Open and Closed Sets definition Does the complement of a set being closed necessarily imply that the set itself is open? Could the set be both/neither open and/nor closed if its complement is strictly closed?
 A: Definition. A set is closed if and only if its complement is open.
In every topological space $(X,\tau)$ there are at least two set which are both open and closed: 
$\varnothing$ and $X$. 
If the discrete topology, where $\tau={\mathcal P}(X)$, every subset of $X$ is both open and closed.
A topological space is connected if and only if the only subsets which are both open and closed are $\varnothing$ and $X$. 
In $\mathbb R$, there are $\boldsymbol{c}=2^{\aleph_0}$ open subsets and as many closed, while there are $2^{\boldsymbol{c}}$ subsets which are neither open nor closed.
A: Yes to the first question, no to the second. There are multiple ways to define a closed set. One of which is that a set is closed iff the complement is open. This is equivalent to saying, "a set is closed iff it contains all it's limit points".
We can see that the complement of an open set is closed iff the complement of an open set contains all its limit points. 
$(\implies)$ Let $U \subset X$ be open and let $x \in U$. Now consider $X\backslash U$ which is closed by assumption. We know $x \notin X\backslash U$, and since $U$ is an open set containing $x$, then $x$ is not a limit point of $X\backslash U$ because $(X\backslash U) \cap [U \backslash \{x \}] = \emptyset$. Since $x$ was an arbitrary point outside of $X\backslash U$, we know there are no limit points of $X\backslash U$ outside of $X\backslash U$. This is to say that if $X\backslash U$ has a limit point, it has to be contained in $X\backslash U$. 
$(\impliedby)$ Let $V \subset X$ be a set such that $V \backslash X$ contains all it's limit points. This means any $y \in V$ cannot be a limit point of $X\backslash V$. Thus, we can find an open set $W \subset X$ such that $y \in W$ and $(X\backslash V) \cap [W \backslash \{y \}] = \emptyset \implies (X\backslash V) \cap W  = \emptyset$. This means that $W \subset V$. We can now form a collection of open sets $\{W_y \}_{y \in V}$ with the property that: $$(X\backslash V) \cap W_y = \emptyset \space \text{and} \space W_y \subset V$$ This is true for all $y \in V$. It follows that $$\bigcup_{y \in V}W_y \subseteq V$$ But we also know for each $y \in V$ that $$y \in \bigcup_{y \in V}W_y$$ because $y \in W_y \subset \bigcup_{y \in V}W_y$. This implies that $$V \subseteq \bigcup_{y \in V}W_y$$ By double containment we conclude that $$V = \bigcup_{y \in V}W_y$$ Now that we have expressed $V$ as an arbitrary union of open sets, we know $V$ is open because any union of open sets is open. Thus, the complement of any set that contains all its limit points is open. $\Box$
A: In topology the definition of a close set is the complement of an open set
A: The complement of a closed set must be open.
Some people define a set to be closed iff its complement is open actually. You can use the defition of a set being closed means that it contains all its limit points to conclude that the complement is open as well.
Suppose a set $A$ is closed.This means all points in $A$ are limit points. Then pick $x\in A^c$, which means that $x\notin
A$. Since $x\notin A$, $x$ is not a limit point of $A$. This mean that $\exists \epsilon >0: B^*_{\epsilon}(x)\cap A=\emptyset$,such that there is an $\epsilon$-ball contained in $A^c$ which means that $x$ is an interior point of $A^c$ . But this means that $A^c$ is open since a set is open iff all its points are interior points.
