NoteNextra-origin/content/Math4201/Math4201_L7.md

Math 4201 Topology I (Lecture 7)

Review from last lecture

Not every open set in subspace topology is open in the original space

Let X=\mathbb{R} with standard topology and Y=[0,1]\cup [2,3]. equipped with subspace topology generated by the standard basis for \mathbb{R}.

so [0,1]=(-1,\frac{3}{2})\cap Y In particular, [0,1] is open set in Y, but not an open set in \mathbb{R}.

New materials

Closed sets in topological space

Proposition of open set in subspace topology

If X is a topological space, then Y\subseteq X is open and is with the subspace topology. If Z\subset Y is open subspace of Y, then Z is also an open subspace of X.

Proof

Since Z\subset Y is open in the subspace topology, there is an open U\subset Y such that Z=U\cap Y.

SInce Z is the intersection of open sets in X, then Z is open in X.

Definition of closed set

For any topology \mathcal{T} on a set X, a subset Z\subseteq X is said to be closed if its complement Z^c is an open set (in X).

Note the complement is defined Z=X\setminus Z.

Example of closed set in standard topology of real numbers

For example, [a,b] is a closed set in the standard topology of real numbers. since \mathbb{R}-[a,b]=(-\infty,a)\cup (b,\infty) is an open set.

Example of closed set in trivial topology

For any set X, the trivial topology is \mathcal{T}_0=\{\emptyset, X\}. Since X^c=\emptyset is an open set, X is a closed set. Since \emptyset^c=X is an open set, \emptyset is a closed set.

Example of closed set in finite complement topology

For any set X, the finite complement topology is \mathcal{T}_1=\{U\subseteq X\mid X\setminus U\text{ is finite}\}.

Then the set of all finite subsets of X is a closed set.

For general, if \mathcal{T} is a topology on X, then:

1. \emptyset, X are closed sets.
2. \mathcal{T} is closed with respect to arbitrary unions.

Let \{Z_\alpha\}_{\alpha \in I} be an arbitrary collection of closed sets in X. Then X-Z_\alpha is open for each \alpha \in I. So \forall \alpha \in I. \bigcup_{\alpha \in I} (X-Z_\alpha)=X-\bigcap_{\alpha \in I} Z_\alpha is open. So \bigcap_{\alpha \in I} Z_\alpha is closed.

So the corollary is: an arbitrary intersection of closed sets is closed.

3. \mathcal{T} is closed with respect to finite intersections. This also implies that a finite union of closed sets is closed.

If \{Z_1, Z_2, \ldots, Z_n\} is a closed subset of X, then X-Z_i is open for each i=1,2,\ldots,n. So \forall i=1,2,\ldots,n. \bigcap_{i=1}^n (X-Z_i)=X-\bigcup_{i=1}^n Z_i is open. So \bigcup_{i=1}^n Z_i is closed.

Note

We can also define the topology using the closed sets instead of the open sets.

1. \emptyset, X are closed sets.
2. \mathcal{T} is closed with respect to arbitrary intersections.
3. \mathcal{T} is closed with respect to finite unions.

This yields the same topology.

Theorem of closed set in subspace topology

Let X is a topological space and Y\subseteq X equipped with the subspace topology.

A subset Z\subseteq Y is closed in Y if and only if Z is the intersection of a closed W\subseteq X and Y. That is Z=W\cap Y.

Proof

\Rightarrow

If Z is closed in Y, then Y-Z is open in Y.

Then, there is open set U\subseteq X such that Y-Z=U\cap Y.

So Z=(X-U)\cap Y, X-U is closed in X because U is open in X.

Take W=X-U.

\Leftarrow

If Z=W\cap Y for some closed W\subseteq X, then Y-Z=Y-(W\cap Y)=(Y-W)\cap Y is open in Y. So Z is closed in Y.

Lemma of closed in closed subspace

Let X is a topological space and Y\subseteq X is closed and is equipped with the subspace topology. Then any closed subset of Y is also closed in X.

Warning

Not any subset of a topological space X is either open or closed.

Example of open and closed subset

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

(a,b) is open, but not closed. [a,b] is closed, but not open. [a,b) is neither open nor closed. \emptyset,\mathbb{R} is both open and closed.

Example of open and closed subset in other topologies

Let X=[0,1]\cup (2,3) induced by the standard topology of \mathbb{R}.

Z=[0,1] is an open subset of X.

Z=[0,1] is also closed subset of X since Z=[0,1]\cap X is open in \mathbb{R}.

We can associate an open and a closed to any subset A of a topological space X.

Interior and closure of a set

The interior of a set A is defined as follows:

\operatorname{Int}(A)=\bigcup_{U\subseteq A, U\text{ is open in }X} U

Also denoted as A^\circ.

The interior of a set A is the largest open subset of A.

That is \forall U\subseteq A, U\text{ is open in }X, then U\subseteq \operatorname{Int}(A). (by definition that U must be in collection of open sets that is a subset of A)

Closure of a set

The closure of a set A is the smallest closed subset of X that contains A.

Note that if we change the definition as the intersection of all closed subsets of X that contained in $A$, we will get the empty set.

\overline{A}=\bigcap_{A\subseteq C, C\text{ is closed in }X} C

The closure of a set A is the smallest closed subset of X that contains A. (follows the same logic as the previous definition)