Interior And Closure

Closed Set

A subset

A

of a topological space

X

is said to be closed in

X

X \ A

is open in

X

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 ${A_{α}}_{α \in I}$ is a collection of closed sets, then we know that $X \ ⋂_{α \in I} A_{α} = ⋃_{α \in I} (X \ A_{α})$ , and since each $A_{β}$ is closed then $X \ A_{β}$ is open, making $⋃_{α \in I} (X \ A_{α})$ an arbitrary union of open sets, and is thus open, making $X \ ⋂_{α \in I} A_{α}$ open

Now suppose that $I$ is finite, and we'll show that $X \ ⋃_{α \in I} A_{α}$ is open, we know this set is equal to $⋂_{α \in I} (X \ A_{α})$ , 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 $X \ C$ is open in $X$ , therefore $Y \cap (X \ C)$ is open in $Y$

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

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

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

\begin{matrix} A & = Y ∖ (Y ∖ A) \\ = Y ∖ (Y \cap U) \\ = (Y ∖ Y) \cup (Y ∖ U) \\ = Y ∖ U \\ = (X \cap Y) ∖ U \\ = Y \cap (X ∖ U) \end{matrix}

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

Transitivity of Closure

Y \subseteq X

is closed in

X

and

B \subseteq Y

is closed in

Y

then

B

is closed in

X

Since

B

is closed in

Y

then we know that

B = C \cap Y

where

C

is open in

X

. Now we have

X ∖ B = X ∖ (C \cap Y) = (X ∖ C) \cup (X ∖ Y)

Since

C, Y

were both closed in

X

then their complements are open showing that

X ∖ B

is the intersectino of two open sets and thus open and therefore

B

is closed in

X

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

Int (A)

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

\overline{A}

The Closure is Closed

\overline{A}

is closed

The closure of

A

is defined as

⋂_{S \supseteq A} S

where each

S

is a closed set, to show it's closed, we must show that

X ∖ ⋂_{S \supseteq A} S

is open which we can see since it equals

⋂_{S \supseteq A} (X ∖ S)

and thus is an arbitrary union of open sets so it is open.

The Interior is Open

I n t (A)

is open

The interior is an arbitrary union of open sets, and is thus open.

Is the Interior of the Closure of an Open Set Itself?

U

is an open set, is it true that

U = Int (\overline{U})

No it's not true, the closure has this healing property to fill in "internal gaps", and if that doesn't make sense consider

U = (- 1, 0) \cup (0, 1)

, then

\overline{U} = [- 1, 1]

but then the interior can't bring the hole back, as

Int (\overline{U}) = (- 1, 1) \neq U

Closed Supersets are Supersets of the Closure

Suppose that

\overline{A}

is the closure of

A

in a topological space, then given an closed set

A \subseteq U

we have

\overline{A} \subseteq U

TODO

Open Subsets are Subsets of the Interior

Suppose that

Int (A)

is the interior of

A

in a topological space, then given an open set

U \subseteq A

we have

U \subseteq Int (A)

By definition, the interior of A is the union of all open sets contained in A. Let $U_{i}$ be an arbitrary open subset of A. Since the interior is the largest open subset of the subset A, $U_{i} \subseteq ⋃_{i}^{} U_{i} \subseteq i n t (A)$ must be true. So, $U_{i} \subseteq i n t (A)$ . Since $U_{i}$ was chosen arbitrary, any open subset is a subset of the interior.

Interior is Smaller, Closure is Bigger

Suppose that

A

is a subset of a topological space, then we have

Int (A) \subseteq A \subseteq \overline{A}

We know that $Int (A)$ is a union sets which are subsets of $A$ thus it is a subset. On the other hand we know that $\overline{A}$ is an intersection of sets which are supersets of $A$ and therefore we know that it is a superset.

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 = Int (A)

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

The other direction is easier, suppose that $A = Int (A)$ , then $Int (A)$ is open, then so is $A$ as needed.

Closed Sets Equal their Closure

A

is a subset of a topological space

X

, then

A

is closed if and only if

A = A

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

Now suppose that $A = A$ , since $A$ 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

\overline{A}

is the closure of

A

X

, then the closure of

A

Y

equals

\overline{A} \cap Y

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

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 $A \subseteq (C \cap Y)$ , therefore $A \subseteq C$ showing that $C$ is a closed set containing $A$ , thus $\overline{A} \subseteq C$ so that $(\overline{A} \cap Y) \subseteq (C \cap Y) = B$ , 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 \overline{A}

if and only if every neighborhood of

x

intersects

A

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

Suppose that $x \notin \overline{A}$ , since $\overline{A}$ is closed, then $U = X ∖ \overline{A}$ must be open and containing $x$ and since $A \subseteq \overline{A}$ , 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 $A \subseteq X ∖ U$ and $X ∖ U$ is closed therefore $\overline{A} \subseteq X ∖ U$ , so that $\overline{A} \cap U = \emptyset$ , since we know $x \in U$ we know that $x \notin \overline{A}$ as needed.

Note that $ℚ = ℝ$ because for every $x \in ℝ$ we have that for any $(a, b)$ that contains $x$ then $(a, b) \cap ℚ \neq \emptyset$

Closure Basis Intersection Equivalence

Suppose that a basis

ℬ

generates a topology

𝒯

, then

x \in \overline{A}

if and only if every basis element

B

containing

x

intersects

A

Suppose that $x \in \overline{A}$ , therefore every open set containing $x$ intersects $A$ , now suppose that $B \in ℬ$ , 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 $x \in \overline{A}$ which is equivalent to every open set containing $x$ intersecting $A$ , so suppose that $U \in 𝒯$ since $ℬ$ generates $𝒯$ , then there is some $B \in ℬ$ such that $x \in B \subseteq U$ since $A$ intersects $B$ then it also intersects $U$ as needed.