Answer to Question #144079 in Differential Geometry | Topology for rzgar ali

Question #144079
Let (X, T) be topological space and A⊂=X. Prove the following
1 (A°) ^c= line over(A^c).
2- A ((A)°)^C = A^COC
3- A= (A").
4- b(A)⊂= A.
5- b(A) = b(A^c).
6- b(A) = (line over A) - A°
1
Expert's answer
2020-11-17T12:33:05-0500

"1.\\text{We should prove that } (A^o)^c=\\overline{A^c}\\\\\n\\text{or } X-A^o=\\overline{X-A}.\\\\\n\\text{Let } \\tau \\text{ be the topology on }X \\text{ and } A\\sube X,\\ A\\text{ is a subset of X}.\\\\\n\\text{Let } \\mathbb{K}=\\{K\\in \\tau: K\\sube A\\}.\\\\\n\\text{Then:}\\\\\nX-A^o=X-\\cup_{K\\in \\mathbb{K}}K\\text{(definition of } A^o)=\\\\\n=\\cap_{K\\in\\mathbb{K}} (X-K)\\text{(De Morgan's Laws: Difference with Union)}.\\\\\n\\text{By the definition of a closed set, } K \\text{ is open in X } \\Leftrightarrow\\\\\nX-K \\text{ is closed in } X.\\\\\n\\text{Also we have } X-A\\sube X-K.\\\\\n\\text{Let } \\mathbb{K}^\\prime:=\\{K^\\prime\\sube X: (X-A)\\sube K^\\prime, K^\\prime \\text{ is closed in X}\\}.\\\\\n\\text{We see that } K\\in \\mathbb{K}\\Leftrightarrow X-K\\in \\mathbb{K}^\\prime.\\\\\n\\text{Thus:}\\\\\nX-A^o=\\cap_{K\\prime\\in \\mathbb{K}^\\prime}K^\\prime=\\overline{X-A}\\text{(definition of closure of } X-A).\\\\\n2.\\ A \\text{ is called coc-compact open set (or coc-open set) if}\\\\\n\\text{for every } x\\in A \\text{ there exists an open set } U\\sube X\\\\\n\\text{and a compact subset } K \\in C(X,\\tau) \\text{ such that } x \\in U-K\\sube A.\\\\\n\\text{The complement of coc-open set is called coc-closed set}.\\\\\n3.\\ A=A^o.\\\\\n\\text{This equality is true if and only if } A\\text{ is an open set}\\\\\n\\text{(definition of an open set)}.\\\\\n6.\\ \\partial A=\\overline{A}-A^o.\\\\\n\\text{This is a definition of the boundary } \\partial A\\text{ of a subset }A\\\\\n\\text{of a topological space } (X, \\tau).\\\\\n4.\\ \\partial A\\sube A.\\\\\n\\partial A \\text{ is a set of all boundary points of } A.\\\\\n5.\\text{ We know that } \\overline{A}=(\\text{int}(A^c))^c.\\\\\n\\text{Then } \\overline{A^c}=(\\text{int}(A))^c \\text{ because } (A^c)^c=A\\\\\n\\partial A=\\overline{A}\\cap (\\text{int}(A))^c=\\overline{A}\\cap\\overline{A^c}\\\\\n\\partial A^c=\\overline{A^c}\\cap\\overline{A}\\\\\n\\partial A=\\partial A^c"


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