Thread: Interior points in a metric space

1. Interior points in a metric space

1. Consider the metric space X = (Q ∩ [0; 3]; dE):

(a) the the point 2 is an interior point of the subset A of X where
A = {x ∈ Q | 1 ≤ x ≤ 3}?
True. Since you can make an open ball around 2.

(b) The the point 2 is an interior point of the subset B of X where
B = {x ∈ Q | 2 ≤ x ≤ 3}?
False. Since you can't make an open ball around 2 that is contained in the set.

(c) The point 3 is an interior point of the subset C of X where
C = {x ∈ Q | 2 < x ≤ 3}?
True. Since you can construct a ball around 3, where all the points in the ball is in the metric space.

(d) Describe the possible forms that an open ball can take in X = (Q ∩ [0; 3]; dE).
I don't really get this question, but I think the possible forms are all balls in [0,3] . IS THAT RIGHT???

2. Originally Posted by Dreamer78692
1. Consider the metric space X = (Q ∩ [0; 3]; dE):
Frankly I do not understand your notation.
Rather than guess and guess wrongly, please clarify what this space is.
How is the metric defined? What is dE?

3. dE = Euclidean Metric.. Sorry about that.

(Q ∩ [0; 3]; Euclidean Metric)

4. Originally Posted by Dreamer78692
dE = Euclidean Metric.. Sorry about that.
(Q ∩ [0; 3]; Euclidean Metric)
So the points of the space are rational numbers, $\displaystyle \mathcal{X}=\{r\in\mathbb{Q}:0\le r \le 3\}.$.
If that is right then a ball will looks like $\displaystyle \mathcal{B}(x;\delta)=(x-\delta,x+\delta)\cap\mathcal{X}$ where $\displaystyle \delta\in \mathbb{R}^+$.

Is that how you understand it?

5. yes that is the space.

I was wondering with reference to Q1. is that whether you can make an open ball around 2, since that ball will contain irrational numbers which are not in the set,since we are working in the world of the rationals, or does this even matter.

6. Originally Posted by Dreamer78692
yes that is the space.
I was wondering with reference to Q1. is that whether you can make an open ball around 2, since that ball will contain irrational numbers which are not in the set,since we are working in the world of the rationals, or does this even matter.
Why do you think it would contain any irrational numbers?
$\displaystyle \mathcal{B}(2;0.2)\subset\mathcal{X}\subset\mathbb {Q}$

7. I get it... Made a stupid mistake. Got confused on the set that was being used.