Limit Points

Limit Point

Suppose that

A

is a subset of a topological space

X

, then a point

x \in X

is called a limit point of $A$ when every neighborhood of

x

intersects

A

in some point other than

x

itself

Limit Point Iff Every Basis Element Containing Intersects at a Different Point

Suppose that

A

is a subset of a topological space

X

which is generated by a basis

ℬ

, then a point

x \in X

is a limit point when every basis element

B

continaing

x

intersects

A

in some point other than

x

itself

TODO: Add the proof here.

Limit Point iff Element of Closure minus a Point

x

is a limit point iff

x \in A ∖ {x}

x

is a limit point iff every neighborhood of

x

intersects

A

in some point other than

x

itself, which is equivalent to every neighborhood intersecting

A ∖ {x}

iff

x \in A ∖ {x}

This now shows the motivation behind the choice "limit point", when we just looked at the closure isolated points we're part of it, but now they are removed.

Limit Points are a Subset of Closure

Suppose that

A

is a subset of a topological space

X

, if

A^{'}

is the set of limit points of

A

, then

A^{'} \subseteq A

Suppose that

x \in A^{'}

, then every neighborhood of

x

intersects

A

therefore

x \in A

as needed.

Closure Equals Limit Points Union Itself

Let

A

be a subset of a topological space

X

. If

A^{'}

is the set of all limit points of

A

, then

A = A^{'} \cup A

Since $A \subseteq A$ and $A^{'} \subseteq A$ , then $A^{'} \cup A \subseteq A$

Suppose that $x \in A$ , if $x \in A$ , then we would be done, so we assume that $x \notin A$ and our goal remains to show that $x \in A^{'}$ , so let $U$ be a neighborhood of $x$ , we want to show that it intersects $A$ at a point different than $x$ , since $x \in A$ , then we know that $U$ intersects $A$ , since $x \notin A$ then $U$ cannot intersect $A$ at that point so we've shown that $x \in A^{'}$ as needed.

The Closure Is the Union of a and the Limits Points of a That Are Not in A

Let

A

be a subset of a topological space

X

, and let

B = X ∖ A

then

A = A ⊔ (A^{'} \cap B)

This boils down to proving that

A \cup A^{'} = A ⊔ (A^{'} \cap B)

. This is the case because if we take an element from the left then if it's in

A

then it's clearly in the right hand side so assume instead its in

A^{'}

then either

x \in A

x \notin A

in either case its an element from right hand side. Taking an element from the right, and disposing of the case of

x \in A

similarly then whenever

x \in A^{'} \cap B

then

x \in A^{'}

still so it's an element of the left hand side.

The above corollary might seem a little useless, but the point of it is to help you be able to find the closure faster, fundamentally this is because you already know that $\overline{A}$ will already contain all of $A$ , and thus to find the rest of the closure you just have to inspect elements which are not in $A$ which are limit points. This reduces your search space when trying to show that a set is closed for example.

A Subset of a Topological Space is Closed if and only if it contains all it's Limit Points

Suppose

A

is a subset of a topological space

X

then it is closed if and only if it contains it's limit points

A set is closed iff

A = A

since

A = A \cup A^{'}

, then this is true iff

A^{'} \subseteq A

Closed Iff Every Point Outside the Set Is Not a Limit Point

Suppose

A

is a subset of a topological space

X

then it is closed if and only for every point

x \in X ∖ A

, then

x

is not a limit point.

Therefore if

p \in X

is a limit point of

A

then it must be that

x \in A

therefore

A^{'} \subseteq A

so it is closed.