# Open Set Proofs

1. Sep 24, 2009

### frankli

1. The problem statement, all variables and given/known data

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.

3. The attempt at a solution

The definition I learnt 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..

Last edited: Sep 24, 2009
2. Sep 25, 2009

### lanedance

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

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

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

Last edited: Sep 25, 2009
3. Sep 26, 2009

### 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.

4. Sep 26, 2009

### lanedance

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

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