Education: An Open set

Metric space:> An Open Set

Definition

A set is said to be open in a metric space if it equals its interior (

A = Int(A)

). When we encounter topological spaces, we will generalize this definition of open. However, this definition of open in metric spaces is the same as that as if we regard our metric space as a topological space.

Properties:

The empty-set is an open set (by definition: $int(\emptyset)=\emptyset$ ).
An open ball is an open set.
For any set B, int(B) is an open set. This is easy to see because: int(int(B))=int(B).
If A,B are open, then $A\cap B$ is open. Hence finite intersections of open sets are open.
If ${A_i: i \in I}$ (for any set if indexes I) are open, then their union $\cup_{i\in I} A_i$ is open.

Proof of 2:
Let

B_r(x)

be an open ball. Let

y \in B_r(x)

. Then

y \in B_{r-d(x,y)}(y) \subseteq B_r(x)

.
In the following drawing, the green line is

d(x,y)

and the brown line is

r-d(x,y)

. We have found a ball to contain

y

inside

B_r(x)

Proof of 4:
A, B are open. we need to prove that

int(A \cap B) = A\cap B

. Because of the proprieties of int, we only need to show that

int(A \cap B) \supseteq A\cap B

. let

x \in A \cap B

. We know also, that

x \in int(A), x \in int(B)

. That means that there are balls:

B_{{\epsilon}_1}(x) \subset A, B_{{\epsilon}_2}(x) \subset B

. Let

\epsilon = \min\{{\epsilon_1, \epsilon_2}\}

, we have that

B_{\epsilon}(x) \subset A, B_{\epsilon}(x) \subset B \Rightarrow B_{\epsilon}(x) \subset A\cap B

. By the definition of an internal point we have that

x\in A\cap B

(

B_{\epsilon}(x)

is the required ball).

Interestingly, this property does not hold necessarily for an infinite intersection of open sets. To see an example on the real line, let

A_n=\{(-1/n,1/n)\}

. We then see that

\cap^\infty_{i=1}A_i=\{0\}

which is closed.

Proof of 5:
Proving that the union of open sets is open, is rather trivial: let

{A_i: i \in I}

(for any set if indexes I) be a set of open sets. we need to prove that

int(\cup_{i\in I} A_i) \supseteq \cup_{i\in I} A_i

: If

x\in A_i

then it has a ball

B_\epsilon(x) \subset A_i \subseteq \cup_{i\in I} A_i

. The same ball that made a point an internal point in

A_i

will make it internal in

\cup_{i\in I} A_i

Proposition: A set is open, if and only if it is a union of open-balls.
Proof: Let A be an open set. by definition, if

x\in A

there there a ball

B_{\epsilon_x}(x) \subseteq A

. We can then compose A:

A = \cup_{x\in A}B_{\epsilon_x}(x)

. The equality is true because:

\cup_{x\in A}B_{\epsilon_x}(x) \subseteq A

because

\forall x \in A: B_{\epsilon_x}(x) \subseteq A

\cup_{x\in A}B_{\epsilon_x}(x) \supseteq A

in each ball we have the element

x

and we unite balls of all the elements of

A

.
On the other hand, a union of open balls is and open set, because every union of open sets is open.

Examples

As we have seen, every open ball is an open set.
For every space $X$ with the discrete metric, every set is open.

Proof: Let

U

be a set. we need to show, that if

x\in U

then

x

is an internal point. Lets use the ball around

x

with radius

\frac{1}{2}

. We have

B_\frac{1}{2}(x) = \{y\mid d(x,y) < \frac{1}{2}\} = \{x\} \subseteq U

. Therefore

x

is an internal point.

The space $\mathbb{R}$ with the regular metric. Every open segment $(a,b)$ is an open set. The proof of that is similar to the proof that $int([a,b]) = (a,b)$ , that we have already seen.