assume that $|X| \geqslant 2$ for any LOS X.

2 Spread, density, weight and character of the lexicographic product of two LOTS

In this section we consider the lexicographic product of two LOTS X and Y.

Lemma 2.1

• (a) , , and .

• (b) If $|Y|>2$, then .

Proof

(a) For $x\in X$, the subspace of is order-isomorphic to Y, the order topology of induces on $_xL$ the order topology of $_xL$. The inequalities , and thus follow. The inequality follows from the equality $d=hd$.

(b) Since , for a non endpoint $y\in Y$, is a discrete subspace of we get . This implies . $\square$

Theorem 2.2

Proof

Suppose that . If , for $a,b\in X$ with $a<b$, then , hence . Moreover, if $D\subseteq X$ is discrete, then the set is a discrete set in and follows. Consequently, . Conversely, consider a cellular family ${{\mathscr {C}}}$ in , one can assume that each $I\in {{\mathscr {C}}}$ is an open interval of the form with

The image of ${{\mathscr {C}}}$ under the mapping is a collection of pairwise disjoint open intervals in X. Noting that defines a jump in X, one concludes that .

If $|Y| > 2$, the inequality follows from Lemma 2.1. Conversely, if is discrete, then for any $x\in X$, is discrete in Y, so that $|D_x| \leqslant s(Y)$. Consequently, and therefore, . $\square$

An analogous formula holds for the density.

Theorem 2.3

Proof

If , then follows from Theorem 2.2. Suppose is dense and . Then the set is a dense set in X. Consequently, and therefore, . Conversely, suppose $D\subseteq X$ is a dense subset of X satisfying $|D| \leqslant d(X)$. Let . Then is dense in , adding to P any possible isolated end points. Note that so that follows.

If $|Y| > 2$, the inequality follows from Lemma 2.1. Suppose $D\subseteq Y$ is dense in Y with $|D| \leqslant d(Y)$, one can assume that $0,1\in D$ if they exist in Y. Then is dense in and follows. $\square$

We next consider the weight.

Theorem 2.4

.

Proof

Suppose and let ${{\mathscr {B}}}$ be a base of open intervals in with . For any $x\in X$ choose $I(x)\in {{\mathscr {B}}}$ with $(x,0)\in I(x)$ and $(x,1)\notin I(x)$. Since $X\ni x\mapsto I(x)$ is injective, the inequality follows. On the other hand and therefore, .

If $|Y| > 2$, the inequality follows from Lemma 2.1. The inequality easily follows if Y does not have endpoints. Now suppose $0\in Y$. Given $x\in X$, a neighbourhood base for (x, 0) must have cardinality so that the collection of neighbourhood bases for (x, 0), for all $x\in X$, must have cardinality . Similarly for the case that $1\in Y$. Consequently, after considering a neighbourhood base for points (x, y) with $y\notin \{0,1\}$, it follows that . $\square$

We end this section by considering . For a LOS X, we will denote the topology generated by sets of the form

$a,b\in X$, and $\{1\}$ (if $1\in X$) by $\tau _{\mathrm{r}}$, and similarly, the topology generated by sets of the form , $a,b\in X$, and $\{0\}$ (if $0\in X$) by $\tau _{\mathrm{l}}$. We will write $\chi _{\mathrm{r}}(x)$ to mean the character of $x \in (X,\tau _{\mathrm{r}})$ and $\chi _{\mathrm{r}}(X)$ for the character of $(X,\tau _{\mathrm{r}})$. Similarly for $\chi _{\mathrm{l}}(x)$ and $\chi _{\mathrm{l}}(X)$. If one needs to specify that the character of x is taken in X, then one writes $\chi (x,X)$ (similarly $\chi _{\mathrm{r}}(x,X)$ and $\chi _{\mathrm{l}}(x,X)$). Recall that as shown in Fig. 1, the character of a LOTS is equal to its pseudo-character. One can also note that for any $x\in X$, .

Proposition 2.5

For two LOTS X and Y,

Proof

For we have

Taking the supremum over all we get the formula for . Dually we get the formula for , while . $\square$

Example 2.6

Let us look at the following examples:

• 1. Consider . Then $s(Z) = \aleph _0$ (since $s({\mathbb {R}}) = {\mathbb {J}}_{{\mathbb {R}}} = \aleph _0$). But . Analogously, $d(Z) = \aleph _0$ and . On the other hand, if one takes , then .

• 2. Let $X = {\mathbb {R}}$ and . Then .

• 3. If and

  

  then . However, if we take and , then .

3 Spread, density, weight and character of the lexicographic product of LOTS

In what follows, is the lexicographic product of $X_\alpha$, where $X_\alpha$ is a LOS for every $\alpha < \mu$ and $\mu$ is a limit ordinal.

Theorem 3.1

If , then

Proof

For every $0< \alpha < \mu$ we have , so that, by Theorem 2.2, . Hence $s(X) \geqslant \sup _{\alpha< \mu }\bigl |\prod _{\gamma < \alpha } X_\gamma \bigr |$.

Now let us look at the weight of X. We show that $w(X) \leqslant \sup _{\alpha< \mu }\bigl |\prod _{\gamma < \alpha } X_\gamma \bigr |$ from which the result follows. Let $z = (z_\alpha )_{\alpha < \mu }\in X$ be defined as follows:

$$\begin{aligned} z_\alpha = {\left\{ \begin{array}{ll}\, \text{ chosen } \text{ arbitrarily } \in X_\alpha &{}\text{ if } X_\alpha \text{ does } \text{ not } \text{ have } 1, \\ \,1 &{} \text{ otherwise. } \end{array}\right. } \end{aligned}$$

For every $\gamma < \mu$ let and let $D = \bigcup _{\gamma < \mu }Z_\gamma$. Then $|Z_\gamma | = \bigl |\prod _{\alpha \leqslant \gamma }X_\alpha \bigr |$ and therefore,

We show that for every $x\in X$ and every $c < x$, there exists $a\in D$ such that $c\leqslant a < x$. Let $x = (x_\alpha )_{\alpha < \mu }$ and $c = (c_\alpha )_{\alpha < \mu }$. Let $\alpha _0$ be the first index such that . If $c_\alpha = 1$ for all $\alpha > \alpha _0$, then let $a = c \in D$, otherwise there exists some $\delta > \alpha _0$ such that $c_\delta < y$ for some $y\in X_\delta$, and we define $a = (a_\alpha )_{\alpha < \mu }\in D$ by

$$\begin{aligned} a_\alpha = {\left\{ \begin{array}{ll} \,c_\alpha &{}\text{ if }\;\;\alpha < \delta , \\ \,y &{}\text{ if }\;\;\alpha = \delta , \\ \,z_\alpha &{}\text{ if }\;\;\alpha > \delta . \end{array}\right. } \end{aligned}$$

Dually, there exists a subset with $|D'| \leqslant \sup _{\gamma< \mu }\bigl |\prod _{\alpha <\gamma }X_\alpha \bigr |$ such that for every $x\in X$ and every $d > x$, there exists $b\in D'$ such that $d\geqslant b > x$. Consequently, there exists a base of cardinality $\sup _{\gamma< \mu }\bigl |\prod _{\alpha <\gamma }X_\alpha \bigr |$ as required to show. $\square$

We are left with the calculation of the character $\chi (X)$ of a lexicographic product , where $\mu$ is a limit ordinal. We first calculate $\chi _{\mathrm{r}}(x,X)$ for $x\in X$ and then $\chi _{\mathrm{r}}(X)$. By duality we obtain $\chi _{\mathrm{l}}(x,X)$ and $\chi _{\mathrm{l}}(X)$. Finally, we get $\chi (x,X)$ and $\chi (X)$ as a maximum of $\chi _{\mathrm{r}}$ and $\chi _{\mathrm{l}}$.

For $x = (x_\alpha )_{\alpha < \mu }\in X$, let

[We use the following convention: $\mu _1(x) = \mu$ when . In other words, $\inf \varnothing = \mu$, which is true when $A_1(x)$ is cofinal in $\mu$. If $A_1(x)$ is not cofinal in $\mu$ then .]

Lemma 3.2

For $x = (x_\alpha )_{\alpha < \mu }\in X$ we have

Proof

Case I: $A_1(x)$ is cofinal in $\mu$. In this case we show that . Choose $z_\alpha > x_\alpha$ for every $\alpha \in A_1(x)$ and let $y^\gamma = (y^\gamma _\alpha )_{\alpha < \mu }\in X$, for every $\gamma \in A_1(x)$, be defined by:

$$\begin{aligned} y^\gamma _\alpha = {\left\{ \begin{array}{ll}\, x_\alpha &{}\text{ if }\;\; \alpha \ne \gamma , \\ \,z_\gamma &{} \text{ if }\;\; \alpha = \gamma . \end{array}\right. } \end{aligned}$$

Take any $A\subseteq A_1(x)$ with , then $y^\gamma > x$ for every $\gamma \in A$ and $\inf _{\gamma \in A}y^\gamma = x$. Hence $\chi _{\mathrm{r}}(x,X)\leqslant |A| = \mathrm{cf}(\mu )$. To prove the converse, suppose $\chi _{\mathrm{r}}(x,X) < \mathrm{cf}(\mu )$. There exists a set B with and elements $b^\beta = (b^\beta _\alpha )_{\alpha < \mu }\in X$ with $b^\beta > x$ for all $\beta \in B$ such that $\inf _{\beta \in B}b^\beta = x$. Let , then . Therefore, there exists $\xi <\mu$ such that $\alpha _\beta < \xi$ for all $\beta \in B$. Take any $\gamma \in A_1(x)$ with $\gamma > \xi$. Then $x< y^\gamma < b^\beta$ for all $\beta \in B$, so that $\inf _{\beta \in B}b^\beta \geqslant y^\gamma > x$, a contradiction.

Case II: $A_1(x)$ is not cofinal in $\mu$. If $x_\alpha = 1$ for all $\alpha < \mu$ then x is the greatest element of X, hence $\chi _{\mathrm{r}}(x,X) = 1$. Otherwise, , see the proof of Proposition 2.5 for . If $\mu _1(x)$ is a limit ordinal, then $\chi _{\mathrm{r}}((x_\alpha )_{\alpha <\mu _1(x)}$, by Case I above. Otherwise, let $\beta <\mu$ satisfy $\beta + 1 = \mu _1(x)$. Then and by Proposition 2.5 one obtains

$\square$

For our next corollary we let

Hence, using the above convention, $\mu _1 = \mu$ if and only if is cofinal in $\mu$. For an ordinal $\alpha$, let us denote by $\alpha ^-$ the immediate predecessor of $\alpha$ if it exists; otherwise, if $\alpha$ is a limit ordinal (or 0) we let $\alpha ^- = \alpha$. Moreover, in line with our interest in infinite cardinals, we let $\sup \varnothing = \aleph _0$ (instead of $\sup \varnothing = 0$).

Corollary 3.3

For we have

To calculate $\chi _{\mathrm{l}}(x,X)$ and $\chi _{\mathrm{l}}(X)$ for $x = (x_\alpha )_{\alpha < \mu }\in X$, let

As above we use the convention: $\mu _0(x) = \mu$ when . By duality we have the following two results:

Lemma 3.4

For $x = (x_\alpha )_{\alpha < \mu }\in X$ we have

Also, if we let , we have the following corollary.

Corollary 3.5

For we have

Using Lemmas 3.2 and 3.4 we can calculate $\chi (x,X)$ by the formula

Finally, by Corollaries 3.3 and 3.5 and the formula , we have the following result for the character of a lexicographic product.

Theorem 3.6

If , then

where .

Remark 3.7

A particular case of Theorem 3.6 is when $\delta = \mu _0 = \mu _1$, then

Example 3.8

Let us look at the following examples.

• 1. Suppose for all $\alpha < \omega _1$ and let . Then $\chi (X) = \aleph _2$.

• 2. Suppose for all $\alpha < \omega _1$ and let . Then $\chi (X) = \aleph _1$.

• 3. Let and

  

  If , where $X_\alpha = Y$ for all $\alpha <\omega _0$ and $X_\alpha = Z$ for all $\omega _0\leqslant \alpha <\omega _1$, then $\chi (X) = \aleph _1$. However if, , where $X_\alpha = Y$ for all $\alpha \leqslant \omega _0$ and $X_\alpha = Z$ for all $\omega _0<\alpha <\omega _1$, then $\chi (X) = \aleph _2$.

• 4. Let for all $\alpha < \omega _\omega$, $\alpha \ne \omega _n$ for all $n < \omega _0$, and for all $\alpha < \omega _\omega$, $\alpha = \omega _n$ for some $n<\omega _0$. If , then .

We end this section by noting that the results of Sect. 2 and of this section give us a way of calculating $\phi (X)$, where , $\mu = \lambda + 1$ is a successor ordinal and $\phi$ is any of the considered cardinal functions.

One needs to look into the following four cases to calculate the spread and use Theorems 2.2 and 3.1:

• (1) $|X_\lambda |>2$: ;

• (2a) $|X_\lambda |=2$, $\lambda$ is a limit ordinal: , because

  ;

• (2b) (i) $|X_\lambda |=2$, $\lambda = \nu + 1$ with $|X_\nu |>2$;

• (2b) (ii) $|X_\lambda |=2$, $\lambda = \nu + 1$ with $|X_\nu |=2$:

  In both cases .

  In (2b) (ii), in general if $|X_\nu |\leqslant \aleph _0$, we can simplify to obtain $s(X) = k(X) = \bigl |\prod _{\alpha < \mu } X_\alpha \bigr |$.

Analogous formulas apply for d(X). If we want to calculate the weight of X we use Theorem 2.4 to conclude

Finally, one uses Proposition 2.5 to calculate the character of X. As above, let $\mu = \lambda + 1 = \lambda _0 + n$, $1\leqslant n <\omega _0$, be a successor ordinal, where $\lambda _0$ is a limit ordinal. Let and let

Analogously define . Put and $Z_k = X_{\lambda _0+k}$ for . Note that $\chi _{\mathrm{r}}(Z_{-1}),\chi _{\mathrm{l}}(Z_{-1})$ and $\chi (Z_{-1})$ can be calculated by Corollaries 3.3, 3.5 and Theorem 3.6, respectively. Then

Example 3.9

Let us look at a simple example. Suppose for all $\alpha < \omega _1$ and let .

• 1. Let and let . Then

  

• 2. Let $X_{\omega _1} = \{0,1\}$ and let . Then