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Prove that the

i) int(A) is a regular open set for every closed set A

ii) closure(U) is a regular close set for every open set U

For i) is this a valid solution. We have to show that int(A)=int[closure(intA)] Since A is closed set, then A=closure[int(A)] then int(A)=int(A) ???

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2 Answers 2

$\newcommand{\cl}{\operatorname{cl}}\newcommand{\int}{\operatorname{int}}$Your argument for (i) doesn’t work, because it’s not true in general that $A=\cl\int A$ when $A$ is a closed set. For example, suppose that $A$ is the middle-thirds Cantor set in $\Bbb R$; then $A$ is closed, but $\int A=\varnothing$, so $\cl\int A=\varnothing\ne A$. For that matter, just let $A=\{0\}$: this is also a closed set in $\Bbb R$, but $\cl\int\{0\}=\cl\varnothing=\varnothing\ne\{0\}$.

Let $A$ be a closed set in a space $X$, and let $U=\int A$. Clearly $U\subseteq A$, so $\cl U\subseteq\cl A=A$, and therefore $\int\cl U\subseteq\int A=U$. It only remains to show that $U\subseteq\int\cl U$. HINT: Start from the fact inclusion $U\subseteq\cl U$ and take interiors on both sides.

For (ii) you can argue in very similar fashion. Start with an open set $U$, and let $A=\cl U$. Then $U\subseteq A$, so $U=\int U\subseteq\int A$, and ... ?

Alternatively, you can start with an open set $U$ and let $A=X\setminus U$. Then $A$ is closed, so by (i) we know that $\int A$ is regular open. If you know that the complement of a regular open set is regular closed, you can conclude that $X\setminus\int A$ is regular closed and try to show that it’s equal to $\cl U$.

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For every open set $O$ we have $O=\text{int}(O)$ and thus $O\subseteq \text{int}(\overline O)$ (Why?)
Similarly, if $A$ is closed, we have $A=\overline A$, thus also $A \supseteq\overline{\text{int}(A)}$.

There is one special property of the closure and interior operators, that you should use throughout the proof.

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