Show that $U = int(cl(U))$ Let $U \subset X$ and $V := X  \backslash  U$. Show that $U = int(cl(U))$ only if, $V = cl(int(V))$.
I found this: $cl(U)=U \cup int(cl(U))$, but i don't know if it can be used here. Is it true that $U \subset int(cl(U)) $? If it is then $U \cup int(cl(U)) = int(cl(U))$ ?. And also i found that $cl(U) = U $ if $U$ is open, but how to check if this set is open?
 A: This is equivalent to showing that if $U = int(cl(U))$ then $V = cl(int(V))$.
Let's first recall 2 properties (which is easy to prove if $X$ is a metric space but also true for topological space):

*

*$cl(A)^c = int(A^c)$ and 2. $int(A)^c = cl(A^c)$ (which is just applying 1. on $A^c$)

Suppose $U = int(cl(U))$. Then $V = U^c = int(cl(U))^c = cl(cl(U)^c)$, by applying 2 on $cl(U)$.
But we also have $cl(U)^c = int(U^c)$ (applying 1) so then $V = cl(int(U^c)) = cl(int(V))$
I am however not sure if you can use $1$ and $2$.
A: Let $U \subseteq X$. Define the interior of $U$ to be the largest open set contained in $U$ (the existence of such is guaranteed by taking the union over all open sets contained in $U$, and this collection is nonempty since the empty set is in there). Denote this by $U^{\mathrm{o}}$. Define the closure of $U$ to be the smallest closed set containing $U$. Denote this by $\overline{U}$ (the existence of such is guaranteed by taking the intersection over all closed sets containing $U$, and this collection is nonempty since $X$ is in there).
Example: Let $U = (0,1]$. Then $\overline{U} = [0,1]$, $U^\mathrm{o} = (0,1)$, $\overline{(U^\mathrm{o})} = [0,1],$ $(\overline{U})^\mathrm{o} = (0,1).$
Claim 1: If $A \subseteq B \subseteq X$, then $B^c \subseteq A^c$, where we define $A^c = X \setminus A$.
Proof: Let $x \in B^c$, then $x \notin B$ per definition. Since $A \subseteq B$, $x \notin B$ implies $x \notin A$, so $x \in A^c$.
Claim 2: Let $U \subseteq X$ be a set. Then $(\overline{U})^c = (U^c)^\mathrm{o}$.
Proof: For notational convenience, let $V = U^c$. We note $U \subseteq \overline{U}$, so $(\overline{U})^c \subseteq V = U^c$ and $(\overline{U})^c$ is open. By maximality, we get $(\overline{U})^c \subseteq V^\mathrm{o} \subseteq V$. Taking complements again, we have $V^c = U \subseteq (V^\mathrm{o})^c$. By minimality of the closure, we get $U \subseteq \overline{U} \subseteq (V^\mathrm{o})^c$. Taking complements again, we have $V^\mathrm{o} \subseteq (\overline{U})^c$. This shows $V^\mathrm{o} = (\overline{U})^c$.
Claim 3: Let $U \subseteq X$ be a set. Then $(U^\mathrm{o})^c = \overline{(U^c)}$.
Proof: Exercise. It's the same kind of argument as in Claim 2.
Claim 4: Let $U \subseteq X$, $V = U^c$. If $U = (\overline{U})^\mathrm{o}$, then $V = \overline{(V^\mathrm{o})}$.
Proof: $V = U^c = ((\overline{U})^\mathrm{o})^c$. By Claim 3, we have $((\overline{U})^\mathrm{o})^c = \overline{((\overline{U})^c)}$. By Claim 2, we have $\overline{((\overline{U})^c)} = \overline{((U^c)^\mathrm{o})} = \overline{(V^\mathrm{o})}$.
Claim 5: Let $U \subseteq X$, $V = U^c$. If $V = \overline{(V^\mathrm{o})}$, then $U = (\overline{U})^\mathrm{o}$.
Proof: Exercise. It's the same argument as in Claim 4.
A: We need the following well known lemma:
Lemma: For any $A$ we have $$int(A)=cl(A^c)^c.$$
Now
$$U=int(cl(U))= cl(cl(U)^c)^c \Leftrightarrow U^c=cl(cl(U)^c)=cl(cl(U^{cc})^c)=cl(int(U^c)).
$$
where second equality follows from Lemma applied with $A=cl(U)$, and last equality follows from Lemma applyied with  $A=U^c$.
