Open Set Boundary: Proving S as the Half Plane with y = -x Line Boundary

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In summary, the conversation discusses finding a proof for showing that the set S = {(x, y) ∈ ℝ^2 : x > y} is open, with the boundary being the line y = x. The attempt at a solution involves finding a radius r, but there seems to be a typo in the equation provided. The correct answer is given as r = (x1 - y1)/√2.
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Homework Statement



Show that the set of [tex]\mathbb{R}^2[/tex] given by [tex]S = \{(x, y) \in \mathbb{R}^2 : x > y\}[/tex] is open.

Homework Equations


The Attempt at a Solution



Why is S the half plane that has boundary given by the line [tex]y = -x?[/tex]
 
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  • #2
tuggler said:

Homework Statement



Show that the set of [tex]\mathbb{R}^2[/tex] given by [tex]S = \{(x, y) \in \mathbb{R}^2 : x > y\}[/tex] is open.

Homework Equations





The Attempt at a Solution



Why is S the half plane that has boundary given by the line [tex]y = -x?[/tex]
It's not - the boundary is the line y = x. I suspect a typo.
 
  • #3
Mark44 said:
It's not - the boundary is the line y = x. I suspect a typo.

This is what I also thought when I read the problem.

S is open if and only if S=S°.
 
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  • #4
Thanks for the clarification. I am still having trouble with this problem.

I know I must determine a radius r such that [tex]\{(x_1, y_2)\in \mathbb{R}^2: \sqrt{|x-y|^2 +|x_1-y_1|^2}<r\}[/tex]

but how can I find such an r?

What I did was:

I located the region of the plane where [tex]x\gt y[/tex] then I place a point [tex]P=(x_1,y_1)[/tex] in it where it is fairly close to the line [tex]y=x[/tex].

Now I moved to the left from the point P until I hit the line y=x, which I marked Q then this Q has coordinates (y_1,y_1).

But I can't find an r. The answer given is [tex]\frac{x_1 - y_1}{\sqrt{2}}.[/tex]

Which I don't know how they got?
 
  • #5
tuggler said:
Thanks for the clarification. I am still having trouble with this problem.

I know I must determine a radius r such that [tex]\{(x_1, y_2)\in \mathbb{R}^2: \sqrt{|x-y|^2 +|x_1-y_1|^2}<r\}[/tex]
That equation doesn't look right to me. Why is it (x1, y2), and why doesn't y2 show up on the right?
tuggler said:
but how can I find such an r?

What I did was:

I located the region of the plane where [tex]x\gt y[/tex] then I place a point [tex]P=(x_1,y_1)[/tex] in it where it is fairly close to the line [tex]y=x[/tex].

Now I moved to the left from the point P until I hit the line y=x, which I marked Q then this Q has coordinates (y_1,y_1).

But I can't find an r. The answer given is [tex]\frac{x_1 - y_1}{\sqrt{2}}.[/tex]

Which I don't know how they got?
 

1. What is an open set boundary?

An open set boundary is a mathematical concept used in topology to describe the boundary of an open set. It is the set of all points that are not contained within the open set but are in close proximity to it. In other words, it is the set of points that are neither inside nor outside the open set.

2. How is an open set boundary different from a closed set boundary?

An open set boundary differs from a closed set boundary in that it does not include the points on the boundary itself. In contrast, a closed set boundary includes the points on the boundary as well as the points inside the set.

3. What is the purpose of an open set boundary in topology?

The open set boundary is used to define the limits of an open set and to distinguish it from other types of sets, such as closed sets. It is also used in the definition of continuity in mathematical functions.

4. Can an open set have a boundary?

Yes, an open set can have a boundary. In fact, all open sets have a boundary, but it is important to note that the boundary is not considered part of the open set itself.

5. How is the open set boundary related to the interior and exterior of a set?

The open set boundary is closely related to the interior and exterior of a set. The interior of a set is the largest open set contained within it, while the exterior is the largest open set containing the set. The open set boundary separates the interior and exterior of a set.

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