Search Results

You are looking at 1 - 3 of 3 items for

  • Author or Editor: Marilyn Breen x
Clear All Modify Search

Abstract  

Let S be an orthogonal polygon in the plane. Assume that S is starshaped via staircase paths, and let K be any component of Ker S, the staircase kernel of S, where KS. For every x in S\K, define W K(x) = {s: s lies on some staircase path in S from x to a point of K}. There is a minimal (finite) collection W(K) of W K(x) sets whose union is S. Further, each set W K(x) may be associated with a finite family U K(x) of staircase convex subsets, each containing x and K, with ∪{U: U in U K(x)} = W K(x). If W(K) = {W K(x 1), ..., W K(x n)}, then KV K ≡ ∩{U: U in some family U K(x i), 1 ≤ in} ⊆ Ker S. It follows that each set V K is staircase convex and ∪{V k: K a component of Ker S} = Ker S. Finally, if S is simply connected, then Ker S has exactly one component K, each set W K(x i) is staircase convex, 1 ≤ in, and ∩{W k(x i): 1 ≤ i ≤ n} = Ker S.

Restricted access

Abstract  

A Krasnosel’skii-type theorem for compact sets that are starshaped via staircase paths may be extended to compact sets that are starshaped via orthogonally convex paths: Let S be a nonempty compact planar set having connected complement. If every two points of S are visible via orthogonally convex paths from a common point of S, then S is starshaped via orthogonally convex paths. Moreover, the associated kernel Ker S has the expected property that every two of its points are joined in Ker S by an orthogonally convex path. If S is an arbitrary nonempty planar set that is starshaped via orthogonally convex paths, then for each component C of Ker S, every two of points of C are joined in C by an orthogonally convex path.

Restricted access

Abstract  

Fix k, d, 1 ≤ kd + 1. Let

\documentclass{aastex} \usepackage{amsbsy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{bm} \usepackage{mathrsfs} \usepackage{pifont} \usepackage{stmaryrd} \usepackage{textcomp} \usepackage{upgreek} \usepackage{portland,xspace} \usepackage{amsmath,amsxtra} \pagestyle{empty} \DeclareMathSizes{10}{9}{7}{6} \begin{document} $$\mathcal{F}$$ \end{document}
be a nonempty, finite family of closed sets in ℝd, and let L be a (dk + 1)-dimensional flat in ℝd. The following results hold for the set T ≡ ∪{F: F in
\documentclass{aastex} \usepackage{amsbsy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{bm} \usepackage{mathrsfs} \usepackage{pifont} \usepackage{stmaryrd} \usepackage{textcomp} \usepackage{upgreek} \usepackage{portland,xspace} \usepackage{amsmath,amsxtra} \pagestyle{empty} \DeclareMathSizes{10}{9}{7}{6} \begin{document} $$\mathcal{F}$$ \end{document}
}. Assume that, for every k (not necessarily distinct) members F 1, …, F k of
\documentclass{aastex} \usepackage{amsbsy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{bm} \usepackage{mathrsfs} \usepackage{pifont} \usepackage{stmaryrd} \usepackage{textcomp} \usepackage{upgreek} \usepackage{portland,xspace} \usepackage{amsmath,amsxtra} \pagestyle{empty} \DeclareMathSizes{10}{9}{7}{6} \begin{document} $$\mathcal{F}$$ \end{document}
,∪{F i: 1 ≤ ik} is starshaped and the corresponding kernel contains a translate of L. Then T is starshaped, and its kernel also contains a translate of L. Assume that, for every k (not necessarily distinct) members F 1, …, F k of
\documentclass{aastex} \usepackage{amsbsy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{bm} \usepackage{mathrsfs} \usepackage{pifont} \usepackage{stmaryrd} \usepackage{textcomp} \usepackage{upgreek} \usepackage{portland,xspace} \usepackage{amsmath,amsxtra} \pagestyle{empty} \DeclareMathSizes{10}{9}{7}{6} \begin{document} $$\mathcal{F}$$ \end{document}
,∪{F i: 1 ≤ ik} is starshaped and there is a translate of L meeting each set ker F i, 1 ≤ ik − 1. Then there is a translate L 0 of L such that every point of T sees via T some point of L 0. If k = 2 or d = 2, improved results hold.

Restricted access