General topology: Prove a Set is Open

[lep11]

Homework Statement

Let A:={x∈ℝ² : 1<x²+y²<2}. Is A open, closed or neither? Prove.

Homework Equations

triangle inequality d(x,y)≤d(x,z)+d(z,y)

The Attempt at a Solution

First I draw a picture with Wolfram Alpha. My intuition is that the set is open.

Let (a,b)∈A arbitrarily and r=min{√(a²+b²)-1, √2 -√(a²+b²)} (by geometry). It is clear that then B_r(a,b)⊂A but it needs to be formally proven. Let z∈B_r(a,b) and now I need to show that ||x-z||=d(x,z)<r.
d(x,z)≤...I am suck here.

And the next thing would be to prove the set is not closed. Its complement A^c={x∈ℝ² : x²+y²≤1}∪{x∈ℝ² : x²+y²≥√2} is clearly closed, though.

 

[andrewkirk]

To prove the set is open you need to show it can be constructed from given open sets using the allowed operations, which are arbitrary unions and finite intersections.

What have you been given as the original set of open sets for the topology of $\mathbb R^2$ (known as the 'basis')?

There are two common bases used for $\mathbb R^2$.. One is the set of all open boxes of form $(a,b)\times(c,d)$ for different combinations $a,b,c,d$ and the other is the set of all open balls of form $\{(x,y)\ :\ \|(x,y)-(x_0,y_0)\|<r\}$ for different combinations of $x_0,y_0,r$. The approach to proving openness will depend which of these bases you are using (or other basis, if you have been given that).

 

[lep11]

To prove the set is open you need to show it can be constructed from given open sets using the allowed operations, which are arbitrary unions and finite intersections.

What have you been given as the original set of open sets for the topology of $\mathbb R^2$ (known as the 'basis')?

There are two common bases used for $\mathbb R^2$.. One is the set of all open boxes of form $(a,b)\times(c,d)$ for different combinations $a,b,c,d$ and the other is the set of all open balls of form $\{(x,y)\ :\ \|(x,y)-(x_0,y_0)\|<r\}$ for different combinations of $x_0,y_0,r$. The approach to proving openness will depend which of these bases you are using (or other basis, if you have been given that).

I am not familiar with that approach. We are using open balls.

 

[S.G. Janssens]

Edited, no longer interested in contributing to this thread.

 

[andrewkirk]

It's easy using balls. Your set is the intersection of two open sets. One is an open ball and the other is the complement of a closed ball.

 

[lep11]

It's easy using balls. Your set is the intersection of two open sets. One is an open ball and the other is the complement of a closed ball.

May be easy but how to prove it formally?

 

[PeroK]

May be easy but how to prove it formally?

If you can assume that an open ball is open and that the intersection of two open sets is open, then the proof should be easy. But, showing that an open ball is open would seem to me to be not much different from showing that the set you have is open.

So, perhaps your original approach is the one that is expected: show that each point in A has a neighbourhood in A. If you continue with this approach, you might consider how far points are from the origin.

 

[lep11]

If you can assume that an open ball is open and that the intersection of two open sets is open, then the proof should be easy. But, showing that an open ball is open would seem to me to be not much different from showing that the set you have is open.

Actually we have proven in class that open ball is open set and that a finite intersection of open sets is open.

Now A=B((0,0),r₁)∩B((0,0),r₂)={y∈ℝ²: d(y,(0,0))≤1)}^c∩{z∈ℝ²: d(z,(0,0))<√2}

{y∈ℝ²: d(y,(0,0))≤1)} is closed ball by definition and therefore its complement {y∈ℝ²: d(y,(0,0))≤1)}^c is open? And intersection of two open sets is open so A is open.

So, perhaps your original approach is the one that is expected: show that each point in A has a neighbourhood in A. If you continue with this approach, you might consider how far points are from the origin.

I think this approach is more elementary.

 

[andrewkirk]

Actually we have proven in class that open ball is open set

That is not consistent with your response in post 3 to my question about what basis (or sometimes written as 'base') you have been given. If you are given the open balls as a basis then there it is redundant to prove the openness of an open ball. It is open by definition of the topology.

Also, it is redundant to prove that a finite intersection of open sets is open. Such an intersection is open by definition.

Any topology must be given either by specifying a set of open sets - which are simply specified, not proven - or it is constructed from other topologies, as in product or box topologies. I'm guessing you haven't dealt with the latter yet, so you must have been given a specified collection of open sets to start with. What is that collection?

 

[lep11]

That is not consistent with your response in post 3 to my question about what basis (or sometimes written as 'base') you have been given. If you are given the open balls as a basis then there it is redundant to prove the openness of an open ball. It is open by definition of the topology.

Also, it is redundant to prove that a finite intersection of open sets is open. Such an intersection is open by definition.

Any topology must be given either by specifying a set of open sets - which are simply specified, not proven - or it is constructed from other topologies, as in product or box topologies. I'm guessing you haven't dealt with the latter yet, so you must have been given a specified collection of open sets to start with. What is that collection?

Now I am even more confused.

 

[micromass]

That is not consistent with your response in post 3 to my question about what basis (or sometimes written as 'base') you have been given. If you are given the open balls as a basis then there it is redundant to prove the openness of an open ball. It is open by definition of the topology.

Also, it is redundant to prove that a finite intersection of open sets is open. Such an intersection is open by definition.

Any topology must be given either by specifying a set of open sets - which are simply specified, not proven - or it is constructed from other topologies, as in product or box topologies. I'm guessing you haven't dealt with the latter yet, so you must have been given a specified collection of open sets to start with. What is that collection?

andrew, the OP is likely only acquainted with metric spaces and not with any topology. So the word topology or bases are likely foreign to him.

 

[lep11]

so, how to proceed?

 

[PeroK]

so, how to proceed?

You need to start the way you started in post #1:

Homework Equations

triangle inequality d(x,y)≤d(x,z)+d(z,y)

The Attempt at a Solution

Let (a,b)∈A arbitrarily and r=min{√(a²+b²)-1, √2 -√(a²+b²)} (by geometry). It is clear that then B_r(a,b)⊂A but it needs to be formally proven. Let z∈B_r(a,b)

And now, think about the distance of $z$ from the origin. Hint: use your relevant equation!

 

[lep11]

You need to start the way you started in post #1:
And now, think about the distance of $z$ from the origin. Hint: use your relevant equation!

x=(x₁, x₂) ∈A
and z=(z₁,z₂)∈B(x,r)

d(x,z)=√((x_1²-z_1²)+(x_2²-z_2²))<r.

d(z,(0,0))=√(z_1²+z_2²)≤√((x_1²-z_1²)+(x_2²-z_2²)) + √(x_1²+x_2²)<r+√(x_1²+x_2²)<...?

(triangle inequality)

 

[PeroK]

d(x,z)=√((x_1²-z_1²)+(x_2²-z_2²))<r.

d(z,(0,0))=√(z_1²+z_2²)≤√((x_1²-z_1²)+(x_2²-z_2²)) + √(x_1²+x_2²)<r+√(x_1²+x_2²)

(triangle inequality)

It might be simpler to work the metric notation throughout. You could even rewrite the definition of $A$:

$A = \{x: 1 < d(0, x) < \sqrt{2} \}$

This is good, because it means you are using the properties of any metric, not just the usual metric for $\mathbb{R}^2$

And you have $r = min\{d(0, x) - 1, \sqrt{2} - d(0, x) \}$

Now, you had: Let $a \in A$ ...

Can you pick it up from there?

 

[lep11]

It might be simpler to work the metric notation throughout. You could even rewrite the definition of $A$:

$A = \{x: 1 < d(0, x) < \sqrt{2} \}$

This is good, because it means you are using the properties of any metric, not just the usual metric for $\mathbb{R}^2$

And you have $r = min\{d(0, x) - 1, \sqrt{2} - d(0, x) \}$

Now, you had: Let $a \in A$ ...

Can you pick it up from there?

d(a,z)<r.

d(0,z)≤d(0,a)+d(a,z)<r + d(a,0)

i am confused and topology is giving me brain cancer soon

 

[PeroK]

d(a,z)<r.

d(0,z)≤d(0,a)+d(a,z)<r + d(a,0)

Well, okay, but what next?

Also, you need to pay attention. My last post wasn't exactly perfect, so take a look and fix it!

 

[lep11]

Well, okay, but what next?

d(a,z)<r.

d(0,z)≤d(0,a)+d(a,z)<r + d(a,0)<√2?

 

[PeroK]

d(a,z)<r.

d(0,z)≤d(0,a)+d(a,z)<r + d(a,0)<√2?

Why is that?

 

[lep11]

Why is that?

That's what I need to show next somehow

 

[PeroK]

That's what I need to show next somehow

Why did you define $r$ the way you did?

A metric is a distance function, so you should be able to think these things through in terms of "the distance to $z$" etc.

That may or may not help. You should also draw these points on a diagram to help you see why $z$ is in $A$.

 

[lep11]

Why did you define $r$ the way you did?

A metric is a distance function, so you should be able to think these things through in terms of "the distance to $z$" etc.

That may or may not help. You should also draw these points on a diagram to help you see why $z$ is in $A$.

By geometry? I need to show d(0,z)>1 as well. Clearly z is in A.

 

[PeroK]

By geometry? I need to show d(0,z)>1 as well. Clearly z is in A.

If you look at the definition of $r$ you'll see that:

$r \le d(0, a) - 1$

$r \le \sqrt{2} - d(0, a)$

 

[lep11]

If you look at the definition of $r$ you'll see that:

$r \le d(0, a) - 1$

$r \le \sqrt{2} - d(0, a)$

$...<r+d(0,a)≤d(0,a)+ \sqrt{2} - d(0, a)=\sqrt{2}$ now I am getting somewhere

how about d(0,z)>1?

 

[lep11]

but how to show d(0,z)>1?

 

[PeroK]

but how to show d(0,z)>1?

Use the other inequality in post #23.

 

[lep11]

Use the other inequality in post #23.

Yes, I have been trying to plug it in d(0,z)≤d(0,a)+d(a,z) with no success.

 

[PeroK]

Yes, I have been trying to plug it in d(0,z)≤d(0,a)+d(a,z) with no success.

You need an inequality for $d(0, z)$ greater than something.

 

[lep11]

You need an inequality for $d(0, z)$ greater than something.

Sure, but unfortunately I still can't figure out how to prove d(0,z)>1.

d(0,z)...>...>1?

 

[PeroK]

Sure, but unfortunately I still can't figure out how to prove d(0,z)>1.

d(0,z)...>...>1?

It's just the triangle inequality again:

$d(0, a) \le d(0, z) + d(z, a)$

 

[lep11]

It's just the triangle inequality again:

$d(0, a) \le d(0, z) + d(z, a)$

$d(0, a) \le d(0, z) + d(z, a)$ ⇔ $d(0, z) ≥ d(0, a)-d(z, a)$

 

[PeroK]

$d(0, a) \le d(0, z) + d(z, a)$ ⇔ $d(0, z) ≥ d(0, a)-d(z, a)$

That's true,but what about using the inequalities you know for $d(0, a)$ and $d(z, a)$?

 

[lep11]

$d(0, z) ≥ d(0, a)-d(z,a) >d(0,a)>1$

 

[PeroK]

$d(0, z) ≥ d(0, a)-d(z,a) > d(0,a)>1$

That middle equality cannot be correct. $d(0, a)-d(z,a) \le d(0,a)$ surely?

 

[lep11]

That middle equality cannot be correct. $d(0, a)-d(z,a) \le d(0,a)$ surely?

What's wrong?