I'm currently painstakingly working my way through the basics of Topology and I'm on interior sets. My textbook provides this basic question for me to prove
Question to solve
Show that interior satisfies $\operatorname{int}A=(\bar{A'})'$
So, the interior of a set $A$ is the largest open set contained inside $A$, and this question wants me to relate that definition to the idea that the closed complement of A forms an interior set when again made to a complement.
My first idea to proving this is to simplify the question by moving the second complement over the equal so the LHS is now of the form $(\operatorname{int}A)'$. After that…I'm not sure how to progress. Any hints would be appreciated.
Best Answer
Observe that the interior of $A$ is also the union of all open sets contained in $A$. So, to check that $x\in \textrm{int}(A)$, it suffices to obtain an open $U$ such that $x\in U\subseteq A$.
If an element $x$ is in the interior, then there is an open set $U$ such that $x\in U\subseteq A$, hence $x$ is not in the closure of the complement of $A$, therefore $\textrm{int}(A)\subseteq (\bar{A^c})^c$. Conversely, if $x\in(\bar{A^c})^c$, then there is a neighbourhood $U$ of $x$ disjoint from $A^c$ for $x$ is not in the closure of $A^c$, therefore $U\subset A$, hence $x\in \textrm{int}(A)$.