Basic problem on topology $( James Dugundji)$ Consider $(X, \tau)$ be a topological space . Then


    
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*$Fr[Fr\{Fr(A) \}] = Fr[Fr(A)]$, where $Fr(A) =\overline A \cap \overline{A^c}$ is the frontier of the set $A$
    
*Assume that $Fr(A) \cap Fr(B) = \phi$. Then $ (A \cup B)^{\circ}= A^{\circ} \cup B^{\circ}$ and $Fr(A \cap B) = [\overline A \cap Fr(B)] \cup [\overline B \cap Fr(A)]$
    
*Assume that $\mathfrak B$ be a subbasis for $X$ and $D \subset X$ such that $U \cap D \neq \phi $ for each $U \in \mathfrak B$. Does this imply $D$ is dense in $X$


I have tried


*

*we know that $Fr(A) =\overline A \cap \overline{A^c}$


$\therefore Fr \{Fr(A)\} =\overline{ \overline A \cap \overline{A^c}} \cap \overline {(\overline A \cap \overline{A^c})^c} =  \overline A \cap \overline{A^c} \cap (\overline A^c \cup \overline {A^c}^c) $
Further how to proceed


*I can easily  prove that $A^{\circ} \cup B^{\circ} \subseteq (A \cup B)^{\circ}$. Please give me hint of its converse  and $Fr(A \cap B) = [\overline A \cap Fr(B)] \cup [\overline B \cap Fr(A)]$

*I think $D$ may not be dense. Please give me counter example.
Any help would be appreciated. Thank you
 A: Some HINTS:
$\newcommand{\cl}{\operatorname{cl}}\newcommand{\fr}{\operatorname{Fr}}\newcommand{\int}{\operatorname{int}}$You know that $\fr A$ is closed. Suppose that $U\subseteq\fr(\fr A)$ is open. Then $U\subseteq\cl(\fr A)=\fr A$, so $U\cap(X\setminus\fr A)=\varnothing$, and therefore $U\cap\cl(X\setminus\fr A)=\varnothing$. But $U\subseteq\cl(X\setminus\fr A)$, so $U=\varnothing$. Thus, $\fr(\fr A)$ is nowhere dense. To finish the first problem, show that if $C$ is a nowhere dense closed set, then $\fr C=C$.
For the first part of the second question suppose that $x\notin(\int A)\cup\int B$. Then $x\notin\int A$, so $x\in\cl(X\setminus A)$, and $x\notin\int B$, $x\in\cl(X\setminus B)$. Now show that if $x\in\int(A\cup B)$, then $x\in\cl A$ and $x\in\cl B$. Conclude that $x\in(\fr A)\cap\fr B$ to get a contradiction.
Note that if $(\fr A)\cap\fr B=\varnothing$, then $\big(\fr(X\setminus A)\big)\cap\fr(X\setminus B)=\varnothing$, so the first part of the second question implies that 
$$\begin{align*}
X\setminus\big(\cl(A\cap B)\big)&=\int\big(X\setminus(A\cap B)\big)\\
&=\int\big((X\setminus A)\cup(X\setminus B)\big)\\
&=\big(\int(X\setminus A)\big)\cup\int(X\setminus B)\\
&=(X\setminus\cl A)\cup(X\setminus\cl B)\\
&=X\setminus\big((\cl A)\cap\cl B\big)\;,
\end{align*}$$
and hence that $\cl(A\cap B)=(\cl A)\cap\cl B$. Now apply this to the second part of the second question
$$\begin{align*}
\cl(A\cap B)\cap\cl\big(X\setminus(A\cap B)\big)&=(\cl A)\cap(\cl B)\cap\cl\big((X\setminus A)\cup(X\setminus B)\big)\\
&=(\cl A)\cap(\cl B)\cap\big(\cl(X\setminus A)\cup\cl(X\setminus B)\big)\\
&=\ldots
\end{align*}$$
Your conjecture for the third question is correct: $D$ need not be dense in $X$. Let
$$\mathfrak{B}=\{(\leftarrow,a):a\in\Bbb R\}\cup\{(a,\to):a\in\Bbb R\}\;,$$
the set of all open rays in $\Bbb R$. Verify that $\mathfrak{B}$ is a subbase for the usual topology on $\Bbb R$. Then find a set $D\subseteq\Bbb R$ such that $D\cap U\ne\varnothing$ for each $U\in\mathfrak{B}$, but $D$ is not dense in $\Bbb R$. (It’s possible to choose $D$ to be a countable, closed, discrete set.)
