Union of semi open sets in generalized topological space

Question

In order to define the semi closure of a set in a generalized topological space one must show first that the union of semi open sets is open. I found articles that cite Csaszar but they did not show how he proved it or what his remarks were. Does anyone here know how?

Answer 1

Let $\mathscr{A}$ be a family of semi-open sets in a space $X$, and let $S=\bigcup\mathscr{A}$; we wish to show that $S$ is semi-open. For each $A\in\mathscr{A}$ there is an open $U_A$ in $X$ such that $U_A\subseteq A\subseteq\operatorname{cl}U_A$. Let

$$U=\bigcup_{A\in\mathscr{A}}U_A\;;$$

then $U$ is open, and

$$U\subseteq S=\bigcup\mathscr{A}\subseteq\bigcup_{A\in\mathscr{A}}\operatorname{cl}U_A\subseteq\operatorname{cl}\bigcup_{A\in\mathscr{A}}U_A=\operatorname{cl}U\;,$$

so $S$ is semi-open.

I’ve used the definition that a set $A$ is semi-open if there is an open $U$ such that $U\subseteq A\subseteq\operatorname{cl}U$; some writers use the definition that $A$ is semi-open if $A\subseteq\operatorname{cl}\operatorname{int}A$. These definitions are easily seen to be equivalent.

• If $A\subseteq\operatorname{cl}\operatorname{int}A$, we simply set $U=\operatorname{int}A$ and have $U\subseteq A\subseteq\operatorname{cl}U$.

• If, on the other hand, there is an open $U$ such that $U\subseteq E\subseteq\operatorname{cl}U$, then $U\subseteq\operatorname{int}A$, so $A\subseteq\operatorname{cl}U\subseteq\operatorname{cl}\operatorname{int}A$.