Proof. Using the Cantor–Bendixson theorem, we write X in the form $Y\cup Z$ , $Y\cap Z=\emptyset $ , such that Y is a compact space without isolated points and Z is countable. Now, to get $\mathscr {K}$ , it suffices to find a relevant cover of Y and then to extend it by the singletons of Z. Instead, we simply assume that X has no isolated points.

There is a continuous surjection $u:\mathcal {P}(\mathbb {N}) \to X$ such that $u^{-1}(d)$ is a singleton, for $d\in D_0\cup D_1$ . Indeed, if we expand each $d\in D=D_0\cup D_1$ to a copy $P_d$ of the Cantor set in X, then [Reference Kuratowski7, Section 45, II, Theorem 4] gives us a continuous surjection $u:\mathcal {P}(\mathbb {N}) \to X$ which is injective on $u^{-1}(\bigcup _{d\in D}P_d)$ . In particular, since d is in $P_d$ , the preimage $u^{-1}(d)$ is a singleton.

Now, the disjoint sets $u^{-1}(D_i)$ , $i=0,1$ , are countable, and by adopting standard arguments concerning countable dense homogeneity of $2^{\mathbb {N}}$ (cf. [Reference van Mill9, Theorem 1.6.9]) one can find a homeomorphism $h:\mathcal {P}(\mathbb {N})\to \mathcal {P}(\mathbb {N})$ such that $u^{-1}(D_0)\subseteq h(\mathsf {Fin})$ and $u^{-1}(D_1)\subseteq h(\mathsf {cFin})$ .

Finally, if $\mathsf K$ is a collection of compact sets in $\mathcal {P}(\mathbb {N})$ satisfying the conditions in the definition of $\mathfrak d^*$ , with $|\mathsf K|=\mathfrak d^*$ , the collection $\mathscr {K}=\{(u\circ h)({\mathcal K}):\ {\mathcal K}\in \mathsf K\}$ satisfies the assertion of the lemma.

Proof. Let a function $f:[0,1]\rightarrow [0,1]$ be non-decreasing. Let D be the set of discontinuity points of f and let X be the closure in the square $[0,1]\times [0,1]$ of the graph of f restricted to the set $[0,1] \setminus D$ . For each $t\in [0,1]$ let $X(t)=X\cap (\{t\}\times [0,1])$ .

Since f is non-decreasing, the set D of discontinuity points of f is countable and for each $d\in D$ we have $X(d)=\{(d,d_0),(d,d_1)\}$ , where $d_0< d_1$ , while $X(t)=\{(t,f(t))\}$ for $t\in [0,1]\setminus D$ .

Let

$$ \begin{align*}Q_i=\{(d,d_i):\ d\in D\},\ i=0,1. \end{align*} $$

Since X is compact, Lemma 2.2 provides a covering $\mathscr {K}$ of X by compact sets such that $|\mathscr {K}|\leq \mathfrak d^*$ and each element of $\mathscr {K}$ intersects at most one of the sets $Q_0,\ Q_1$ .

Let $\pi : X\rightarrow [0,1]$ be the projection onto the first coordinate. Since each $K\in \mathscr {K}$ contains at most one point from every pair $X(d)=\{(d,d_0),(d,d_1)\}$ , $d\in D\cap \pi (K)$ , K is the graph of a continuous function $f_K: \pi (K)\rightarrow [0,1]$ . Since K is compact, $f_K$ is uniformly continuous. Moreover, f and $f_K$ coincide on $\pi (K)\setminus D$ , so f is uniformly continuous on $\pi (K)\setminus D$ .

We are ready to address statement $(*)$ . Let $E\subseteq [0,1]$ have cardinality $\mathfrak c$ . Since the sets $\pi (K)$ , $K\in \mathscr {K}$ , cover $[0,1]$ and $|\mathscr {K}|\leq \mathfrak d^* <\mathrm {cf}(\mathfrak c)$ (cf. Lemma 2.2), there is $K\in \mathscr {K}$ with $|\pi (K)\cap E|=\mathfrak c$ , and since f is uniformly continuous on the set $(\pi (K)\cap E)\setminus D$ , we get $(*)$ .

Proof. It is enough to find h satisfying the statement of the lemma for singletons $A\subseteq \kappa $ , that is, A of the form $\{\alpha \}$ for some $\alpha \in \kappa $ , since without loss of generality we may assume that for every finite $A\subseteq \kappa $ there exists $\gamma \in \kappa $ such that

$$ \begin{align*}\forall j\in\mathbb{N}\, \forall \alpha\in A\,\exists i\in\mathbb{N}\: \big([n^{\alpha}_{i}, n^{\alpha}_{i+1})\subseteq[n^{\gamma}_{j}, n^{\gamma}_{j+1}) \big).\end{align*} $$

For every $\alpha \in \kappa $ and $i\in \mathbb {N}$ set $f_\alpha (i)=n^\alpha _{2i}$ and find a strictly increasing $h\in \mathbb {N}^{\mathbb {N}}$ such that $h(0)=0$ and $h\not \leq ^* f_\alpha $ for all $\alpha \in \kappa $ . We claim that $\langle h(j):j\in \mathbb {N}\rangle $ is as required. Indeed, let us fix $i>0$ such that $h(i)\geq f_\alpha (i)=n^\alpha _{2i}$ . Thus,

$$ \begin{align*}\{n^\alpha_k:k\leq 2i\}\subseteq \bigcup_{l<i}[h(l),h(l+1)),\end{align*} $$

which gives that $ \big |\big \{k\ { < }\ 2i : n^\alpha _k,n^\alpha _{k+1}$ belong to the same interval $ [h(l),h(l+1)) \big \}\big |\geq i $ , and consequently $J_{\{\alpha \}}$ defined in the formulation must be infinite because i with $h(i)\geq f_\alpha (i)=n^\alpha _{2i}$ may be taken arbitrarily large.

Proof. For every $\mathcal K\in \mathsf K$ , $\mathcal K$ being compact, there is an increasing number sequence $\langle n^{\mathcal K}_{i}:i\in \mathbb {N}\rangle $ such that $X\cap [n^{\mathcal K}_{i},n^{\mathcal K}_{i+1})\neq \emptyset $ for any $X\in \mathcal K$ and $i\in \mathbb N$ . Since $|\mathsf K|<\mathfrak d$ , by Lemma 3.2 there exists a strictly increasing function $h:\mathbb {N}\to \mathbb {N}$ such that $h(0)=0$ and for any finite $\mathsf K'\subseteq \mathsf K$ the set $C(\mathsf K')$ of those $j\in \mathbb {N}$ such that for every $\mathcal K\in \mathsf K'$ there exists $i\in \mathbb {N}$ with

(3.1) $$ \begin{align} [h(j),h(j+1))\supseteq [n^{\mathcal K}_{i},n^{\mathcal K}_{i+1}) \end{align} $$

is infinite. Note that the family

$$ \begin{align*}\mathcal C=\{C(\mathsf K'):\mathsf K'\text{ is a finite subset of }\mathsf K\}\end{align*} $$

is centered because by the definition of $C(\mathsf K')$ we have

$$ \begin{align*}C(\mathsf K^{\prime}_0\cup \mathsf K^{\prime}_1\cup\cdots\cup \mathsf K^{\prime}_n)=\bigcap_{i\leq n}C(\mathsf K^{\prime}_i)\end{align*} $$

for any finite family $\{\mathsf K^{\prime }_0,\mathsf K^{\prime }_1,\ldots ,\mathsf K^{\prime }_n\}$ of finite subsets of $\mathsf K$ .

We claim that $\mathcal C$ and $\phi :\mathbb {N}\to \mathbb {N}$ such that $\phi ^{-1}(j)=[h(j),h(j+1))$ for all $j\in \mathbb {N}$ , are as required. Indeed, given $X\in \bigcup \mathsf K$ , let $\mathcal K\in \mathsf K$ be such that $X\in \mathcal K$ . Then

$$ \begin{align*} \phi[X]&=\{j\in\mathbb{N}: X\cap [h(j),h(j+1))\neq\emptyset\}\supseteq \\ &\supseteq \big\{j\in\mathbb{N}\: :\: \exists i\in\mathbb{N}\: \big([h(j),h(j+1))\supseteq [n^{\mathcal K}_{i},n^{\mathcal K}_{i+1})\big)\big\}\supseteq C(\{\mathcal K\})\in\mathcal C, \end{align*} $$

which completes the proof.

Proof. As we have already noticed, since both $\mathcal P(\mathbb N)\setminus \mathsf {Fin}$ and $\mathcal P(\mathbb N)\setminus \mathsf {cFin}$ are homeomorphic to the Baire space $\mathbb N^{\mathbb N}$ which can be covered by $\mathfrak d$ many compact subspaces, we have $\mathfrak d^*\leq \mathfrak d$ .

Let $\mathcal B$ be a reaping family of cardinality $\mathfrak r$ . For every $B\in \mathcal B$ set $\mathcal K(B)_d=\downarrow \!(\mathbb N\setminus B)$ and $\mathcal K(B)_u=\uparrow \!B$ .

It follows that

$$ \begin{align*}\mathcal P(\mathbb N)=\bigcup\{\mathcal K(B)_d:B\in\mathcal B\}\cup\bigcup \{\mathcal K(B)_u:B\in\mathcal B\}\end{align*} $$

and $\mathcal K(B)_d\cap \mathsf {cFin}=\mathcal K(B)_u\cap \mathsf {Fin}=\emptyset $ for all $B\in \mathcal B$ , hence $\mathfrak d^*\leq \mathfrak r$ , which together with the first paragraph yields $\mathfrak d^*\leq \min \{\mathfrak r,\mathfrak d\}$ .

To show that $\mathfrak d^*\geq \min \{\mathfrak u,\mathfrak d\}$ , suppose, towards a contradiction, that there is a family $\mathsf K=\mathsf K_d\cup \mathsf K_u$ of compact subspaces of $\mathcal P(\mathbb N)$ of size $\kappa =|\mathsf K|<\min \{\mathfrak u,\mathfrak d\}$ such that $\bigcup \mathsf K=\mathcal P(\mathbb {N})$ and

(3.2) $$ \begin{align} \mathcal K_0\cap\mathsf{cFin}=\mathcal K_1\cap\mathsf{Fin}=\emptyset \end{align} $$

for any $\mathcal K_0\in \mathsf K_d$ and $\mathcal K_1\in \mathsf K_u$ .

Replacing each $\mathcal K\in \mathsf K_d$ (resp. $\mathcal K\in \mathsf K_u$ ) with a countable family of compact subspaces of $\mathcal P(\mathbb {N})$ covering $\downarrow ^*\!\mathcal K$ (resp. $\uparrow ^*\!\mathcal K$ ), we can assume that $\mathcal U_u=\bigcup \mathsf K_u$ as well as $\mathcal U_d=\sim \!\bigcup \mathsf K_d$ are semifilters on $\mathbb {N}$ . Set $\mathcal U=\mathcal U_u\cup \mathcal U_d$ . It follows that

$$ \begin{align*}\mathcal U_d^+=\mathcal P(\mathbb N)\setminus\sim\!\mathcal U_d= \mathcal P(\mathbb N)\setminus\sim\!(\sim\!\bigcup\mathsf K_d)= \mathcal P(\mathbb N)\setminus \bigcup\mathsf K_d\subseteq \bigcup\mathsf K_u=\mathcal U_u,\end{align*} $$

and hence

(3.3) $$ \begin{align} \mathcal U^+\subseteq \mathcal U_d^+\subseteq \mathcal U_u\subseteq \mathcal U. \end{align} $$

By the construction, the semifilter $\mathcal U$ can be covered by $\kappa $ many compact subspaces. Since $\kappa <\mathfrak d$ , by Corollary 3.3, there exists a free filter $\mathcal F$ generated by at most $\kappa $ many subsets of $\mathbb {N}$ and a monotone surjection $\phi :\mathbb {N}\to \mathbb {N}$ with $\mathcal F\supseteq \{\phi [X]: X \in \mathcal U\}$ . Moreover, the family $\{\phi [X]: X \in \mathcal U\}$ is a semifilter and one easily checks that

$$ \begin{align*} \mathcal F^+\subseteq \{\phi[X]: X \in \mathcal U\}^+ = \{\phi[X]: X \in \mathcal U^+\}\subseteq \{\phi[X]: X \in \mathcal U\}\subseteq\mathcal F, \end{align*} $$

the second inclusion being a consequence of (3.3). This shows that $\mathcal F$ is an ultrafilter generated by at most $\kappa $ many sets, which is impossible because $\kappa <\mathfrak u$ .

It follows from the above that

$$ \begin{align*}\min\{\mathfrak u,\mathfrak d\}\leq \mathfrak d^* \leq \min\{\mathfrak r,\mathfrak d\},\end{align*} $$

and therefore these two inequalities must actually be equalities because $\mathfrak r\leq \mathfrak u$ .