Question

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°

Answer 1

$1.\text{We should prove that } (A^o)^c=\overline{A^c}\\ \text{or } X-A^o=\overline{X-A}.\\ \text{Let } \tau \text{ be the topology on }X \text{ and } A\sube X,\ A\text{ is a subset of X}.\\ \text{Let } \mathbb{K}=\{K\in \tau: K\sube A\}.\\ \text{Then:}\\ X-A^o=X-\cup_{K\in \mathbb{K}}K\text{(definition of } A^o)=\\ =\cap_{K\in\mathbb{K}} (X-K)\text{(De Morgan's Laws: Difference with Union)}.\\ \text{By the definition of a closed set, } K \text{ is open in X } \Leftrightarrow\\ X-K \text{ is closed in } X.\\ \text{Also we have } X-A\sube X-K.\\ \text{Let } \mathbb{K}^\prime:=\{K^\prime\sube X: (X-A)\sube K^\prime, K^\prime \text{ is closed in X}\}.\\ \text{We see that } K\in \mathbb{K}\Leftrightarrow X-K\in \mathbb{K}^\prime.\\ \text{Thus:}\\ X-A^o=\cap_{K\prime\in \mathbb{K}^\prime}K^\prime=\overline{X-A}\text{(definition of closure of } X-A).\\ 2.\ A \text{ is called coc-compact open set (or coc-open set) if}\\ \text{for every } x\in A \text{ there exists an open set } U\sube X\\ \text{and a compact subset } K \in C(X,\tau) \text{ such that } x \in U-K\sube A.\\ \text{The complement of coc-open set is called coc-closed set}.\\ 3.\ A=A^o.\\ \text{This equality is true if and only if } A\text{ is an open set}\\ \text{(definition of an open set)}.\\ 6.\ \partial A=\overline{A}-A^o.\\ \text{This is a definition of the boundary } \partial A\text{ of a subset }A\\ \text{of a topological space } (X, \tau).\\ 4.\ \partial A\sube A.\\ \partial A \text{ is a set of all boundary points of } A.\\ 5.\text{ We know that } \overline{A}=(\text{int}(A^c))^c.\\ \text{Then } \overline{A^c}=(\text{int}(A))^c \text{ because } (A^c)^c=A\\ \partial A=\overline{A}\cap (\text{int}(A))^c=\overline{A}\cap\overline{A^c}\\ \partial A^c=\overline{A^c}\cap\overline{A}\\ \partial A=\partial A^c$