The set G = {X ∈ R^2 : d(X, 0) ≤ 1} is open in R^2

[G-X]

Prove: The set [Math]G = [/Math] { [Math] X ∈ R^2 : d(X, 0) ≤ 1[/Math] } is open in [Math]R^2[/Math] .(Here [Math]X = (x, y)[/Math], [Math]0 = (0, 0)[/Math])

Let [Math]y = 0, x = \sqrt{1/2}[/Math] by the distance formula of G [Math]\sqrt{((\sqrt{1/2} - 0) - 0)^2 + ((\sqrt{1/2} - 0) - 0)^2} = \sqrt{1} = 1[/Math]
Thus by the restrictions of G [Math](\sqrt{1/2}, 0) ∈ G[/Math]. Let [Math]a = (\sqrt{1/2}, 0)[/Math].

If G is open, by definition, there exists an [Math]ε > 0[/Math] such that [Math]Bε(a) ∈ G[/Math].

Let [Math]ε >0, r > 0[/Math] and [Math]r < ε[/Math] then [Math](\sqrt{1/2} + r, 0) ∈ Bε(a)[/Math]. Let [Math]b = (\sqrt{1/2} + r, 0)[/Math].

By the distance formula of G:
[Math]\sqrt{(\sqrt{1/2} + r)^2 + (\sqrt{1/2} + r)^2} = \sqrt{1 + 2r^2 + 4\sqrt{1/2}r} = *[/Math].

Since r > 0 then * > 1 by the restrictions of [Math]G, b ∉ G, Bε(a) ∉ G[/Math].

This is a contradiction meaning G is not open in [Math]R^2[/Math].

 

[pka]

Prove: The set [Math]G = [/Math] { [Math] X ∈ R^2 : d(X, 0) ≤ 1[/Math] } is open in [Math]R^2[/Math] .(Here [Math]X = (x, y)[/Math], [Math]0 = (0, 0)[/Math])

The set \(G=\{X\in\Re^2: d(X,0)\le 1\}\) is not an open set in \(\Re^2\).
To see that consider the point \(P: (0,1)\). \(P\in G\) but there is no basic open ball centered at \(P\) which is a subset of \(G\).
However the set \(H=\{X\in\Re^2: d(X,O)< 1\}\) is an open set. To see that suppose that \(Q\in H\setminus\{O\}\)
Let \(\delta=\frac{1}{2}\min\{d(Q,O),~1-d(Q,O)\}\) The ball \(B(Q;\delta)\subset H\), showing \(H\) is open.

 

[G-X]

A set [Math]A ⊆ X[/Math] is open if it contains an open ball about each of its points. By definition of an open ball [Math]r > 0[/Math]. We define an [Math]ε > 0[/Math] and subsequently make the radius be between 0 and ε because if we used a point ε away from the center that would be a closed ball, so I denote r. So there must exist a single ball and thus a single r around each point.

If [Math]b∉G[/Math], thus [Math]Bε(a)⊄G[/Math]

Maybe I have some confusion with my justification?

 

[pka]

A set [Math]A ⊆ X[/Math] is open if it contains an open ball about each of its points. By definition of an open ball [Math]r > 0[/Math]. We define an [Math]ε > 0[/Math] and subsequently make the radius be between 0 and ε because if we used a point ε away from the center that would be a closed ball, so I denote r.
If [Math]b∉G[/Math], thus [Math]Bε(a)⊄G[/Math] Maybe I have some confusion with my justification?

What course are you doing where these questions/concepts come up?
You are really confused about the concept of an open set.

 

[G-X]

Do you mind explaining where the confusion is occurring. For me, it appears my reasoning is correct. If a point of a ball is not contained in a set then that ball cannot be contained in the set.

 

[pka]

Do you mind explaining where the confusion is occurring. For me, it appears my reasoning is correct. If a point of a ball is not contained in a set then that ball cannot be contained in the set.

Well do you mind telling us the nature of the course in which these concepts/problems occur. Unless we know what sort of material you know it is impossible to point to your confusion. You do not seem to understand the role basic open sets play in the topology of metric spaces. It possible that you have only been exposed to these topics in a basic Euclidean metric space. If that is the case then that is O.K. It just means that proofs are much more complicated. So tell us the context.

 

[G-X]

You do not seem to understand the role basic open sets play in the topology of metric spaces.

I appreciate the help. But, I do not understand this statement. Clearly I must be in error, so there must be an error somewhere in my proof that one should be able to point out and for I to understand.

 

[G-X]

Updated Proof:

Prove: The set [Math]G = [/Math] { [Math] X ∈ R^2 : d(X, 0) ≤ 1[/Math] } is open in [Math]R^2[/Math] .(Here [Math]X = (x, y)[/Math], [Math]0 = (0, 0)[/Math])

Let [Math]y = 0, x = 1[/Math] by the distance formula of G [Math]\sqrt{(1 - 0)^2 + (0 - 0)^2} = \sqrt{1} = 1[/Math]
Thus by the restrictions of G [Math](1, 0) ∈ G[/Math]. Let [Math]a = (1, 0)[/Math].

If G is open, by definition, there exists an [Math]ε > 0[/Math] such that [Math]Bε(a) ∈ G[/Math].

Let [Math]ε >0, r > 0[/Math] and [Math]r < ε[/Math] then [Math](1 + r, 0) ∈ Bε(a)[/Math]. Let [Math]b = (1 + r, 0)[/Math].

By the distance formula of G:
[Math]\sqrt{((1 + r) - 0)^2 + (0 - 0)^2} = \sqrt{1 + r^2 + 2r} = *[/Math].

Since r > 0 then * > 1 by the restrictions of [Math]G, b ∉ G, Bε(a) ∉ G[/Math].

This is a contradiction meaning G is not open in [Math]R^2[/Math].

 

[G-X]

Updated Proof Again:

Prove: The set [Math]G = [/Math] { [Math] X ∈ R^2 : d(X, 0) ≤ 1[/Math] } is open in [Math]R^2[/Math] .(Here [Math]X = (x, y)[/Math], [Math]0 = (0, 0)[/Math])
We will attempt to prove G is not open in [Math]R^2[/Math] by contradiction

Let [Math]x = 1, y = 0[/Math] by the restrictions imposed on G:
[Math]d(X, 0) = \sqrt{(1 - 0)^2 + (0 - 0)^2} = \sqrt{1} = 1[/Math]
Thus [Math](1, 0) ∈ G[/Math]. Let [Math]a = (1, 0)[/Math].

Let us assume G is open, by definition, there exists an [Math]ε > 0[/Math] such that [Math]Bε(a) ∈ G[/Math].

Let [Math]ε >0, r > 0[/Math] and [Math]r < ε[/Math] then [Math](1 + r, 0) ∈ Bε(a)[/Math]. Let [Math]b = (1 + r, 0)[/Math].

Again by the restrictions imposed on G:
[Math]d(b, 0) = \sqrt{((1 + r) - 0)^2 + (0 - 0)^2} = \sqrt{1 + r^2 + 2r} = *[/Math].

Since r > 0 then * > 1 by the restrictions of [Math]G, b ∉ G, Bε(a) ∉ G[/Math].

This is a contradiction meaning G is not open in [Math]R^2[/Math].

 

[G-X]

Minor Update:

Prove: The set [Math]G = [/Math] { [Math] X ∈ R^2 : d(X, 0) ≤ 1[/Math] } is open in [Math]R^2[/Math] .(Here [Math]X = (x, y)[/Math], [Math]0 = (0, 0)[/Math])
We will attempt to prove G is not open in [Math]R^2[/Math] by contradiction

Let [Math]x = 1, y = 0[/Math] by the restrictions imposed on G:
[Math]d(X, 0) = \sqrt{(1 - 0)^2 + (0 - 0)^2} = \sqrt{1} = 1[/Math]
Thus [Math](1, 0) ∈ G[/Math]. Let [Math]a = (1, 0)[/Math].

Let us assume G is open, by definition, there exists an [Math]ε > 0[/Math] such that [Math]Bε(a) ∈ G[/Math].

Let [Math]ε >0, r > 0[/Math] and [Math]r < ε[/Math] then [Math](1 + r, 0) ∈ Bε(a)[/Math]. Let [Math]b = (1 + r, 0)[/Math].

Again by the restrictions imposed on G:
[Math]d(b, 0) = \sqrt{((1 + r) - 0)^2 + (0 - 0)^2} = \sqrt{(1 + r)^2} = 1 + r[/Math].

Since r > 0 then d(b, 0) > 1 by the restrictions of [Math]G, b ∉ G, Bε(a) ∉ G[/Math].

This is a contradiction meaning G is not open in [Math]R^2[/Math].

 

[pka]

Prove: The set [Math]G = [/Math] { [Math] X ∈ R^2 : d(X, 0) ≤ 1[/Math] } is open in [Math]R^2[/Math] .(Here [Math]X = (x, y)[/Math], [Math]0 = (0, 0)[/Math])FALSE
We will attempt to prove G is not open in [Math]R^2[/Math] by contradiction

Let \(P: \left(\dfrac{\sqrt2}{2},\dfrac{\sqrt2}{2}\right)\) Note that \(d(P,0)=1,~~\therefore P\in G\).
The ball \(\mathcal{B}_{\delta}(P)\) centered at \(P\) with radius \(\delta>0\) contains points not in \(G\).
That is impossible if \(G\) is an open set. Maybe that you do not have a grasps of open sets.
A set is open if for every point \(p\) in the set any point close to \(p\) is also in the set.
If \(G\) is a open set and \(p\in G\) then \(\exists c>0\) such \(\mathcal{B}_{c}(p)\subset G\).

 

[G-X]

Sorry, I mentioned that I am proving the opposite through contradiction. G is in fact not open as it has boundary points on 1 - How I see it.