NoteNextra-origin/content/Math4201/Math4201_L8.md

Math4201 Lecture 8

Recall from real analysis, a set is closed if and only if it has limit points.

New materials

Limit points

Let (X,\mathcal{T}) be a topological space. A is a subset of X, then we say x\in X is a limit point of A if any open set U\subset X containing x has another point y\in A-\{x\}.

Example of limit points

Let X=\mathbb{R} with standard topology.

Let A=(0,1), then set of limit points of A is [0,1].

Let A=\left{\frac{1}{n}\right}_{n\in \mathbb{N}}, then set of limit points of A is \{0\}.

Let A=\{0\}\cup (1,2), then set of limit points of A is [1,2]

Let A=\mathbb{Z}, then set of limit points of A is \emptyset.

Proposition of limit points and closed sets

A set is close if and only if it has limit points.

Theorem: For any subset A of a topological space X, the closure of A is \overline{A}=A\cup A'.

Proof

First we want to prove the theorem implies the proposition,

\Rightarrow

Let A be a close set in X, then \overline{A}=A because A in the intersection of all closed subsets Z\subseteq A in X that contains A.

So Z=A is such a closed subset of X that contains A.

By the theorem, \overline{A}=A\cup A'. Combining this with the fact that A is closed, we have A=A\cup A'.

So A'\subseteq A.

\Leftarrow

Suppose A\subseteq X is a set that includes all its limit points, then A'\subseteq A.Then A'\cup A=A.

By the theorem, \overline{A}=A\cup A'=A.

Since \overline{A} is the smallest closed subset of X that contains A, we have A is closed.

Definition of neighborhood

Let (X,\mathcal{T}) be a topological space. A neighborhood of a point x\in X is an open set U\in \mathcal{T} such that x\in U.

Lemma of intersection of neighborhoods for closure of a set

x\in \overline{A} if and only if any neighborhood of U of x non-trivial intersects A. (A\cap U\neq \emptyset)

Proof of Lemma

\Leftarrow

We proceed by contradiction.

Suppose A\notin \overline{A}, then x\notin \overline{A}.

Then \overline{A}=\bigcap_{A\subseteq Z, Z\text{ is closed}} Z

So, there is A\subseteq Z\subset X and Z is closed.

So this implies that x\in X-Z\coloneq U and U is open since it a complement of a closed set Z.

Since A\subseteq Z, we have A\cap U= \emptyset. (disjoint)

So U and A are disjoint. So U is an open neighborhood of x that is disjoint from A.

This contradicts the assumption that x\in \overline{A}.

\Rightarrow

Let x\in \overline{A}, and we want to show that any neighborhood of U of x non-trivial intersects A. (A\cap U\neq \emptyset)

By contradiction, suppose that there is an open neighborhood of x that is disjoint from A. Then Z\coloneq X-U is closed and A\subseteq Z because U\cap A= \emptyset.

Also x\notin Z.

By the definition of closure, \overline{A}=\subset Z.

Since x\notin Z, we have x\notin \overline{A}.

This contradicts the assumption that x\in \overline{A}.

Proof of theorem

For any subset A of a topological space X, the closure of A is \overline{A}=A\cup A'.

Proof

First we show A\subseteq \overline{A}.

If x\in A', then any open neighborhood U of x has a non-trivial intersection with A by the lemma.

So x\in \overline{A}.

We already know A\subseteq \overline{A}.

Therese two inductions implies A\cup A'\subseteq \overline{A}.

Next we show that \overline{A}\subseteq A\cup A'.

If x\in \overline{A}, then by the lemma, any open neighborhood U of x has a non-trivial intersection with A.

If x\in A, then x\in A\cup A'.

If x\notin A, then the intersection of any open neighborhood U of x with A does not contain x.

This implies that this intersection has to include a point y that is not x.

Since this holds for any open neighborhood U of x, we have x\in A'. (x is a limit point of A)

So x\in A'.

Therese two inductions implies \overline{A}\subseteq A\cup A'.

Tip

Now the three definition of closure are equivalent.

1. The smallest closed subset of X that contains A.
2. A\cup A'.