Closed sets

The opposite of an open set, as one may have guessed, is a closed set. However these properties are not mutually exclusive.

Definition

Definition

A set S \subseteq \mathbb{R} is closed if its compliment S^C = \{p \in \mathbb{R} \mid p \not \in S\} is open.

Lets use our same example from the last page.

Let A = (0,1) and B = [0,1]. We have that A^C = (-\infty,0] and {B^C = (-\infty,0) \cup (1,\infty)}.

It is clear to see that A is not a closed set, as it’s compliment A^C is not open.

We do see however, that B is a closed set. This is because B^C is the union of two open sets, which is of course open by Theorem 2.1.

We can now prove some important sets are closed. Do note that both \mathbb{R} and \varnothing are both closed and open as these properties are not mutually exclusive. Sets that are both open and closed are sometimes called clopen.

Theorem

The set of real numbers \mathbb{R} is closed.

The compliment of \mathbb{R}, \{p \in \mathbb{R} \mid p \not \in \mathbb{R}\} is the empty set \varnothing which is open by Theorem [num]. Thus \mathbb{R} is closed.

Theorem 1.2

The empty set \varnothing is closed.

The compliment of \varnothing, \{p \in \mathbb{R} \mid p \not \in \varnothing\} is the set of all real numbers \mathbb{R} which is open by Theorem [num]. Thus \varnothing is closed.

These sets are the only sets in \mathbb{R} that are both open and closed, a fact we will later prove.

Theorem 1.3

Let a,b \in \mathbb{R}. Then [a,b] is closed.

The compliment of [a,b] is (-\infty,a) \cup (b,\infty). Both of these sets are open, and the union of two open intervals is open, so [a,b] is closed.

Union and Intersection of Closed Sets

Similar to open sets, we can preserve the property of being closed with unions and intersections. The main two properties are

  • Any finite union of closed sets is closed

  • Any arbitrary intersection of closed sets is closed.

Notice the switch in the terms ‘arbitrary’ and ‘finite’ as compared to the similar properties of closed sets.

Theorem 2.1

Let \{A_i \subset \mathbb{R} \mid i \in I\} be closed sets for any finite index set I. Then \bigcup_{i \in I} A_i is open.

Since \{A_i \subset \mathbb{R} \mid i \in I\} are closed, each A_i^C is open. We use De Morgan’s laws to state

\left(\bigcup_{i\in I} A_i\right)^{C} = \bigcap_{i\in I}A^C_i.

The intersection of finite many open sets is open by Theorem [num], thus \cup_{i\in I} A_i is closed.

Note 1: Theorem 2.2

Let \{A_i \subset \mathbb{R} \mid i \in I\} be open sets for a finite index set I. Then \bigcap_{i \in I} A_i is open.

Since \{A_i \subset \mathbb{R} \mid i \in I\} are closed, each A_i^C is open. Similar to the last proof, we use De Morgan’s laws to state

\left(\bigcup_{i\in I} A_i\right)^{C} = \bigcap_{i\in I}A^C_i.

The union of arbitrarily many closed sets may not be closed, as shown below.

Example

To see an example of why an infinite intersection of open sets may not be open, let take p to be any singleton in (a,b) (A singleton being a set containing only one element).

The set p is closed, as it’s compliment consist of (-\infty,p) \cup (p,\infty). But the union of all \bigcup p is (a,b), which is not closed as it’s compliment is not open.

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