On globs (Q800507)

Any subset of R expressible as \(\cup\{A(\alpha):\alpha\in <1,\infty)\}\), where each \(A(\alpha)\) is a closed subset of R and whenever \(\alpha,\beta\in <1,\infty)\), \(\alpha <\beta,\) each point of \(A(\alpha)\) is a bilateral c-point of \(A(\beta)\), will be called a linear glob. \textit{S. Agronsky}, in his Ph.D. dissertation (1974) proved: (1) Each \(F_{\sigma}\) bilaterally c-dense-in-itself subset of R is a linear glob and (2) A subset A of R is a linear glob iff there exists a real upper semicontinuous Darboux function of a real variable such that \(A=f^{- 1}((0,\infty)).\) The author defined globs in \(R^ 2\) as follows: A point \(x\in R^ 2\) is a panoramic c-limit point of a subset A of \(R^ 2\) iff each nondegenerate closed triangle containing x contains \(2^{\aleph_ 0}\) points of A. A system \(\{A(\alpha):\alpha\in <1,\infty)\}\) of closed subsets of \(R^ 2\) is called a hierarchy if each point of \(A(\alpha)\) is a panoramic c-limit point of \(A(\beta)\) whenever \(\alpha,\beta\in <1,\infty)\) and \(\alpha <\beta.\) A subset of \(R^ 2\) will be called a glob in \(R^ 2\) iff it is a union of some hierarchy. The main results are as follows: 1. The globness is not a topological invariant in \(R^ 2\), since there exists a glob A in \(R^ 2\) and a homeomorphism g of \(R^ 2\) onto \(R^ 2\) such that g(A) is not a glob. 2. Each panoramically c-dense-in-itself \(F_{\sigma}\) subset of \(R^ 2\), of which each nonempty relatively open set contains a glob in \(R^ 2\), is a glob in \(R^ 2\). 3. Each glob in \(R^ 2\) is the inverse image of a non void open set under a Darboux upper semicontinuous function. 4. The product of two globs in R is not necessarily a glob in \(R^ 2\). It is a glob in \(R^ 2\) if one of these globs is open in density topology. 5. The difference of each open subset of \(R^ 2\) and of a subset of \(R^ 2\) which is of Lebesgue measure zero, contains a glob in \(R^ 2\). 6. For each subset A of \(<0,1>\times <0,1>\) of Lebesgue measure zero (of the first category), there exists a perfect subset P of \(<0,1>\) and a subset Q of full Lebesgue measure in \(<0,1>\) (residual in \(<0,1>)\) such that \((P\times Q)\cap A=\emptyset.\)