Is (a,b) Open in R^n? Proof and Extension to n Dimensions

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In summary, the author is saying that (a,b) in Rn is not open. However, they mention that it might not be open in R^2 even if it is open in R.
  • #1
rumjum
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Homework Statement



We know that (a,b) is open in R. But is it open R^n?

Homework Equations





The Attempt at a Solution



I don't think (a,b) is open in R^n even if it is open in R. Let's take for example n=2, then

E = {(x,y) | x^2+y^2 < r^2} , where r^2 = |b|^2 + Y^2 , for all y belongs to R. However , we can see that even -b satisfies the above equation but -b is not part of (a,b). Hence, we have a point in R^2 which is not internal point of the segment (a,b). Hence, in R^2 (a,b) is not open.

It looks like (a,b) is not closed either for the neighborhood (-a-e, -a+e), for e >0 for any y belongs to R for any limit point (-a,y) in R^2 has no intersection with (a,b).

Any comments on the proof. Can this be extended to the n dimensional space?

Thanks.
 
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  • #2
What do you mean by (a,b)in Rn? In R, (a, b) means the interval from a to b on the number line, not including the endpoints a and b. But there is no one "number line" in Rn. I suspect you mean the straight line interval from (a, 0,0,..., 0) to (b, 0,0,...,0)- i.e. on the 'x' axis, but you should say that clearly. Or do you intend "a" and "b" to be, not numbers put points in Rn? Unfortunately, in that case, your "proof" below still makes no sense.

Your example:
E = {(x,y) | x^2+y^2 < r^2} , where r^2 = |b|^2 + Y^2 , for all y belongs to R.
makes no sense. Again, there is no "(x,y)" in Rn- points don't have only two coordinates. If, as I said above, you mean these as points, then "x^2+ y^2< r^2" makes no sense. You seem to be trying to construct circles (or n-spheres?) around the origin with varying radii. But the origin may be nowhere near your original segment.

What you want to do is choose a point on your interval and construct a neighborhood (an n-sphere) with that point as center and show that, no matter how small the radius is, it will contain some points that are not in the interval. That will show that the point is not an interior point and the interval is not only not open but has empty interior.
 
  • #3
HallsofIvy said:
What do you mean by (a,b)in Rn? In R, (a, b) means the interval from a to b on the number line, not including the endpoints a and b. But there is no one "number line" in Rn. I suspect you mean the straight line interval from (a, 0,0,..., 0) to (b, 0,0,...,0)- i.e. on the 'x' axis, but you should say that clearly. Or do you intend "a" and "b" to be, not numbers put points in Rn? Unfortunately, in that case, your "proof" below still makes no sense.

What I meant was what I copied from Rudin "Principle of mathematical analysis". It is quoted in the second chapter that "(a,b) is open in R but it is not open in R^2". Period.

I only extended my problem to n dimensional. If my question does not makes sense to you, then I suppose you have a better way of interpreting the statement from Rudin's book

HallsofIvy said:
Your example:

makes no sense. Again, there is no "(x,y)" in Rn- points don't have only two coordinates. If, as I said above, you mean these as points, then "x^2+ y^2< r^2" makes no sense. You seem to be trying to construct circles (or n-spheres?) around the origin with varying radii. But the origin may be nowhere near your original segment.

That I think I should have clarified. My example was for R^2 , which is why I was constructing circles and not n-spheres.

HallsofIvy said:
What you want to do is choose a point on your interval and construct a neighborhood (an n-sphere) with that point as center and show that, no matter how small the radius is, it will contain some points that are not in the interval. That will show that the point is not an interior point and the interval is not only not open but has empty interior.

Thanks. Yeah, I was thinking in those lines myself.
That was actually my first post but I am a bit surprised. Are people on this post usually a bit cynical?
 
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FAQ: Is (a,b) Open in R^n? Proof and Extension to n Dimensions

What does it mean for a set (a,b) to be open in R^n?

In mathematics, a set (a,b) is considered open in R^n if for every point within the set, there exists a small neighborhood around that point that is also contained within the set. In other words, there is no boundary or edge between the points in the set and the points outside the set.

How do you determine if a set (a,b) is open in R^n?

To determine if a set (a,b) is open in R^n, you can use the open set definition. This means that for every point in the set, you can find a small neighborhood around that point that is also contained within the set. Another way to determine if a set is open is to check if its complement (the set of all points not in the set) is closed. If the complement is closed, then the set is open.

What is the difference between an open set and a closed set in R^n?

An open set in R^n is a set where every point has a small neighborhood around it that is also contained within the set. In contrast, a closed set in R^n is a set where all of its limit points (points that are arbitrarily close to the set) are contained within the set itself. In other words, there are no points on the "edge" of a closed set that are not included in the set.

Can a set (a,b) be both open and closed in R^n?

No, a set (a,b) cannot be both open and closed in R^n. By definition, an open set and a closed set are mutually exclusive. A set can either have no boundary (open) or contain all of its limit points (closed), but it cannot have both properties simultaneously.

How does the concept of open sets relate to topology and continuity?

In topology, open sets are used to define the concept of continuity. A function is continuous if the preimage of every open set is also open. In other words, if a function maps an open set to another open set, it is considered continuous. This idea of continuity is important in many areas of mathematics, including analysis, differential equations, and geometry.

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