and any $l\in\\{1,\ldots,n\\}$, it is enough to show that
$s_{1}\coloneqq(\bar{t}\setminus t_{1})\cup\left(t(l)\setminus
t(1)\right)\in\mathcal{F}_{\beta^{\prime}}$. Let us denote $r_{0}=t\setminus
t_{1}$, $r_{1}=t(l)$, $r_{2}=t$, $s_{0}=\bar{t}\setminus t_{1}$,
$s_{2}=\bar{t}$. This way we obtain a pack
$(r_{0},r_{1},r_{2},s_{0},s_{1},s_{2})$, and we can use Theorem 3.3.18. ∎

Proof of Theorem 3.3.16 The theorem follows by a straightforward application
of Theorem 3.3.17. ∎

#### C. On concepts of strict convexity

In the construction of nets demanded in condition (**) of Theorem 3.3.16, we
will use an equivalent definition of the class $\mathcal{F}_{\varrho}$ of
$\varrho$-convex sets, for the fixed set $\mathbb{T}$

We will fix the number $\varrho\geq 1$ and use the following notation.

  * We set $\mathcal{F}=\left\\{t\in\mathcal{T}\colon t\subset\mathbb{T}\right\\}$.

  * For a convex, closed set $s\subset\mathbb{R}^{2}$, and $\boldsymbol{x}\in\mathbb{R}^{2}\setminus s$, $P_{s}\boldsymbol{x}$ will be the minimal distance projection: $P_{s}\boldsymbol{x}=\boldsymbol{y}$, if $\boldsymbol{y}\in s$ and $\|\boldsymbol{x}-\boldsymbol{y}\|=\min{\left\\{\|\boldsymbol{x}-\boldsymbol{z}\|\colon\boldsymbol{z}\in s\right\\}}$;

  * we also write $B(s,\boldsymbol{x})=B$, if $B\in\mathcal{B}_{\varrho}$ and $P_{B}\boldsymbol{x}=P_{s}\boldsymbol{x}$.

  * If additionally $s\in\mathcal{F}$ and $\boldsymbol{x}\in\mathbb{T}\setminus s$, we define $E(s,\boldsymbol{x})$ to be the connected component of $\mathbb{T}\setminus B(s,\boldsymbol{x})$ containing $\boldsymbol{x}$ as a member.

  * We set $C(s,\boldsymbol{x})=\mathbb{T}\setminus E(s,\boldsymbol{x})$, if $E(s,\boldsymbol{x})$ is defined.

  * Lastly, we denote

$\mathcal{C}_{\varrho}=\Big{\\{}\mathbb{T}\setminus E\colon E\text{ is a
connected component of }\mathbb{T}\setminus B\text{,}\\\ \text{ for some
}B\in\mathcal{B}_{\varrho}\text{, }\operatorname{int}(\mathbb{T}\cap
B)\neq\emptyset\text{, }\mathbb{T}\setminus B\neq\emptyset\Big{\\}}\text{.}$

######  Theorem 3.3.22.

For any set $t\in\mathcal{F}$, the following conditions are equivalent:

  * $\alpha$)

$t\in\mathcal{F}_{\varrho}$,

  * $\beta$)

$t=\bigcap_{C\in\mathcal{D}\cup\\{\mathbb{T}\\}}C$, for some family
$\mathcal{D}\subset\mathcal{C}_{\varrho}$,

  * $\gamma$)

$t\subset C(t,\boldsymbol{x})$, for every
$\boldsymbol{x}\in\mathbb{T}\setminus t$.

Our proof of the theorem will follow from a sequence of lemmas concerning the
conditions $\alpha$), $\beta$), $\gamma$).

######  Lemma 3.3.23.

For any $t\in\mathcal{F}$, condition $\gamma$) implies $\beta$), and condition
$\beta$) implies $\alpha$).

Proof. To show that $\gamma)\implies\beta)$, it is enough to take
$\mathcal{D}=\left\\{C(t,\boldsymbol{x})\colon\boldsymbol{x}\in\mathbb{T}\setminus
t\right\\}$.

To prove that $\beta)\implies\alpha)$, consider $t\neq\mathbb{T}$. We define a
function $f\colon\mathcal{C}_{\varrho}\to\mathcal{B}_{\varrho}$, continuous
with respect to the Hausdorff metric, by the relation: $f(\mathbb{T}\setminus
E)=B$, when $E$ is a connected component of $\mathbb{T}\setminus B$. Then, for
$\boldsymbol{x}_{0}\in\partial t\cap\operatorname{int}{(\mathbb{T})}$ we can
find sets $C_{1},C_{2},\ldots\in\mathcal{D}$ satisfying
$\boldsymbol{x}_{0}\in\partial\bigcap_{i\in\mathbb{N}}C_{i}$. By a subsequence
argument, we can assume that $d\left(f(C_{i}),B_{0}\right)\to 0$, when
$i\to\infty$, for some ball $B_{0}\in\mathcal{B}_{\varrho}$ satisfying
$\boldsymbol{x}_{0}\in\partial B_{0}$. This, in turn, implies the
$\varrho$-convexity of $t$ at $\boldsymbol{x}_{0}$. ∎

The proof of the implication $\alpha)\implies\gamma)$ is somewhat more
laborious.

######  Lemma 3.3.24.

If $t\in\mathcal{F}_{\varrho}$, $\boldsymbol{x},\boldsymbol{y}\in t$,
$\boldsymbol{x}\in\operatorname{int}{(\mathbb{T})}$, then we have
$[\boldsymbol{x},\boldsymbol{y}]\setminus\\{\boldsymbol{x},\boldsymbol{y}\\}\subset\operatorname{int}{(t)}$.

Proof. In view of convexity of $\mathbb{T}$, the assumption
$\boldsymbol{x}\in\operatorname{int}{(\mathbb{T})}$ implies
$I\subset\operatorname{int}{(\mathbb{T})}$ with
$I=[\boldsymbol{x},\boldsymbol{y}]\setminus\\{\boldsymbol{x},\boldsymbol{y}\\}$.
For any $\boldsymbol{z}\in I$ and any ball $B\in\mathcal{B}_{\varrho}$
satisfying $\partial B\ni\boldsymbol{z}$, we have $V\cap I\not\subset B$, for
any neighborhood $V$ of $\boldsymbol{z}$. Thus, supposition of
$\boldsymbol{z}\in\partial t$ contradicts the assumption that
$t\in\mathcal{F}_{\varrho}$. ∎

######  Lemma 3.3.25.

If a set $t\in\mathcal{T}$, for some ball $B\in\mathcal{B}_{\varrho}$ and
point $\boldsymbol{x}\in\partial B\cap\partial t$, satisfies the conjunction

$\exists{\varepsilon_{1}>0}\colon B(\boldsymbol{x},\varepsilon_{1})\cap
B\subset t\quad\text{and}\quad\forall{\varepsilon>0}\colon
B(\boldsymbol{x},\varepsilon)\cap t\not\subset B\text{,}$

then $t$ cannot be $\varrho$-convex at $\boldsymbol{x}$.

Proof. Let $\varepsilon_{1}$ satisfy $B(\boldsymbol{x},\varepsilon_{1})\cap
B\subset t$. For any ball $B_{1}\in\mathcal{B}_{\varrho}$ and any number
$\varepsilon\in(0,\varepsilon_{1})$, we have the following two implications.
If $B_{1}\neq B$, $\boldsymbol{x}\in\partial B_{1}$, then
$B(\boldsymbol{x},\varepsilon)\cap B\not\subset B_{1}$, so
$B(\boldsymbol{x},\varepsilon)\cap t\not\subset B_{1}$. If $B_{1}=B$, then
$B(\boldsymbol{x},\varepsilon)\cap t\not\subset B_{1}$ by assumptions. Since
$\varepsilon\in(0,\varepsilon_{1})$ has been arbitrarily chosen, the set $t$
is not $\varrho$-convex at $\boldsymbol{x}$. ∎

For $\boldsymbol{x},\boldsymbol{y}\in\mathbb{T}$,
$\boldsymbol{x}\neq\boldsymbol{y}$, we call $l$ a _bow with endpoints
$\boldsymbol{x},\boldsymbol{y}$_, if $l$ is an arc of some circle $\partial
B$, $B\in\mathcal{B}_{\varrho}$, with endpoints
$\boldsymbol{x},\boldsymbol{y}$, and $d(l)=\|\boldsymbol{x}-\boldsymbol{y}\|$.
Thus, we always have $d(l)\leq 1$, as $d(\mathbb{T})\leq 1$.

######  Lemma 3.3.26.

For any $t\in\mathcal{F}_{\varrho}$ and any bow $l$ with endpoints
$\boldsymbol{x},\boldsymbol{y}$, if $\boldsymbol{x},\boldsymbol{y}\in t$ and
$\boldsymbol{x}\in\operatorname{int}{(\mathbb{T})}$, $l\subset\mathbb{T}$,
then $l\subset t$.

Proof. Changing coordinate system if necessary, we have

$l=\left(\lvert x\rvert\leq\alpha,\
y=\sqrt{\varrho^{2}-x^{2}}-\sqrt{\varrho^{2}-\alpha^{2}}\right)\text{,}$

for some $\alpha\in(0,1)$. Let us denote

$l_{\lambda}=\left(\lvert x\rvert\leq\lambda,\
y=\sqrt{\varrho^{2}-x^{2}}-\sqrt{\varrho^{2}-\lambda^{2}}\right)\text{,}$

for $\lambda\in[0,\alpha]$. Assume that $l\not\subset t$. Since
$l=l_{\alpha}$, we have $l_{\lambda}\not\subset t$ for $\lambda$ close to
$\alpha$. On the other hand, by Lemma 3.3.24, we have
$\boldsymbol{0}\in\operatorname{int}{(t)}$, thus $l_{\lambda}\subset t$ for
$\lambda$ close to $0$. Let $\lambda_{0}$ be the highest number in
$(0,\alpha)$ for which $l_{\lambda_{0}}\subset t$. Then, the set $k\coloneqq
l_{\lambda_{0}}\cap\partial t$ is nonempty and closed. Let us consider point
$\boldsymbol{x}_{0}=(x_{0},y_{0})\in k$ of the smallest possible value of
$x_{0}$. In view of Lemma 3.3.24, the line segment $\left(\lvert
x\rvert<\alpha,y=0\right)$, is contained in $\operatorname{int}{(t)}$,
therefore the endpoints of $l_{\lambda_{0}}$ are not elements of $k$. We have
then $\boldsymbol{x}_{0}\in\operatorname{int}{(\mathbb{T})}$, by
$\lambda_{0}\in(0,\alpha)$, moreover, for the ball $B_{0}$ satisfying
$l_{\lambda_{0}}\subset\partial B_{0}$, we have

$\forall{\varepsilon>0}\colon B(\boldsymbol{x}_{0},\varepsilon)\cap
t\not\subset B_{0}\quad\text{and}\quad\exists{\varepsilon>0}\colon
B(\boldsymbol{x}_{0},\varepsilon)\cap B_{0}\subset t\text{.}$

According to Lemma 3.3.25 we have obtained a contradiction with the assumption
that $t\in\mathcal{F}_{\varrho}$, which completes the proof. ∎

######  Lemma 3.3.27.

Let $t\in\mathcal{F}_{\varrho}$, $\boldsymbol{x}\in\mathbb{T}\setminus t$. We
then have
$\partial^{-}E(t,\boldsymbol{x})\cap\operatorname{int}{(t)}=\emptyset$.

Proof. Suppose that there exists a point
$\boldsymbol{z}\in\partial^{-}E(t,\boldsymbol{x})\cap\operatorname{int}{(t)}$.
Changing coordinate system if necessary, we can assume that
$\boldsymbol{x}=(0,\chi)$, for some $\chi>0$,
$P_{t}\boldsymbol{x}=\boldsymbol{0}$,
$B(t,\boldsymbol{x})=B((0,-\varrho),\varrho)$,
$\boldsymbol{z}=(\xi,\sqrt{\varrho^{2}-\xi^{2}}-\varrho)$, for some
$\xi\in(0,1)$.

For $\varepsilon>0$ let us denote line segment
$I_{\varepsilon}\coloneqq\left(x=0,0<y\leq\varepsilon\right)$. The arc

$l=\left(0\leq x\leq\xi,y=\sqrt{\varrho^{2}-x^{2}}-\varrho\right)\text{,}$

with endpoints $\boldsymbol{0}$ and $\boldsymbol{z}$, satisfies
$l\subset\partial^{-}E(t,\boldsymbol{x})$ with $E(t,\boldsymbol{x})\supset
I_{\varepsilon}\cup(\boldsymbol{z}+I_{\varepsilon})$, for some
$\varepsilon>0$. Then, by connectedness of $E(t,\boldsymbol{x})$, we have
$l+I_{\varepsilon_{1}}\subset E(t,\boldsymbol{x})$, for some
$\varepsilon_{1}\in(0,\varepsilon)$.

Now we can see that in the strip $\operatorname{clos}(l+I_{\varepsilon_{1}})$
we can find bows with endpoints $\boldsymbol{0}$ and $\boldsymbol{v}$, which
we may denote $l_{\boldsymbol{v}}$, satisfying
$\boldsymbol{v}\in\boldsymbol{z}+I_{\varepsilon_{1}}$. Adjusting
$\varepsilon_{1}$ to be suitably small, we can assure that
$\boldsymbol{z}+I_{\varepsilon_{1}}\subset\operatorname{int}(t)$. By Lemma
3.3.26, we have $l_{\boldsymbol{v}}\subset t$, for
$\boldsymbol{v}\in\boldsymbol{z}+I_{\varepsilon_{1}}$. Lastly, let us note
that for those arcs we also have $l_{\boldsymbol{v}}\not\subset(y\leq 0)$,
which contradicts the fact that $P_{t}(0,\chi)=\boldsymbol{0}$, with some
$\chi>0$. ∎

Proof of Theorem 3.3.22 In view of Lemma 3.3.23, we will show that
$\alpha)\implies\gamma)$. For $t\in\mathcal{F}_{\varrho}$ let us suppose that
there exists $\boldsymbol{z}\in E(t,\boldsymbol{x})\cap t$. Then,
$B(\boldsymbol{z},\varepsilon)\cap B(t,\boldsymbol{x})=\emptyset$, for some
$\varepsilon>0$. Remark 3.3.5 yields existence of a ball
$B(\boldsymbol{z}_{1},\varepsilon_{1})\subset
B(\boldsymbol{z},\varepsilon)\cap\operatorname{int}{(t)}$. As a result,
$B(\boldsymbol{z}_{1},\varepsilon_{1})$ and $\\{\boldsymbol{z}\\}$ are
contained in the same connected component of $\mathbb{T}\setminus
B(t,\boldsymbol{x})$, and we have
$B(\boldsymbol{z}_{1},\varepsilon_{1})\subset E(t,\boldsymbol{x})\cap t$.

In a suitable coordinate system we have $\boldsymbol{x}=(0,\chi)$,
$P_{t}\boldsymbol{x}=\boldsymbol{0}$, $t\subset(y\leq 0)$, thus
$\boldsymbol{z}_{1}\in(y<0)$, $B(t,\boldsymbol{x})=B((0,-\varrho),\varrho)$
and $\boldsymbol{z}_{1}\notin B((0,-\varrho),\varrho)$. Therefore, there
exists a point $\boldsymbol{y}\in\partial
B((0,-\varrho),\varrho)\cap\left([\boldsymbol{0},\boldsymbol{z}_{1}]\setminus\\{\boldsymbol{0},\boldsymbol{z}_{1}\\}\right)$.
Then, $\boldsymbol{y}\in\operatorname{int}{(t)}$, as
$\boldsymbol{z}_{1}\in\operatorname{int}(t)$, and
$\boldsymbol{y}\in\partial^{-}E(t,\boldsymbol{x})$, as $\boldsymbol{z_{1}}\in
E(t,\boldsymbol{x})$, $\boldsymbol{0}\in\partial^{-}E(t,\boldsymbol{x})$, with
$E(t,\boldsymbol{x})$ being a connected component of $\mathbb{T}\setminus
B((0,-\varrho),\varrho)$. We have obtained a contradiction with Lemma 3.3.27,
which ends the proof. ∎

#### D. Constructions of nets

Our proof of the condition (**) in Theorem 3.3.16 will be based on the
following two theorems.

######  Theorem 3.3.28.

There are numbers $\delta>0$ and $j\in\mathbb{N}$, independent on
$\mathbb{T}$, such that for any soul $(t,t_{1})\in\tau_{\beta^{\prime}/2}$
satisfying $d(t,t\setminus t_{1})\leq\delta$, the family
$\mathcal{N}_{j}^{(t,t_{1})}$ is not empty.

######  Theorem 3.3.29.

For any $\delta>0$ there is a number $k\in\mathbb{N}$, independent on
$\mathbb{T}$, such that for any soul $(t,t_{1})\in\tau_{\beta^{\prime}/2}$,
there are sets $t(i)\in\mathcal{F}_{\beta^{\prime}/2}$, $1\leq i\leq k$,
satisfying $t\setminus t_{1}=t(1)\subset\ldots\subset t(k)=t$ and
$d(t(i),t(i-1))\leq\delta$, for $2\leq i\leq k$.

Using these theorems it is easy to see the validity of the following.

######  Theorem 3.3.30.

Condition (**) of Theorem 3.3.16 is satisfied.

Proof. Let us fix numbers $\delta>0$, $j\in\mathbb{N}$ as in Theorems 3.3.28.
For the number $\delta$ we choose $k\in\mathbb{N}$ as in Theorem 3.3.29. Then
we set $n^{*}=(k-1)j$. In particular numbers $\delta$, $j$, $k$, and $n^{*}$
do not depend on $\mathbb{T}$. With this notation, for any soul
$(t,t_{1})\in\tau_{\beta^{\prime}/2}$, we can find sets
$t(1),\ldots,t(k)\in\mathcal{F}_{\beta^{\prime}/2}$, satisfying $t\setminus
t_{1}=t(1)\subset\ldots\subset t(k)=t$, $d(t(i),t(i+1))\leq\delta$, for $2\leq
i\leq k$. Next, for $i\in\\{2,\ldots,k\\}$ we can find nets
$\boldsymbol{t}(i)\in\mathcal{N}_{j}^{(t(i),t(i)\setminus t(i-1))}$. Their
concatenation
$\boldsymbol{t}=\boldsymbol{t}(2)\circ\ldots\circ\boldsymbol{t}(k)$ is a
member of $\mathcal{N}_{n^{*}}^{(t,t_{1})}$. ∎

The proof of Theorem 3.3.1, given as a consequence of Theorem 3.3.4 in part A.
of this section, is thus completed, as we have the following.

Proof of Theorem 3.3.4 According to Theorem 3.3.30, condition (**) is
satisfied. By Theorem 3.3.16, so is the condition (*). ∎

Our proofs of Theorems 3.3.28 and 3.3.29 are based on the global
characterization of the class $\mathcal{F}_{\varrho}$ given in Theorem 3.3.22.

Proceeding to the proof of Theorem 3.3.28, we fix number
$\bar{\varrho}>\varrho$ and introduce some additional notation.

For any $\boldsymbol{z}\in\mathbb{R}^{2}$, $s\subset\mathbb{R}^{2}$,
$s\neq\emptyset$, we set
$d(\boldsymbol{z},s)=\inf{\left\\{\|\boldsymbol{z}-\boldsymbol{y}\|\colon\boldsymbol{y}\in
s\right\\}}$.

  * $\bullet$ With $\delta(\varrho,\bar{\varrho})$ we denote the smallest positive number for which conditions $\|\boldsymbol{z}\|\geq\gamma^{\prime}/3-\delta(\varrho,\bar{\varrho})$, $d(\boldsymbol{z},B((0,-\varrho),\varrho))\leq\delta(\varrho,\bar{\varrho})$, for $\boldsymbol{z}\in\mathbb{R}^{2}$, imply $\boldsymbol{z}\in B((0,-\bar{\varrho}),\bar{\varrho})$.

For $s\in\mathcal{T}$ and $\boldsymbol{x}\in\mathbb{T}\setminus s$,
analogically to the sets $B(s,\boldsymbol{x})$, $E(s,\boldsymbol{x})$,
$C(s,\boldsymbol{x})$, we define:

  * $\bar{B}(s,\boldsymbol{x})=B$, if $B\in\mathcal{B}_{\bar{\varrho}}$, $P_{B}\boldsymbol{x}=P_{s}\boldsymbol{x}$,

  * $\bar{E}(s,\boldsymbol{x})$, to be the connected component of $\mathbb{T}\setminus\bar{B}(s,\boldsymbol{x})$ containing $\boldsymbol{x}$,

  * $\bar{C}(s,\boldsymbol{x})=\mathbb{T}\setminus\bar{E}(s,\boldsymbol{x})$, thus

$\bar{C}(s,\boldsymbol{x})\supset C(s,\boldsymbol{x})\text{,}\quad\text{for
}\boldsymbol{x}\in\mathbb{T}\setminus s\text{.}$ (3.3.11)

  * We obviously put

$\mathcal{C}_{\bar{\varrho}}=\Big{\\{}\mathbb{T}\setminus E\colon
E\neq\emptyset\text{ is a connected component of }\mathbb{T}\setminus B\\\
\text{for some ball }B\in\mathcal{B}_{\bar{\varrho}}\text{,
}\operatorname{int}(\mathbb{T}\cap B)\neq\emptyset\text{, }\mathbb{T}\setminus
B\neq\emptyset\Big{\\}}\text{.}$

Let us note that Theorem 3.3.22 is also valid when $\varrho$ is replaced with
$\bar{\varrho}$, therefore, for any set $t\in\mathcal{T}$, the following
conditions are equivalent:

  * $\bar{\alpha}$)

$t\in\mathcal{F}_{\bar{\varrho}}$,

  * $\bar{\beta}$)

$t=\bigcap_{C\in\mathcal{D}\cup\\{\mathbb{T}\\}}C$, for some family
$\mathcal{D}\in\mathcal{C}_{\bar{\varrho}}$.

We will use the following observation.

######  Lemma 3.3.31.

For $t\in\mathcal{F}_{\varrho}$, $\boldsymbol{x}\in\mathbb{T}\setminus t$,
$\boldsymbol{z}\in E(t,\boldsymbol{x})$, we have
$d(\boldsymbol{z},B(t,\boldsymbol{x}))\leq
d(\boldsymbol{z},C(t,\boldsymbol{x}))$.

Proof. It is enough to notice that $\|\boldsymbol{z}-\boldsymbol{v}\|\geq
d(\boldsymbol{z},B(t,\boldsymbol{x}))$, for $\boldsymbol{v}\in
C(t,\boldsymbol{x})\cap F$, for some connected component $F$ of
$\mathbb{T}\setminus B(t,\boldsymbol{x})$ that is disjoint with
$E(t,\boldsymbol{x})$. For such points $\boldsymbol{v}$ we have
$[\boldsymbol{z},\boldsymbol{v}]\cap B(t,\boldsymbol{x})\neq\emptyset$, which
completes the proof. ∎

######  Lemma 3.3.32.

For sets $r\subset t$, $\Delta\subset t\setminus r$, satisfying
$r,t\in\mathcal{F}_{\varrho}$, $d(r,t)\leq\delta(\varrho,\bar{\varrho})$,
$d(\Delta)\leq\gamma^{\prime}/3$, there exists a set
$\bar{r}\in\mathcal{F}_{\bar{\varrho}}$ satisfying
$r\cup\Delta\subset\bar{r}\subset t$ and $d(\bar{r}\setminus
r)\leq\gamma^{\prime}$.

Proof. Denote $\sigma=\gamma^{\prime}/3$. In the case when $\Delta=\emptyset$,
we set $\bar{r}=r$. If $\Delta\neq\emptyset$ and $d(t\setminus
r,\Delta)\leq\sigma$, we use the relation
$\mathcal{F}_{\varrho}\subset\mathcal{F}_{\bar{\varrho}}$. We take
$\bar{r}=t$, and the bound $d(\bar{r}\setminus r)\leq 3\sigma=\gamma^{\prime}$
follows from the triangle inequality. Otherwise, we set

$\bar{r}=t\cap\bigcap_{\begin{subarray}{c}\boldsymbol{x}\in t\setminus r\\\
d(\boldsymbol{x},\Delta)>\sigma\end{subarray}}\bar{C}(r,\boldsymbol{x})\text{.}$

Since $t\in\mathcal{F}_{\varrho}\subset\mathcal{F}_{\bar{\varrho}}$, and
$\bar{\beta}\textup{)}\implies\bar{\alpha}\textup{)}$, we have
$\bar{r}\in\mathcal{F}_{\bar{\varrho}}$ and
$d(\boldsymbol{x},\Delta)\leq\sigma$, for $\boldsymbol{x}\in\bar{r}\setminus
r$, thus $d(\bar{r}\setminus r)\leq 3\sigma=\gamma^{\prime}$, by triangle
inequality. In accordance with (3.3.11), we also have $r\subset\bar{r}$.

The proof will be completed by showing that $\Delta\subset\bar{r}$. For any
$\boldsymbol{x}\in t\setminus r$ satisfying $d(\boldsymbol{x},\Delta)>\sigma$
and its projection $\boldsymbol{y}=P_{r}\boldsymbol{x}$, we have
$d(\boldsymbol{y},\Delta)>\sigma-\delta(\varrho,\bar{\varrho})$. Moreover, for
$\boldsymbol{z}\in\Delta$ we have

$\displaystyle d(\boldsymbol{z},B(r,\boldsymbol{x}))$ $\displaystyle\leq
d(\boldsymbol{z},C(r,\boldsymbol{x}))\quad$ (cf. Lemma 3.3.31)
$\displaystyle\leq d(\boldsymbol{z},r)\quad$ (as $r\subset
C(r,\boldsymbol{x})$) $\displaystyle\leq\delta(\varrho,\bar{\varrho})\quad$
(since $\boldsymbol{z}\in\Delta\subset t$).

In a suitable coordinate system we then have $\boldsymbol{y}=\boldsymbol{0}$,
$B(r,\boldsymbol{x})=B((0,-\varrho),\varrho)$,
$\bar{B}(r,\boldsymbol{x})=\bar{B}((0,-\bar{\varrho}),\bar{\varrho})$, and

$\displaystyle\|\boldsymbol{z}\|=\|\boldsymbol{z}-\boldsymbol{y}\|\geq
d(\boldsymbol{y},\Delta)>\sigma-\delta(\varrho,\bar{\varrho})\text{,}$
$\displaystyle
d(\boldsymbol{z},B((0,-\varrho),\varrho))\leq\delta(\varrho,\bar{\varrho})\text{.}$

Therefore, $\boldsymbol{z}\in\bar{B}(r,\boldsymbol{x})$, by the definition of
$\delta(\varrho,\bar{\varrho})$. Moreover,
$\boldsymbol{z}\in\bar{C}(r,\boldsymbol{x})$, by the fact that
$\boldsymbol{z}\in\Delta\subset\mathbb{T}$. The inclusion
$\Delta\subset\bar{r}$ is thus shown. ∎

In the forthcoming proof we use the assumption that
$\mathbb{T}\subset[-1,1]\times[-1,1]$.

Proof of Theorem 3.3.28 We can find a number $j\in\mathbb{N}$ such that there
exists a covering $\bigcup_{2\leq h\leq j}\Delta_{h}$ of the square
$[-1,1]\times[-1,1]$ satisfying $d(\Delta_{h})\leq\gamma/3$, for $2\leq h\leq
j$. For this $j$ we further fix numbers $\varrho(h)$ such that
$\beta^{\prime}/2=\varrho(1)<\ldots<\varrho(j)=\beta^{\prime}$, and define
$\delta=\min{\left\\{\delta(\varrho(h-1),\varrho(h))\colon 2\leq h\leq
j\right\\}}$. Clearly, $j$ and $\delta$ do not depend on $\mathbb{T}$.

Let $(t,t_{1})\in\tau_{\beta^{\prime}/2}$ be a soul. We set $t(1)=t\setminus
t_{1}$. Having defined $t(h-1)$, for $2\leq h\leq j$, satisfying $t\setminus
t_{1}\subset t(h-1)\subset t$, $t(h-1)\in\mathcal{F}_{\varrho(h-1)}$, let us
denote $r=t(h-1)$, $\varrho=\varrho(h-1)$, $\bar{\varrho}=\varrho(h)$, and
$\Delta=\Delta_{h}\cap(t\setminus t(h-1))$. Then,
$\delta(\varrho,\bar{\varrho})\geq\delta\geq d(t,r)$,
$d(\Delta)\leq\gamma^{\prime}/3$, and we can apply Lemma 3.3.32. For thus
obtained $\bar{r}$, we set $t(h)=\bar{r}$, which also assures that $t\setminus
t_{1}\subset t(h)\subset t$ and $t(h)\in\mathcal{F}_{\varrho(h)}$. The
resulting sequence of sets $t(1)\subset\ldots\subset t(j)$ satisfies
$t(1)=t\setminus t_{1}$,
$t(h)\in\mathcal{F}_{\varrho(h)}\subset\mathcal{F}_{\beta^{\prime}}$,
$d(t(h)\setminus t(h-1))\leq\gamma^{\prime}$, for $2\leq h\leq j$ and
$t(j)\supset(t\setminus t_{1})\cup\bigcup_{2\leq h\leq j}(\Delta_{h}\cap
t_{1})=t$. This ends the proof. ∎

Now we proceed to the proof of Theorem 3.3.29. To that end we fix number
$\varrho=\beta^{\prime}/2$ and for
$\delta>0,\boldsymbol{x},\boldsymbol{y}\in\mathbb{T},\boldsymbol{x}\neq\boldsymbol{y}$,
we define

  * $\bullet$

$\varepsilon(\delta)=\inf{\left\\{\|\boldsymbol{\bar{z}}-\boldsymbol{z}\|\colon\boldsymbol{\bar{z}}\in
B((0,-\varrho),\varrho)\text{, }\boldsymbol{z}\in B(\boldsymbol{0},1)\setminus
B((0,\delta-\varrho),\varrho)\right\\}}$,

  * $B(\\{\boldsymbol{y}\\},\boldsymbol{x})=B$ if $B\in\mathcal{B}_{\varrho}$, $P_{B}\boldsymbol{x}=\boldsymbol{y}$,

  * $C(\\{\boldsymbol{y}\\},\boldsymbol{x})=\mathbb{T}\setminus E$ if $E$ is connected component of the set $\mathbb{T}\setminus B(\\{\boldsymbol{y}\\},\boldsymbol{x})$ and $\boldsymbol{x}\in E$.

In the forthcoming proof we use the assumption that $d(\mathbb{T})\leq 1$,
which somewhat simplifies the argument.

######  Lemma 3.3.33.

For $r\in\mathcal{F}_{\varrho}$, $\delta>0$, there exists set
$s\in\mathcal{F}_{\varrho}$ satisfying $r\subset s$, $d(r,s)\leq\delta$, and
the implication $\boldsymbol{x}\in\mathbb{T}\setminus r\text{,
}d(\boldsymbol{x},r)\leq\varepsilon(\delta)\implies\boldsymbol{x}\in s$.

Proof. If $d(\mathbb{T},r)\leq\delta$, it is enough to take $s=\mathbb{T}$.
Assume then that $d(\mathbb{T},r)>\delta$. For any
$\boldsymbol{x}\in\mathbb{T}$ satisfying $d(\boldsymbol{x},r)>\delta$, we can
find $\boldsymbol{y}\in[P_{r}\boldsymbol{x},\boldsymbol{x}]$ for which
$\|\boldsymbol{y}-P_{r}\boldsymbol{x}\|=\delta$ (thus
$\boldsymbol{y}\neq\boldsymbol{x}$ and
$C(\\{\boldsymbol{y}\\},\boldsymbol{x})\supset C(r,\boldsymbol{x})$).

For $\boldsymbol{z}\in\mathbb{T}$ such that
$d(\boldsymbol{z},r)\leq\varepsilon(\delta)$ we show that $\boldsymbol{z}\in
C(\\{\boldsymbol{y}\\},\boldsymbol{x})$.

If $\boldsymbol{z}\notin E(r,\boldsymbol{x})$, then $\boldsymbol{z}\in
C(r,\boldsymbol{x})\subset C(\\{\boldsymbol{y}\\},\boldsymbol{x})$.

However, if $\boldsymbol{z}\in
E(r,\boldsymbol{x}),d(\boldsymbol{z},r)\leq\varepsilon(\delta)$, we can use
Lemma 3.3.31 to obtain inequality
$d(\boldsymbol{z},B(r,\boldsymbol{x}))\leq\varepsilon(\delta)$. With suitably
adjusted coordinate system we have $P_{r}\boldsymbol{x}=\boldsymbol{0}$,
$\boldsymbol{y}=(0,\delta)$, $B(r,\boldsymbol{x})=B((0,-\varrho),\varrho)$,
$B(\\{\boldsymbol{y}\\},\boldsymbol{x})=B((0,\delta-\varrho),\varrho)$.
Suppose that $\boldsymbol{z}\in\mathbb{T}\setminus
C(\\{\boldsymbol{y}\\},\boldsymbol{x})$. As $d(\mathbb{T})\leq 1$, we have
both $\boldsymbol{z}\in B(\boldsymbol{0},1)\setminus
B((0,\delta-\varrho),\varrho)$ and
$d(\boldsymbol{z},B((0,-\varrho),\varrho))\leq\varepsilon(\delta)$.
Contradiction with the definition of $\varepsilon(\delta)$ shows that
$\boldsymbol{z}\in\mathbb{T}\cap B(\\{\boldsymbol{y}\\},\boldsymbol{x})\subset
C(\\{\boldsymbol{y}\\},\boldsymbol{x})$.

Lastly, setting $s=\bigcap_{\begin{subarray}{c}\boldsymbol{x}\in\mathbb{T}\\\
d(\boldsymbol{x},r)>\delta\end{subarray}}C(\\{\boldsymbol{y}\\},\boldsymbol{x})$,
in view of the implication $\beta\text{)}\implies\alpha\text{)}$ in Theorem
3.3.22, completes the proof. ∎

Proof of Theorem 3.3.29 Let us fix $k\geq 1/{\varepsilon(\delta)}+1$. For
$(t,t_{1})\in\tau_{\beta^{\prime}/2}$, let us denote $t(1)=t\setminus t_{1}$,
and having defined $t(i)$, for $i\in\\{1,\ldots,k-1\\}$, we apply Lemma 3.3.33
with $r=t(i)$ to obtain $s$ described therein. We set $t(i+1)=s\cap t$. Then,
$d(t(i),t(i+1))\leq\delta$.

Next, let us consider $\boldsymbol{x}\in t$, for which we choose a point
$\boldsymbol{x}_{1}\in t\setminus t_{1}$, together with points
$\boldsymbol{x}_{i}\in[\boldsymbol{x}_{1},\boldsymbol{x}]$ satisfying
$\|\boldsymbol{x}_{i}-\boldsymbol{x}_{i-1}\|<\varepsilon(\delta)$, for $2\leq
i\leq k$, $\boldsymbol{x}_{k}=\boldsymbol{x}$. By induction,
$\boldsymbol{x}_{i}\in t(i)$ for $1\leq i\leq k$, therefore $\boldsymbol{x}\in
t(k)=t$. Finally, $t(k)=t$, by the arbitrariness of $\boldsymbol{x}\in t$, and
$k$ clearly does not depend on $\mathbb{T}$. ∎

Finally, the proof of Theorem 3.3.1 is completed.

###  3.4 Existence of convex Peano curve filling convex compact set
$\mathbb{T}$

Recall that $\mathcal{T}$ is a family of convex, compact subsets of
$\mathbb{R}^{2}$ with non-empty interiors. We can now demonstrate that the
announced Fundamental Construction for any $\mathbb{T}\in\mathcal{T}$ is
feasible. A sequence of sets $\mathbf{t}=(t(K);1\leq K\leq M)$ forms a
population if $t([K,L])\in\mathcal{T}$ for $1\leq K\leq L\leq N$ (for unions
$t([K,L])=\bigcup_{K\leq J\leq L}t(J)$). We also denote (as previously)
$\bigcup\mathbf{t}=t([1,M])$, $d(\mathbf{t})=\text{max}\\{d(t(K)):1\leq K\leq
M\\}$, $I\mathbf{t}=(t(M+1-K);1\leq K\leq M)$ and
$I^{0}\mathbf{t}=\mathbf{t},I^{n}\mathbf{t}=I(I^{n-1}\mathbf{t})$ for
$n\in\mathbb{N}$. For a sequence of sequences of sets $(\mathbf{t}(K);1\leq
K\leq M)$ we denote anti-ordering $\mathcal{A}(\mathbf{t}(K);1\leq K\leq
M)=I^{0}\mathbf{t}(1)\circ I^{1}\mathbf{t}(2)\circ\dots\circ
I^{M-1}\mathbf{t}(M)$. (cf. (3.2.1)).

The existence of the Fundamental Construction for $\mathbb{T}$ is stated by
the following theorem:

######  Theorem B.

For $\mathbb{T}\in\mathcal{T}$, there exist numbers $M(j),m(j)\in\mathbb{N}$
and populations of sets $\mathbf{t}(j)=(t(j;K);1\leq K\leq
M(j))\in\mathcal{T}^{M(j)}$, for $j\in\mathbb{N}$, satisfying the following
conditions:

  1. 1)

$\bigcup\mathbf{t}(1)=\mathbb{T}$;

  2. 2)

for $j\in\mathbb{N}$ we have $M(j+1)=M(j)m(j)$ and
$t(j;K)=t(j+1;[(K-1)m(j)+1,Km(j)])$ for $K\in\\{1,\dots,M(j)\\}$;

  3. 3)

$d(\mathbf{t}(j))\rightarrow 0$ as $j\rightarrow\infty$.

Proof. Without loss of generality, we assume $d(\mathbb{T})\leq 1$,
$\mathbb{T}\subset[-1,1]\times[-1,1]$. We will base our argument on Theorem
3.3.1 and Corollary 3.3.2. We fix numbers $\gamma(j)\in(0,\frac{1}{4})$ for
$j\in\mathbb{N}$ such that $\gamma(j)\rightarrow 0$ as $j\rightarrow\infty$
and $\beta(1)\geq 1$,
$\beta(j+1)=\text{max}\\{2\beta(j),\gamma(j)+\frac{1}{\gamma(j)}\\}$ for
$j\in\mathbb{N}$. Then we fix numbers $m^{\prime}(j)\in 2\mathbb{N}$ for
$j\in\mathbb{N}$ satisfying the following conditions:

  1. I)

For odd $j$, there exists offspring of souls

$\tau_{\beta(j)\gamma(j)}\ni(t,t_{1})\rightarrow\mathopen{<}\mathbf{t}^{\prime}(j;t,t_{1}),\mathbf{t}^{\prime}_{1}(j;t,t_{1})\mathclose{>}\in(\tau_{\beta(j+1),\gamma(j+1)})^{m^{\prime}(j)}$
(3.4.1)

satisfying $d_{x}(\mathbf{t}^{\prime}(j;t,t_{1}))\leq 11\gamma(j)$, in
notation (3.3.1), according to Theorem 3.3.1, for the fixed $\mathbb{T}$.

  2. II)

For even $j$, there exists offspring of souls (3.4.1) satisfying
$d_{y}(\mathbf{t}^{\prime}(j;t,t_{1}))\leq 11\gamma(j)$, in notation (3.3.3),
according to Corollary 3.3.2.

We define numbers $M(j)$ and sequences of souls
$\mathopen{<}\mathbf{t}(j),\mathbf{t}_{1}(j)\mathclose{>}\in(\tau_{\beta(j)\gamma(j)})^{M(j)}$
for $j\geq 1$ according to the following induction. For $j=1$, we set $M(1)=2$
and
$\mathbf{t}(1)=(\mathbb{T},\mathbb{T}),\mathbf{t}_{1}(1)=(\emptyset,\emptyset)$.
Assume now that a sequence of souls

$\mathopen{<}\mathbf{t}(j),\mathbf{t}_{1}(j)\mathclose{>}=((t(j;K),t_{1}(j;K));1\leq
K\leq M(j))\in(\tau_{\beta(j)\gamma(j)})^{M(j)}$

is defined. For $1\leq K\leq M(j)$ denote by
$\mathopen{<}\mathbf{t}^{\prime}(j;K),\mathbf{t}^{\prime}_{1}(j;K)\mathclose{>}$
the offspring of a parent $(t(j;K),t_{1}(j;K))$, according to I) for $j\in
2\mathbb{N}-1$ and according to II) for $j\in 2\mathbb{N}$. Then, using anti-
ordering of a sequence of sequences of sets, we define
$\mathopen{<}\mathbf{t}(j+1),\mathbf{t}_{1}(j+1)\mathclose{>}=\mathopen{<}\mathcal{A}(\mathbf{t}^{\prime}(j;K);1\leq
K\leq M(j)),\mathcal{A}(\mathbf{t}^{\prime}_{1}(j;K);1\leq K\leq
M(j))\mathclose{>}$. Thus also we have
$\mathopen{<}\mathbf{t}(j+1),\mathbf{t}_{1}(j+1)\mathclose{>}\in(\tau_{\beta(j+1)\gamma(j+1)})^{M(j+1)}$.

Notice that the proposed two-term sequence
$\mathopen{<}\mathbf{t}(1),\mathbf{t}_{1}(1)\mathclose{>}=((\mathbb{T},\emptyset),(\mathbb{T},\emptyset))$
is a population of souls in $\tau_{\beta(1)\gamma(1)}$, according to
Definition 3.1.2. According to Theorem 3.2.3, the sequences
$\mathopen{<}\mathbf{t}(j),\mathbf{t}_{1}(j)\mathclose{>}$ are also
populations of souls, recursively for $j=2,3,\dots$. The obtained bases
$\mathbf{t}(j)$ are therefore populations of sets, according to Theorem 3.1.6.
Condition 1) is fulfilled, as well as condition 2) (according to (3.2.3)).

According to notations (3.3.1) and (3.3.3), we also have
$d_{x}(\mathbf{t}(1))\geq d_{x}(\mathbf{t}(2))\geq\ldots$ and
$d_{y}(\mathbf{t}(1))\geq d_{y}(\mathbf{t}(2))\geq\ldots$, together with the
inequalities $d_{x}(\mathbf{t}(j+1))\leq 11\gamma(j)$, for $j\in 2N-1$, and
$d_{y}(\mathbf{t}(j+1))\leq 11\gamma(j)$, for $j\in 2N$, thus condition 3) is
satisfied as well. ∎

Proof of Theorem A. The case when int$(\mathbb{T})=\emptyset$ is trivial. When
int$(\mathbb{T})\neq\emptyset$, we apply Theorems 2.0.1 and B. ∎

The relationships between convex Peano curves will be investigated in our next
publication. Here, we will confine ourselves to the following remark
concerning affine mappings.

Let’s consider any convex Peano curve $f$ filling a given compact convex set
$\mathbb{T}\subset\mathbb{R}^{2}$, and denote the function
$g:\mathbb{R}\rightarrow\mathbb{T}$ as follows:

$g(u)=\left\\{\begin{array}[]{ccc}f(0)&\text{for}&u<0\\\
f(u)&\text{for}&u\in[0,1]\\\ f(1)&\text{for}&u>1.\end{array}\right.$

Next, for a convex set $K$ in a real or complex linear topological space
$\mathbb{X}$ and a continuous functional $\Lambda$ on $\mathbb{X}$, assume
there exist numbers $a,b\in\Lambda(K)$ such that $a\neq b$. Without loss of
generality, one can assume that $\operatorname{Re}a<\operatorname{Re}b$. Then
the function $h:\mathbb{X}\rightarrow\mathbb{T}$, defined as
$h(\mathbf{x})=g\left(\frac{\operatorname{Re}(\Lambda(\mathbf{x})-a)}{\operatorname{Re}(b-a)}\right)$,
is a continuous surjection, where the images $h(L)$ are convex for all convex
subsets $L\subset K$. Thus, we have:

######  Corollary 3.4.1.

If for a convex set $K$ in a real or complex linear topological space
$\mathbb{X}$, there exists a continuous linear functional on $\mathbb{X}$ that
is non-constant on $K$, then, for a compact convex set
$\mathbb{T}\subset\mathbb{R}^{2}$, there exists a continuous surjection
$h:\mathbb{X}\rightarrow\mathbb{T}$, where the images of convex subsets $L$ of
$K$ are convex, and $h(K)=\mathbb{T}$.

It is therefore evident that convex mappings can differ drastically from
affine mappings.

Acknowledgments

Sincere gratitude is due to Professor Jalanta Misiewicz, whose assistance has
fundamentally influenced the shape of this work. I would also like to thank
Professor Andrzej Komisarski for valuable discussions, as well as Professors
Stanisław Goldstein and Władysław Wilczyński for their valuable information on
the literature. Additional thanks go to Małgorzata Kozioł, Jakub Olejnik, and
Damian Prusinowski for their technical and linguistic assistance, and for
their patience.

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Adam Paszkiewicz, Faculty of Mathematics and Computer Science, University of
Lodz, Poland, email<EMAIL_ADDRESS>