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

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.