True or False: Boundary of Sets in R2

[Justhanging]

Homework Statement

True or false:
Let S be any set in R2. The boundary of S is the set of points contained in S which are not in the interior of S.

Homework Equations

The Attempt at a Solution

Common sense tells me true. I don't really understand it though, if S is an open set than it doesn't have a boundary? Maybe if set S contains only one point than its boundary is also its interior. Can someone clarify and give relevant theroms associated with this.

 

[Dick]

I don't think it's true. Open sets can have a boundary. What's the definition of 'boundary'?

 

[Justhanging]

I don't think it's true. Open sets can have a boundary. What's the definition of 'boundary'?

Am not exactly sure, but I seem to remember that interior points have a disk where all the area around the disk are within the set but boundary points have disk where some of the area of the disk is not within the set. The disk being so arbitrary disk with a very small radius. By this definition it seems true as you can only be either a boundary or an interior point.

 

[Dick]

Am not exactly sure, but I seem to remember that interior points have a disk where all the area around the disk are within the set but boundary points have disk where some of the area of the disk is not within the set. The disk being so arbitrary disk with a very small radius. By this definition it seems true as you can only be either a boundary or an interior point.

Trying to answer a question about 'boundary' without knowing the exact definition is probably not optimal. That's vaguely correct. If (0,1) is an open interval in R, is 1 a boundary point? I'd encourage to look up the definition so you can convince me one way or another.

 

[Justhanging]

Trying to answer a question about 'boundary' without knowing the exact definition is probably not optimal. That's vaguely correct. If (0,1) is an open interval in R, is 1 a boundary point? I'd encourage to look up the definition so you can convince me one way or another.

All the definitions I see are the same. The neighborhood of the point needs to have a point in the set and a point outside the set for it to be a boundary. For your quesition I would say 1 is a boundary because it has points outside of the set in its neighborhood. When you say an open set, are you including 0 and 1 in the set?

Are the boundary of points of S not included in S? Like 1 in your set, its a boundary point but not included in your set. Am still confused and the definition in my book doesn't really help solving the problem.

 

[Dick]

All the definitions I see are the same. The neighborhood of the point needs to have a point in the set and a point outside the set for it to be a boundary. For your quesition I would say 1 is a boundary because it has points outside of the set in its neighborhood. When you say an open set, are you including 0 and 1 in the set?

Are the boundary of points of S not included in S? Like 1 in your set, its a boundary point but not included in your set. Am still confused and the definition in my book doesn't really help solving the problem.

(0,1) means the set of all points x such that 0<x<1. Yes, it is open. 1 and 0 are not included. If they were, it wouldn't be open, would it? You seem to be a little fuzzy on what sets are 'open' as well. And yes, 1 is a boundary point. So open sets can have boundary. What do you think now about the original true/false question?

 

[Gib Z]

Some definitions:

Let X be a non-empty subset of [tex] \mathbb{R}^n [/tex] (for simplicities sake) . Then [itex] x \in X [/itex] is called an interior point of X if there exists [itex] \epsilon > 0 [/itex] such that the open ball centered at x, radius [itex] \epsilon [/itex] is entirely contained in X. i.e [itex] B(x,\epsilon) \subseteq X [/itex].

We define the interior of A as such: [tex] int A = \left\{ x \in X : x \mbox{ is an interior point of X } \right\} [/tex].

We define the closure as such: [tex] \overline{A} = \left\{ x \in \mathbb{R}^n : \forall \epsilon > 0 , B(x,\epsilon) \cap X \neq \varnothing \right\} [/tex].

The Boundary of X is defined as such: [tex] \partial X = \overline{A} \backslash int A [/tex]
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Now that the definitions are clear, finding the boundary of a set is almost algorithmic.

Example: [itex] X = (0,1) [/itex]. Every point in X is clearly an interior point. However you want to see it (by the definition, characterization as the set of all limits of convergent sequences in X, whatever you like), its closure is [itex] \overline{X} = [0,1] [/itex]. Then [tex] \partial X = [0,1] \backslash (0,1) = \{ 0, 1 \} [/tex]

"True or false:
Let S be any set in [itex]\mathbb{R}^2[/itex]. The boundary of S is the set of points contained in S which are not in the interior of S."

Take a look at the example. It should give you the idea.

A general hint: Don't build intuition too "far away" from the defined terms, gain your intuition from the definitions. When you are too far away, you may think of terms in a sense that they are not defined as (in this case the question is trying to do that with "Boundary"). It
s better if you mix up having intuition for certain concepts (which sets are open, what is their closure), and then apply definitions to find the boundary.

 

[Justhanging]

(0,1) means the set of all points x such that 0<x<1. Yes, it is open. 1 and 0 are not included. If they were, it wouldn't be open, would it? You seem to be a little fuzzy on what sets are 'open' as well. And yes, 1 is a boundary point. So open sets can have boundary. What do you think now about the original true/false question?

So your saying the answer is false because the boundary of S need not be in S? But this then would only apply to open sets?

 

[Dick]

So your saying the answer is false because the boundary of S need not be in S? But this then would only apply to open sets?

Yes. False, for that reason. Why would you think only open sets could have boundary points not included in the set? [0,1)=(0,1)U{0} isn't open. 1 is still a boundary point that isn't in the set. Give me another different example.

 

[Justhanging]

Yes. False, for that reason. Why would you think only open sets could have boundary points not included in the set? [0,1)=(0,1)U{0} isn't open. 1 is still a boundary point that isn't in the set. Give me another different example.

Open sets would have none of its boundary points included in the set. Wow I made this question a lot harder than it needed to be. Thanks Dick.