Education: Metric space: Interior Point

16/10/2014

Metric space: Interior Point

METRIC SPACE: Interior Point:

Definitions

Definition: We say that x is an interior point of A iff there is an

\epsilon > 0

such that:

B_\epsilon(x) \subseteq A

. This intuitively means, that x is really 'inside' A - because it is contained in a ball inside A - it is not near the boundary of A.

Illustration:

Interior Point	Not Interior Points

Definition: The interior of a set A is the set of all the interior points of A. The interior of a set A is marked

int(A)

. Useful notations:

\operatorname{int}(S)=\{x\in S|x\text{ is an interior point of }S\}

and

\operatorname{int}(S)=\cup \{A\subseteq S|A\text{ is open }\!\!\}\!\!\text{ }

.

Properties

Some basic properties of int (For any sets A,B):

$int(A) \subseteq A$
$int(int(A)) = int(A)\,$
$int(A \cap B) = int(A) \cap int(B)$
$A \subseteq B \Rightarrow int(A) \subseteq int(B)$
$A\text{ is open}\Leftrightarrow A=\operatorname{int}(A)$

Proof of the first:
We need to show that:

x\in int(A) \Rightarrow x \in A

. But that's easy! by definition, we have that

x\in B_\epsilon(x)\subseteq A

and therefore

x\in A

Proof of the second:
In order to show that

int(int(A)) = int(A)\,

, we need to show that

int(int(A)) \subseteq int(A)

and

int(int(A)) \supseteq int(A)

.
The "

\subseteq

" direction is already proved: if for any set A,

int(A) \subseteq A

, then by taking

int(A)

as the set in question, we get

int(int(A)) \subseteq int(A)

.
The "

\supseteq

" direction:
let

x \in int(A)

. We need to show that

x \in int(int(A))

.
If

x\in int(A)

then there is a ball

B_\epsilon(x) \subseteq A

. Now, every point y, in the ball

B_\frac{\epsilon}{2}(x)

an internal point to A (inside

int(A)

), because there is a ball around it, inside A:

y \in B_\frac{\epsilon}{2}(x) \Rightarrow B_\frac{\epsilon}{2}(y) \subset B_\epsilon(x) \subset A

.
We have that

x\in B_\frac{\epsilon}{2}(x) \subset int(A)

(because every point in it is inside

int(A)

) and by definition

x \in int(int(A))

.
Hint: To understand better, draw to yourself

x, B_\epsilon(x), B_\frac{\epsilon}{2}(x), y, B_\frac{\epsilon}{2}(y)

.

Proof of the rest is left to the reader.

Reminder

[a, b] : all the points x, such that $a \leq x \leq b$
(a, b) : all the points x, such that $a < x < b$

Example

For the metric space

\mathbb{R}

(the line), we have:

$int([a,b]) = (a,b)$
$int((a,b]) = (a,b)$
$int([a,b)) = (a,b)$
$int((a,b)) = (a,b)$

Let's prove the first example (

int([a,b]) = (a,b)

). Let

x\in (a,b)

(that is:

a < x < b

) we'll show that

x

is an internal point.
Let

\epsilon = \min\{x - a , b - x \}

. Note that

x + \epsilon \leq x + b - x = b

and

x - \epsilon \geq x - x + a = a

. Therefore

B_\epsilon(x) = (x - \epsilon, x + \epsilon) \subset (a,b)

.
We have shown now that every point x in

(a,b)

is an internal point. Now what about the points

a,b

? let's show that they are not internal points. If

a

was an internal point of

[a,b]

, there would be a ball

B_\epsilon(a) \subset [a,b]

. But that would mean, that the point

a - \frac{\epsilon}{2}

is inside

[a, b]

. but because

a - \frac{\epsilon}{2} < a

that is a contradiction. We show similarly that b is not an internal point.
To conclude, the set

(a,b)

contains all the internal points of

[a,b]

. And we can mark

int([a,b]) = (a,b)