Proving Openness of Subset of S in R2

[frankli]

Homework Statement

Using the definition of an open set, prove that the subset of S ={(x, y)∈ R² | 0 <x< 1， 1< y<2} the Euclidean space R² is open.

The Attempt at a Solution

The definition I learned is a set S \subset Rⁿ is said to be open in Rⁿ if for all x\in S, \exists r greater than 0, such that every point y \in Rⁿ satisfying ||x-y|| < r also belongs to S.
So, what I think is choose r =x for all x ∈ S, there exists r greater than 0, such that if |x1 - x | < r then B(r,x)\subset S, S is open.

but I think it is too simple, I really have no idea how to prove a set open with all the inequalities and functions going on..

 

[lanedance]

I think you need to show that r exists, as you defined, for any point in your set...

so say you take
\textbf{s} = (s,t) \in S = \left\{ (x, y) \in \mathbb{R}^2 | 0 &lt; x &lt; 1, 1&lt; y&lt;2 \right\}

now find r, such that for any p = (p,q) with |p-s| < r, then p is also in S

if you can demonstrate this is true for any arbitrary point s of S, then you have shown it is true for every point in S, and so S is open

 

[anlys]

I don't think what lanedance said about the part where if |p-s| < r, then p is also in S, is true. That's what you're trying to prove. So, you can't assume that. Well, it's true that to show the set S is open, you need to find r such that the open ball about s = (s,t) with radius r is totally contained in S.
One instance of r that I have in mind right now is by letting r = min{s, 1-s, t-1, 2-t} >0. Then, once you've define your r, you need to prove that the open ball is contained in S. To show this, let p = (p1,p2) be an arbitrary point in B(s, r). So, this is implies that d(p, s) < r . Now, with that information, you need to prove that p also belongs to the set S. To do this, you've got to show that 0<p1<1 and 1<p2<2. Once you show this, it implies that p belongs to S by definition. Therefore, the open ball B(s, r) is contained in S.

 

[lanedance]

I don't think what lanedance said about the part where if |p-s| < r, then p is also in S, is true.

yes it is, that is the definition of open in terms of open balls, which seems to be exactly what you have done below

That's what you're trying to prove. So, you can't assume that. Well, it's true that to show the set S is open, you need to find r such that the open ball about s = (s,t) with radius r is totally contained in S.
One instance of r that I have in mind right now is by letting r = min{s, 1-s, t-1, 2-t} >0. Then, once you've define your r, you need to prove that the open ball is contained in S. To show this, let p = (p1,p2) be an arbitrary point in B(s, r). So, this is implies that d(p, s) < r . Now, with that information, you need to prove that p also belongs to the set S. To do this, you've got to show that 0<p1<1 and 1<p2<2. Once you show this, it implies that p belongs to S by definition. Therefore, the open ball B(s, r) is contained in S.