Let *κ* ≥ *ω* be a regular cardinal. In the set ^{κ}κ we consider the following relation, if *f*, *g* ∈ ^{κ}κ then *g* ⪯ *f* iff |{*α* ∈ *κ* : *g*(*α*) > *f*(*α*)}| < *κ*.

A set *K* ⊆ ^{κ}κ is said to be *closed* iff for all *f* ∈ *K* and for all *g* ∈ ^{κ}κ if *f* ⪯ *g* then *g* ∈ *K*. A closed set *K* ⊆ ^{κ}κ is *κ*-*compact* iff there is a function *f* ∈ ^{κ}κ such that *K* ⊆ {*g* ∈ ^{κ}κ : *g* ⪯ *f*}.

#### Definition 2.1

*A set* *X* *of size* 2^{κ} *is a* *generalized K*-*Lusin set if* |*X* ∩ *K*| < 2^{κ} *for every* *κ*-*compact set* *K* ⊆ ^{κ}κ.

Now let 𝓕 ⊆ ^{κ} *κ* then

$$\begin{array}{}{\displaystyle {\lambda}_{\kappa}(\mathcal{F})=min\{\tau :{\mathrm{\forall}}_{g\in {}^{\kappa}\kappa}\phantom{\rule{thinmathspace}{0ex}}|\{f\in \mathcal{F}:f\u2aafg\}|<\tau \},}\\ \phantom{\rule{1em}{0ex}}{\displaystyle {\mathfrak{l}}_{\kappa}=min\{{\lambda}_{\kappa}(\mathcal{F}):\mathcal{F}\subseteq {}^{\kappa}\kappa \wedge |\mathcal{F}|={2}^{\kappa}\}.}\end{array}$$

Assuming GCH we obtain 𝔩_{κ} = 2^{κ}. It follows from the next lemma.

#### Lemma 2.2(GCH)

*There exists a family* 𝓕 ⊆ ^{κ}κ of cardinality κ^{+} *such that*

$$\begin{array}{}{\displaystyle {\lambda}_{\kappa}(\mathcal{F})={2}^{\kappa}.}\end{array}$$

#### Proof

Let

$$\begin{array}{}{\displaystyle {}^{\kappa}\kappa =\{{g}_{\alpha}:\alpha <{\kappa}^{+}\}.}\end{array}$$

We will construct a sequence {*f*_{α} ∈ ^{κ}κ : *α* < *κ*^{+}} such that

*f*_{β} ⪯ *f*_{α} for *β* < *α* < *κ*^{+}

*g*_{α} ⪯ *f*_{α} for *α* < *κ*^{+}

by transfinite recursion.

Assume that for *α* < *κ*^{+} the sequence {*f*_{β} ∈ ^{κ}κ : *β* < *α*} fulfilling (i) and (ii) has been defined. Since *κ*^{+} is regular, we have ^{κ}κ ∖ {*f*_{β} ∈ ^{κ}κ : *β* < *α*} ≠ ∅.

Let *α* be a successor. Take *g*_{α} ∈ ^{κ}κ and suppose that for each *f*_{α} ∈ ^{κ}κ ∖ {*f*_{β} ∈ ^{κ}κ : *β* < *α*} at least one of the conditions (i) or (ii) does not hold.

If (*ii*) does not hold then |{*f* ∈ ^{κ}κ : *f* ⪯ *g*_{β}}| = 2^{κ} for *β* < *α* . A contradiction. The argumentation for proving that (i) holds is similar.

If *α* is limit, then we take *f*_{α} = ⋃_{β<α} *f*_{β}. □

The following lemma is an easy consequence of the above definitions and Lemma 2.2.

#### Lemma 2.3

(GCH). *The following are equivalent*

𝔩_{κ} = 2^{κ}

*there exists a generalized K*-*Lusin set of cardinality* 2^{κ}.

Now we define a generalized strong sequence, (see also [6]). Let *α* be a cardinal number. We say that *A* ⊆ ^{κ}κ is *an* *α*-*directed set* iff for each *B* ⊂ *A* of cardinality less than *α* there exists *f* ∈ ^{κ}κ such that *g* ⪯ *f* for all *g* ∈ *B*.

#### Definition 2.4

*Let* *α* *and* *η* *be cardinals*. *A sequence* (*H*_{ϕ})_{ϕ<η}, *where* *H*_{ϕ} ⊂ *X*, *is called an* *α*-*strong sequence if*:

The following result is true.

#### Theorem 2.5

(GCH). *The following are equivalent*

𝔩_{κ} = 2^{κ}

*there exists a* *κ*^{+}-*strong sequence* (*H*_{ξ})_{ξ<κ+} *with* |*H*_{ξ}| < *κ*^{+} *for all* *ξ* < *κ*^{+}.

#### Proof

By Lemma 2.2 there exists a family 𝓕 ⊆ ^{κ}κ of cardinality *κ*^{+} with λ_{κ}(𝓕) = 2^{κ}. Choose *f*_{0} ∈ ^{κ}κ arbitrarily. Let *H*_{0} = {*f* ∈ 𝓕 : *f* ⪯ *f*_{0}}. Hence |*H*_{0}| < 2^{κ}. Let *H*_{0} be the first element of a *κ*^{+}-strong sequence.

Assume that the *κ*^{+}-strong sequence (*H*_{ξ})_{ξ<ψ} for *ψ* < *κ*^{+} such that

$$\begin{array}{}{\displaystyle {H}_{\xi}=\{f\in \mathcal{F}\setminus \bigcup _{\eta <\xi}{H}_{\eta}:f\u2aaf{f}_{\xi}\},}\end{array}$$

where *f*_{ξ} ∈ ^{κ}κ ∖ {*f*_{η} : *η* < *ξ*}, *f*_{ξ} ⊥ *f*_{η}, (*f*_{ξ} ⊥ *f*_{η} means *f*_{η} ⪯ *f*_{ξ} and *f*_{ξ} ⪯ *f*_{η} for *η* < *ξ*) has been defined.

Since *ψ* < *κ*^{+}, |𝓕| = *κ*^{+} and *κ*^{+} is regular, we have that there exists *f* ∈ ^{κ}κ ∖ {*f*_{ξ} : *ξ* < *κ*^{+}} such that *f* ⊥ *f*_{ξ} for any *ξ* < *ψ*. Name such *f* by *f*_{ψ}. Let *H*_{ψ} = {*f* ∈ 𝓕 ∖ ⋃_{ξ < ψ} *H*_{ξ} : *f* ⪯ *f*_{ψ}. Obviously |*H*_{ψ}| < *κ*^{+}, (because of our assumptions).
By the construction *H*_{ξ} ∪ *H*_{ψ}, *ξ* < *ψ* is not *κ*^{+}-directed, (because *f*_{ψ} ⊥ *f*_{ξ} for any *ξ* < *ψ*).

Now let (*H*_{ξ})_{ξ<κ+} be a *κ*^{+}-strong sequence with *H*_{ξ} ⊆ ^{κ}κ and |*H*_{ξ}| < *κ*^{+} for each *ξ* < *κ*^{+}. According to Definition 2.4 we have that for all *ξ* < *ψ* there exist *h*_{ξ} ∈ *H*_{ξ}, *h*_{ψ} ∈ *H*_{ψ} such that *h*_{ξ} ⊥ *h*_{ψ}. For any *ψ* < *κ*^{+} let

$$\begin{array}{}{\displaystyle {H}_{\psi}^{\prime}=\{{h}_{\psi}\in {H}_{\psi}:\xi <\psi \Rightarrow {\mathrm{\exists}}_{{h}_{\xi}\in {H}_{\xi}}\phantom{\rule{thinmathspace}{0ex}}{h}_{\xi}\perp {h}_{\psi}\}.}\end{array}$$

Since *H*_{ψ}, *ψ* < *κ*^{+} is *κ*^{+}-directed, there exists *h̄*_{ψ} ∈ ^{κ}κ such that *h* ⪯ *h̄*_{ψ}, for all *h*_{ψ} ∈
$\begin{array}{}{\displaystyle {H}_{\psi}^{\prime}}\end{array}$
. Thus we can describe *H*_{ψ} as follows

$$\begin{array}{}{\displaystyle {H}_{\psi}=\{f\in {}^{\kappa}\kappa :f\u2aaf{\overline{h}}_{\psi}\}.}\end{array}$$

Let 𝓕 = ⋃{*H*_{ξ} : *ξ* < *κ*^{+}}. For any *g* ∈ ^{κ}κ consider {*f* ∈ 𝓕 : *f* ⪯ *g*}. Suppose that there exists *g*_{o} ∈ ^{κ}κ such that |{*f* ∈ 𝓕 : *f* ⪯ *g*_{0}}| = *κ*^{+}. But by definition of 𝓕, there exists *ξ*_{0} such that {*f* ∈ 𝓕 : *f* ⪯ *g*_{0}} ⊆ *H*_{ξ0}. Hence |*H*_{ξ0}| = *κ*^{+}. A contradiction. □

The next corollary follows from Theorem 2.5 and Lemma 2.3.

#### Corollary 2.6

(GCH). *The following are equivalent*

*there exists a generalized K*-*Lusin set of cardinality* *κ*^{+}

*there exists an* *κ*^{+}-*strong sequence* (*H*_{ξ})_{ξ<κ+}, *H*_{ξ} ⊆ ^{κ}κ *with* |*H*_{ξ}| < *κ*^{+} *for any* *ξ* < *κ*^{+}.

For a cardinal *α* we introduce the following notation, (see [7])

$$\begin{array}{}{\displaystyle {\hat{s}}_{\alpha}=sup\{\eta :\phantom{\rule{thinmathspace}{0ex}}\text{\hspace{0.17em}there exists an}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\alpha \text{-strong sequence of cardinality\hspace{0.17em}}\eta \}.}\end{array}$$

Then by Theorem 2.5 we have

#### Corollary 2.7

(GCH). 𝔩_{κ} ≤ *ŝ*_{κ+}.

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