Closed Set
A subset $$A$$ of a topological space $$X$$ is said to be closed in $$X$$ if $$X \backslash A$$ is open in $$X$$

Note: When $$X$$ is known by context then we may say $$A$$ is closed to mean that $$A$$ is closed in $$X$$

Closed Topology
Suppose that $$X$$ is a topological space, then the following conditions hold:
• $$\emptyset$$ and $$X$$ are closed
• arbitrary intersections of closed sets are closed
• finite unions of closed sets are closed

The complement of $$\emptyset , X$$ in $$X$$ are $$X , \emptyset$$ respectively, which are both open in $$X$$, making the original sets closed

Suppose that $$\left \lbrace A_{\alpha} \right \rbrace_{\alpha \in I}$$ is a collection of closed sets, then we know that $$X \backslash \bigcap_{\alpha \in I} A_{\alpha} = \bigcup_{\alpha \in I} \left ( X \backslash A_{\alpha} \right )$$, and since each $$A_{\beta}$$ is closed then $$X \backslash A_{\beta}$$ is open, making $$\bigcup_{\alpha \in I} \left ( X \backslash A_{\alpha} \right )$$ an arbitrary union of open sets, and is thus open, making $$X \backslash \bigcap_{\alpha \in I} A_{\alpha}$$ open

Now suppose that $$I$$ is finite, and we'll show that $$X \backslash \bigcup_{\alpha \in I} A_{\alpha}$$ is open, we know this set is equal to $$\bigcap_{\alpha \in I} \left ( X \backslash A_{\alpha} \right )$$, and as a finite intersection of open sets, we know it is open as well showing the original set is open

Closed in a Subspace if and only if it's an Intersection
Let $$Y$$ be a subspace of $$X$$, then a set $$A$$ is closed in $$Y$$ if and only if it equals the intersection of a closed set of $$X$$ with $$Y$$

Let's start by assuming that we have a set $$A = C \cap Y$$ where $$C$$ is closed in $$X$$, therefore $$X \backslash C$$ is open in $$X$$, therefore $$Y \cap \left ( X \backslash C \right )$$ is open in $$Y$$

Note that $$Y \cap \left ( X \backslash C \right )$$ $$=$$ $$\left ( X \cap Y \right ) \backslash C$$ $$=$$ $$\left ( X \cap Y \right ) \backslash \left ( C \cap Y \right )$$ $$=$$ $$Y \backslash A$$ (where the last step is justified since we know $$Y \subseteq X$$ ), this shows that $$Y \setminus A$$ is open, meaning that $$A$$ is closed in $$Y$$

Now let's assume that $$A$$ is closed in $$Y$$, therefore $$A \subseteq Y$$ and $$Y \backslash A$$ is open in $$Y$$, meaning that there is some set $$U$$ open in $$X$$ such that $$Y \backslash A = Y \cap U$$

Let's note that $$Y \setminus \left ( Y \setminus A \right )$$ $$=$$ $$\left ( Y \cap A \right ) \cup \left ( Y \setminus Y \right ) = A \cup \emptyset = A$$ so that

\begin{align*} A &= Y \setminus \left ( Y \setminus A \right ) \\ &= Y \setminus \left ( Y \cap U \right ) \\ &= \left ( Y \setminus Y \right ) \cup \left ( Y \setminus U\right ) \\ &= Y \setminus U \\ &= \left ( X \cap Y \right ) \setminus U \\ &= Y \cap \left ( X \setminus U \right) \end{align*}

Since $$X \setminus U$$ is closed in $$X$$ this shows that $$A$$ equals the intersection of a closed set with $$Y$$ as needed.

Interior
Suppose that $$A$$ is a subset of a topological space, then the interior of $$A$$ is defined to as the union of all open sets contained in $$A$$ and is denoted by $$\operatorname{Int}\left( A \right)$$
Closure
Suppose that $$A$$ is a subset of a topological space, then the closure of $$A$$ is defined to as the intersection of all closed sets containing $$A$$ and is denoted by $$\bar{A}$$
The Closure is Closed
$$\operatorname{Int}\left( A \right)$$ is closed
The closure of $$A$$ is defined as $$\bigcap _ { S \supseteq A } S$$ where each $$S$$ is a closed set, to show it's closed, we must show that $$X \setminus \bigcap _ { S \supseteq A } S$$ is open which we can see since it equals $$\bigcap _ { S \supseteq A } \left( X \setminus S \right)$$ and thus is an arbitrary union of open sets so it is open.
The Interior is Open
$$\bar{A}$$ is open
The interior is an arbitrary union of open sets, and is thus open.
Closed Supersets are Supersets of the Closure
Suppose that $$\bar{A}$$ is the closure of $$A$$ in a topological space, then given an closed set $$A \subseteq U$$ we have $$\bar{A} \subseteq U$$
TODO
Open Subsets are Subsets of the Interior
Suppose that $$\operatorname{Int}\left( A \right)$$ is the interior of $$A$$ in a topological space, then given an open set $$U \subseteq A$$ we have $$U \subseteq \operatorname{Int}\left( A \right)$$
TODO
Interior is Smaller, Closure is Bigger
Suppose that $$A$$ is a subset of a topological space, then we have $\operatorname{Int}\left( A \right) \subseteq A \subseteq \bar{A}$
TODO
Open Sets Equal their Interior
Suppose that $$A$$ is a subset of a topological space $$X$$, then $$A$$ is open if and only if $$A = \operatorname{ Int } \left( A \right)$$

Suppose that $$\mathcal { A }$$ is the collection of open sets which are also subsets of $$A$$, since $$A$$ is open then $$A \in \mathcal { A }$$, then $$\overline{ A } = \bigcup \mathcal{ A } = \bigcup \left( \mathcal{ A } \setminus \left\{ A \right\} \right) \cup A$$, since each element in $$\mathcal{ A } \setminus \left\{ A \right\}$$ is a subset of $$A$$, then so is the union, thus $$\bigcup \left( \mathcal{ A } \setminus \left\{ A \right\} \right) \cup A = A$$ as needed.

The other direction is easier, suppose that $$A = \operatorname{ Int } \left( A \right)$$, then $$\operatorname{ Int } \left( A \right)$$ is open, then so is $$A$$ as needed.

Closed Sets Equal their Closure
If $$A$$ is a subset of a topological space $$X$$, then $$A$$ is closed if and only if $$A = \overline{ A }$$

Suppose that $$\mathcal{ A }$$ is the collection of closed sets which are also supersets of $$A$$, then $$\overline{ A } = \bigcap \mathcal{ A } = \bigcap \left( \mathcal{ A } \setminus \left\{ A \right\} \right) \cap A$$, and since each element in $$\mathcal{ A } \setminus \left\{ A \right\}$$ is a superset of $$A$$, then so is it's intersection, therefore $$\bigcap \left( \mathcal{ A } \setminus \left\{ A \right\} \right) \cap A = A$$

Now suppose that $$A = \overline{ A }$$, since $$\overline{ A }$$ is closed then we know $$A$$ is closed

Closure in a Subspace is an Intersection
Let $$Y$$ be a subspace of $$X$$ and $$A \subseteq Y$$, then suppose that $$\bar{A}$$ is the closure of $$A$$ in $$X$$, then the closure of $$A$$ in $$Y$$ equals $$\bar{A} \cap Y$$

Set $$B$$ equal to the closure of $$A$$ in $$Y$$, since $$\bar{A}$$ is closed in $$X$$, then we know that $$\bar{A} \cap Y$$ is closed in $$Y$$, also note that $$A \subseteq \bar{A} \cap Y$$ because we know $$A \subseteq \bar{A}$$ and $$A \subseteq Y$$ so we can conclude $$B \subseteq \left( \bar{A} \cap Y \right)$$

On the other hand, we know that $$B$$ is closed in $$Y$$, thus $$B = C \cap Y$$ for some set $$C$$ closed in $$X$$, we recall that $$A \subseteq B$$ so that $$A \subseteq \left( C \cap Y \right)$$, therefore $$A \subseteq C$$ showing that $$C$$ is a closed set containing $$A$$, thus $$\bar{A} \subseteq C$$ so that $$\left( \bar{A} \cap Y \right) \subseteq \left( C \cap Y \right) = B$$, as needed.

Neighborhood
In a topological space $$X$$ and a point $$x \in X$$ then if $$U$$ is an open set containing $$x$$ then we say that $$U$$ is a neighborhood of $$x$$
Closure Intersection Equivalence
$$x \in \bar{A}$$ if and only if neighborhood of $$x$$ intersects $$A$$

We prove the contrapositive, that is $$x \notin \bar{A}$$ iff there is some open set $$U$$ containing $$x$$ that doesn't intersect $$A$$

Suppose that $$x \notin \bar{A}$$, since $$\bar{A}$$ is closed, then $$U = X \setminus \bar{A}$$ must be open and containing $$x$$ and since $$A \subseteq \bar{A}$$, we know $$U$$ doesn't intersect $$A$$, as needed.

Now the reverse direction: assuming that we have some open set $$U$$ containing $$x$$ that doesn't intersect $$A$$, we have $$A \subseteq X \setminus U$$ and $$X \setminus U$$ is closed therefore $$\bar{A} \subseteq X \setminus U$$, so that $$\bar{A} \cap U = \emptyset$$, since we know $$x \in U$$ we know that $$x \notin \bar{A}$$ as needed.

Closure Basis Intersection Equivalence
Suppose that a basis $$\mathcal{B}$$ generates a topology $$\mathcal{T}$$, then $$x \in \bar{A}$$ if and only if every basis element $$B$$ containing $$x$$ intersects $$A$$

Suppose that $$x \in \bar{A}$$, therefore every open set containing $$x$$ intersects $$A$$, now suppose that $$B \in \mathcal{B}$$, since it is open, then we know that $$B$$ intersects $$A$$

Now suppose that every basis element containing $$x$$ intersects $$A$$, and we'd like to show that $$x \in \bar{A}$$ which is equivalent to every open set containing $$x$$ intersecting $$A$$, so suppose that $$U \in \mathcal{T}$$ since $$\mathcal{B}$$ generates $$\mathcal{T}$$, then there is some $$B \in \mathcal{B}$$ such that $$x \in B \subseteq U$$ since $$A$$ intersects $$B$$ then it also intersects $$U$$ as needed.