On weakly $1$-convex and weakly $1$-semiconvex sets

Let $O\in\mathbb{R}^{n}$ be the origin. Consider the homeomorphism $\phi:\mathbb{R}^{n}\setminus\{O\}\rightarrow S^{n-1}\times(0,+\infty)$ defined by the formula

Let also $\sigma:S^{n-1}\times(0,+\infty)\rightarrow S^{n-1}$ be the central projection on the sphere, i.e.,

Then $\sigma$ is open.

Fix an arbitrary point $y\in E^{\diamondsuit}$ ($y\in E^{\triangle}$). Without loss of generality, suppose that $y=O$.

Let $z_{\alpha}$ be an arbitrary fixed point of $\eta_{\alpha}(y)\cap E$ (of $\gamma_{\alpha}(y)\cap E$), $\alpha\in S^{n-1}$. Since $E$ is open, there exist open balls $U_{\alpha}:=U\left(z_{\alpha},\varepsilon_{\alpha}\right)$, $\alpha\in S^{n-1}$, such that $\overline{U_{\alpha}}\subset E$. Then the images $\sigma(U_{\alpha})$, $\alpha\in S^{n-1}$, are open subsets of $S^{n-1}$ (open subsets of the projective space $\mathbb{R}\mathbf{P}^{n-1}$). Moreover,

is a cover of the unit sphere $S^{n-1}$ (of the projective space $\mathbb{R}\mathbf{P}^{n-1}$). By the Heine-Borel theorem, there exists a subcover

of $S^{n-1}$ (of $\mathbb{R}\mathbf{P}^{n-1}$).

Let $E_{i}$, $i\in N\subseteq\mathbb{N}$, be the components of $E$. Then for any $j\in M$ there exists $i(j)\in N$ such that $\overline{U_{\alpha_{j}}}\subset E_{i(j)}$.

Consider the distance functions

They are continuous in the domains $E_{i}$, $i\in N$. Then their restrictions to the compacts $\overline{U_{\alpha_{j}}}$ attain their minimum values $d_{j}>0$ on that compacts, i.e.,

Since $M$ is finite, there exists

Then $U\left(x,d\right)\subset E$ for any point $x\in\overline{U_{\alpha_{j}}}$, $j\in M$; see Figure 3. And for any $\alpha\in S^{n-1}$, there exists $j\in M$ such that $\eta_{\alpha}(y)\cap\overline{U_{\alpha_{j}}}\neq\varnothing$ ($\gamma_{\alpha}(y)\cap\overline{U_{\alpha_{j}}}\neq\varnothing$) by the construction. ∎

Fix an arbitrary point $y\in E^{\diamondsuit}$ and show that it is an inner point of $E^{\diamondsuit}$.

Since $E$ is weakly $1$-semiconvex, it follows that $y\not\in\partial E$. Then there exists a number $\varepsilon_{1}>0$ such that $U(y,\varepsilon_{1})\subset(\mathbb{R}^{n}\setminus\overline{E}$).

By Lemma 9, for the fixed $y$ there exist points $x_{\alpha}\in\eta_{\alpha}(y)\cap E$, $\alpha\in S^{n-1}$, and a constant $d>0$ such that $U(x_{\alpha},d)\subset E$.

Let $\varepsilon:=\min\{\varepsilon_{1},d\}$. Consider the neighborhood $U(y,\varepsilon)$ of the point $y$. Let $z\in U(y,\varepsilon)$ and let $\eta_{\alpha}(z)$, $\alpha\in S^{n-1}$, be an arbitrary ray with initial point at $z$. Draw the ray $\eta_{\alpha}(y)$ parallel to the ray $\eta_{\alpha}(z)$. Since $U(x_{\alpha},\varepsilon)\subseteq U(x_{\alpha},d)\subset E$ for the point $x_{\alpha}$ correspondent to $\eta_{\alpha}(y)$, it follows that $\eta_{\alpha}(z)\cap U(x_{\alpha},\varepsilon)\neq\varnothing$ and, therefore, $\eta_{\alpha}(z)\cap E\neq\varnothing$ for any $\alpha\in S^{n-1}$; see Figure 4 a). Thus, $z$ is a $1$-nonsemiconvexity point of $E$. Since $z$ is arbitrary, it implies that all points of $U(y,\varepsilon)$ are $1$-nonsemiconvexity points of $E$. Hence, $y$ is an inner point of $E^{\diamondsuit}$. ∎

Since $E^{\diamondsuit}$ is open, then for any point $y\in\partial E^{\diamondsuit}$, there exists a ray $\eta_{\alpha^{\prime}}(y)$, $\alpha^{\prime}\in S^{n-1}$, not intersecting $E$. Then $\eta_{\alpha^{\prime}}(y)\cap E^{\diamondsuit}=\varnothing$ by Definition 3. Thus, $E^{\diamondsuit}$ is weakly $1$-semiconvex. ∎

The scheme of proving this theorem is exactly the same as for Theorem 1. We fix an arbitrary point of $E^{\triangle}$ and show that there exists a neighborhood of this point which belongs to $E^{\triangle}$. To do so, we use Lemma 9 with respect to the points $y\in E^{\triangle}$ and the straight lines $\gamma_{\alpha}(y)$, $\alpha\in S^{n-1}$, and we also replace the rays with the straight lines everywhere in the proof of Theorem 1. ∎

Suppose that $\mathrm{Int}\,E$ is not weakly $1$-semiconvex. Then there exists a $1$-nonsemiconvexity point $y\in\partial E$ of the set $\mathrm{Int}\,E$.

By Lemma 9, for the point $y$, there exist points $x_{\alpha}\in\eta_{\alpha}(y)\cap\mathrm{Int}\,E$, $\alpha\in S^{n-1}$, and a constant $d>0$ such that $U(x_{\alpha},d)\subset\mathrm{Int}\,E$.

Consider the neighborhood $U(y,d)$ of the point $y$; see Figure 4 b). Since $E$ is weakly $1$-semiconvex, there exists a family of open weakly $1$-semiconvex sets $G_{k}$, $k=1,2,\ldots$, approximating $E$ from the outside. Then there exists an index $k_{0}$ such that $\partial G_{k}\cap U(y,d)\neq\varnothing$ for all $k\geqslant k_{0}$. For each $k\geqslant k_{0}$, choose a point $z_{k}\in\partial G_{k}\cap U(y,d)$ and draw an arbitrary ray $\eta_{\alpha}(z_{k})$, $\alpha\in S^{n-1}$, with initial point at $z_{k}$. Consider the ray $\eta_{\alpha}(y)$ parallel to $\eta_{\alpha}(z_{k})$. Since $U(x_{\alpha},d)\subset\mathrm{Int}\,E\subset E$ for the point $x_{\alpha}$ correspondent to $\eta_{\alpha}(y)$, it follows that $\eta_{\alpha}(z_{k})\cap U(x_{\alpha},d)\neq\varnothing$ and, therefore, $\eta_{\alpha}(z_{k})\cap E\neq\varnothing$. Since $G_{k}\supset E$, $k=1,2,\ldots$, then $\eta_{\alpha}(z_{k})\cap G_{k}\neq\varnothing$, $k\geqslant k_{0}$.

Since the ray $\eta_{\alpha}(z_{k})$ is arbitrary, the point $z_{k}\in\partial G_{k}$ is a $1$-nonsemiconvexity point of $G_{k}$ for all $k\geqslant k_{0}$, which gives a contradiction. ∎

The proof of this theorem is the same as the proof of Theorem 3. We only consider weakly $1$-convex sets instead of weakly $1$-semiconvex and replace the rays with the straight lines everywhere in the proof of Theorem 3. ∎

Since $E^{\diamondsuit}$ is open, for any point $x\in\partial E^{\diamondsuit}$, there exists a ray $\eta(x)$ such that $\eta(x)\cap E=\varnothing$. Moreover, $\bigcup\limits_{x\in\partial E^{\diamondsuit}}\eta(x)\cup E^{\diamondsuit}$ does not contain any ray emanating from $E^{\diamondsuit}$, otherwise, a ray $\eta(y)\subset\bigcup\limits_{x\in\partial E^{\diamondsuit}}\eta(x)\cup E^{\diamondsuit}$, $y\in E^{\diamondsuit}$, does not intersect $E$, which contradicts the definition of $1$-nonsemiconvexity point. ∎

The statements are similar to the proof of Proposition 1. We only consider straight lines instead of rays. ∎

Let

where $D\subset\mathbb{R}^{2}$ is an open bounded convex subset, $P\subset\mathbb{R}^{2}$ is an open convex polygon such that $\overline{P}\subset D$, $\left\{\gamma^{k}\right\}_{k\in M}$, $M\subseteq\mathbb{N}$, is the finite collection of lines passing through the polygon sides. Then $E^{2}\in\left(\mathbf{WS^{2}_{1}}\setminus\mathbf{S^{2}_{1}}\right)\cap\left(\mathbf{WC^{2}_{1}}\setminus\mathbf{C^{2}_{1}}\right)$ and $(E^{2})^{\diamondsuit}=(E^{2})^{\triangle}=P$.

Consider the line segment $Oa_{+}^{3}$, where $a_{+}^{3}\in\mathbb{R}^{3}$ is such that the angle between the vector $a_{+}^{3}$ and the unit vector $u$ of the axis $Ox_{3}$ belongs to the interval $(0,\pi/2)$.

Let $E_{+}^{3}$ be a bounded oblique cylinder with the set $E^{2}$ at the base and elements parallel to $Oa_{+}^{3}$, i.e.,

Let also $E_{-}^{3}$ be the oblique cylinder symmetric to $E_{+}^{3}$ with respect to the coordinate plane $x_{1}Ox_{2}$; see Figure 5 a).

Let $\rho>0$ be the height of $E_{+}^{3}$, and $D_{+}$ be the orthogonal projection of the set $\{z\in\mathbb{R}^{3}:z=x+a_{+}^{3},\,\,x\in D\}$ onto $x_{1}Ox_{2}$. Consider the following cylinders:

and $D_{-}^{3}$ that is the cylinder symmetric to $D_{+}^{3}$ with respect to the coordinate plane $x_{1}Ox_{2}$.

Let

see Figure 5 a).

Consider also the following polygonal oblique prisms:

where $a_{-}^{3}$ is the vector symmetric to $a_{+}^{3}$ with respect to the coordinate plane $x_{1}Ox_{2}$. Prove that

First, show that

Consider an arbitrary point $x\in P_{-}^{3}\cup P_{+}^{3}$. Then $x\in P_{q}^{3}$, $q\in\{-,+\}$. Let $\eta(x)$ be an arbitrary ray emanating from $x$. Show that $\eta(x)\cap E^{3}\neq\varnothing$.

1. 1.

   If $\eta(x)$ intersects a lateral face of $P_{q}^{3}$, then consider the projection, parallel to $Oa_{q}^{3}$, of $\eta(x)$ onto the coordinate plane $x_{1}Ox_{2}$. It is a ray that we define by $\eta(x_{0})$. The ray $\eta(x_{0})$ emanates from the point $x_{0}\in x_{1}Ox_{2}$ which is the projection of $x$ onto $P\subset x_{1}Ox_{2}$. Since $E^{2}$ is a flat weakly $1$-semiconvex set, it implies that $\eta(x_{0})\cap E^{2}\neq\varnothing$, which gives that $\eta(x)\cap E^{3}_{q}\neq\varnothing$, therefore, $\eta(x)\cap E^{3}\neq\varnothing$.

2. 2.

   If $\eta(x)$ intersects a base of $P_{q}^{3}$, then it intersects either $D_{-}^{3}\cup D_{+}^{3}$, which immediately gives that $\eta(x)\cap E^{3}\neq\varnothing$, or it intersects a lateral face of the other prism $P_{q^{\prime}}^{3}$, ${q^{\prime}}\in\{-,+\}$, ${q^{\prime}}\neq q$, by the construction, and further considerations are the same as in item 1, but for $P_{q^{\prime}}^{3}$, a point $x^{\prime}\in\eta(x)\cap P_{q^{\prime}}^{3}$, and the ray $\eta(x^{\prime})\subset\eta(x)$. Then $\eta(x^{\prime})\cap E^{3}\neq\varnothing$, therefor, $\eta(x)\cap E^{3}\neq\varnothing$.

Moreover, if $x\in(E^{3})^{\diamondsuit}$, then $x\in(E^{3})^{\triangle}$.

Now, prove that $E^{3}\in\mathbf{WC^{3}_{1}}$ and, thus, $E^{3}\in\mathbf{WS^{3}_{1}}$, and

It is enough to show that if $z\not\in E^{3}\cup P_{-}^{3}\cup P_{+}^{3},$ then $z\not\in(E^{3})^{\triangle}$. Let $L$ be the plane passing through $z$ parallel to the coordinate plane $x_{1}Ox_{2}$. Then the intersection $L\cap E^{3}$ is either 1) empty or 2) congruent to $D$, or 3) congruent to $E^{2}$.

1) Any straight line passing through $z$ in $L$ does not intersect $E^{3}$.

2) Since $L\cap E^{3}$ is convex in $L$, there exists a straight line passing through $z$ in $L$ and not intersecting $L\cap E^{3}$, therefore, not intersecting $E^{3}$.

3) $L\cap E^{3}\in\mathbf{WC^{2}_{1}}\setminus\mathbf{C^{2}_{1}}$ and $L\cap\left(P_{-}^{3}\cup P_{+}^{3}\right)=(L\cap E^{3})^{\triangle}$ with respect to $L$. Since $z\not\in(L\cap E^{3})^{\triangle}$, there exists a straight line passing through $z$ in $L$ and not intersecting $L\cap E^{3}$, therefore, not intersecting $E^{3}$.

The set $(E^{3})^{\triangle}$ is bounded, connected, and non-convex, obviously.

∎

Consider the oblique cylinders

where

Now make sure that the unbounded open set

belongs to the class $\left(\mathbf{WS^{3}_{1}}\setminus\mathbf{S^{3}_{1}}\right)\cap\left(\mathbf{WC^{3}_{1}}\setminus\mathbf{C^{3}_{1}}\right)$; see Figure 5 b).

Consider the following polygonal oblique prisms.

where

At this point, we choose $a_{+}^{3}$ such that the set $\bigcup\limits_{k=0}^{\infty}P_{k}^{3}$ does not contain any ray.

Prove that

First, show that

Define the bottom base of each prism $P^{3}_{k}$, $k=0,1,2,\ldots$, by $P^{-}_{k}$. Consider an arbitrary point $x\in\bigcup\limits_{k=0}^{\infty}P_{k}^{3}$. Then $x\in P^{3}_{q}$, $q\in\{0,1,2,\ldots\}$. Let $\eta(x)$ be an arbitrary ray emanating from $x$.

1. 1.

   If $\eta(x)$ intersects a lateral face of the prism $P^{3}_{q}$, then consider the projection, parallel to the lateral edges of $P^{3}_{q}$, of $\eta(x)$ onto the plane $L\supset P^{-}_{q}$. It is a ray that we define by $\eta(x_{0})$. The ray $\eta(x_{0})$ emanates from the point $x_{0}\in P^{-}_{q}$ which is the projection of $x$ onto $L$. Since $L\cap E^{3}$ is a flat weakly $1$-semiconvex set as a set congruent to $E^{2}$, it implies that $\eta(x_{0})\cap(L\cap E^{3})\neq\varnothing$, which gives that $\eta(x)\cap E^{3}_{q}\neq\varnothing$, therefore, $\eta(x)\cap E^{3}\neq\varnothing$.

2. 2.

   If $\eta(x)$ intersects a base of $P^{3}_{q}$, then it intersects either $D_{-}^{3}$, which immediately gives that $\eta(x)\cap E^{3}\neq\varnothing$, or it intersects a lateral face of another prism $P^{3}_{q^{\prime}}$, $q^{\prime}\in\{0,1,2,\ldots\}$, by the construction, and further considerations are the same as in item 1, but for a point $x^{\prime}\in\eta(x)\cap P^{3}_{q^{\prime}}$ and the ray $\eta(x^{\prime})\subset\eta(x)$. Then $\eta(x^{\prime})\cap E^{3}\neq\varnothing$, therefor, $\eta(x)\cap E^{3}\neq\varnothing$.

Moreover, if $x\in(E^{3})^{\diamondsuit}$, then $x\in(E^{3})^{\triangle}$.