Unable to understand Open Set

[phiby]

Unable to understand "Open Set"

I keep reading the definition of an open set & neighborhood, but I just don't seem to get it.

This is the defn - "a set U is open if any point x in U can be "moved" a small amount in any direction and still be in the set U."

I don't see how this condition can ever be satisfied.

For eg, after this, it typically says - "an open set is a solid region minus its boundary".

I don't understand this because a solid region minus the boundary does have a boundary. It's just that the new boundary is slightly smaller as compared to the previous boundary i.e. I am not able to grok the concept of how a point at the edge of a region can be moved outwards and still remain in the region.

Can someone explain this?

Same thing with understanding "Neighbourhood of a point" also?

Is there something else I need to study to understand this?

 

[SW VandeCarr]

I keep reading the definition of an open set & neighborhood, but I just don't seem to get it.

This is the defn - "a set U is open if any point x in U can be "moved" a small amount in any direction and still be in the set U."

One example of an open set is {(0,1)} which is the set of points on the open interval of real numbers where the points 0 and 1 are not in the set. This is indicated by the use of parenthesis. If we write {[0,1]} using brackets, we've indicated the points 0 and 1 are in the set. This is a closed interval.

In point set topology, the same relationship holds for open and closed balls. In both cases interior points are infinite in number, so for any interior point in an open ball, its neighborhood contains only interior points. However, closed balls include points whose neighborhood, no matter how small, include exterior points.

A neighborhood may be defined by a radius r centered on a point where the magnitude of r approaches zero as a limit.

 

[HallsofIvy]

I keep reading the definition of an open set & neighborhood, but I just don't seem to get it.

This is the defn - "a set U is open if any point x in U can be "moved" a small amount in any direction and still be in the set U."

I don't see how this condition can ever be satisfied.

For eg, after this, it typically says - "an open set is a solid region minus its boundary".

I don't understand this because a solid region minus the boundary does have a boundary. It's just that the new boundary is slightly smaller as compared to the previous boundary

No, that's not true. In the real line, the closed interval, [0, 1] has the two points 0 and 1 as boundary. The open interval, (0, 1) ([0,1] minus its boudary) does NOT have a "slightly smaller" boundary, precisely because there is no "next larger number" to 0 and no "next smaller number" to 1. The boundary of the set (0, 1) is still the two points 0 and 1. [0, 1] and (0, 1) have exactly the same boundary. The only difference between [0, 1] and (0, 1) is that [0, 1] contains all of its boundary while (0, 1) contains none of its boundary. You are saying (perhaps without realizing it) that "The set $\{y| y< 1\}$ is the same as $\{y| y\le x\}$ for some number x< 1" which is not true.

i.e. I am not able to grok the concept of how a point at the edge of a region can be moved outwards and still remain in the region.

Can someone explain this?

Same thing with understanding "Neighbourhood of a point" also?

Is there something else I need to study to understand this?

You need to review the distinction between "$<$" and "$\le$".

 

[kith]

This is the defn - "a set U is open if any point x in U can be "moved" a small amount in any direction and still be in the set U."

I don't see how this condition can ever be satisfied.

Just apply the definition to the open interval (0,1). For every point x near the boundary (x=0.99 for example), you can find other points which lie between x and the boundary. This is not true for the closed interval [0,1].

 

[Stephen Tashi]

phiby,

Where did you get the definition of "open set" in your original post? It isn't the correct mathematical definition. It's someone's attempt to rephrase the correct definition into language that appeals to the intuition.

 

[phiby]

Ok - I got it now.

I guess I misunderstood the original defn. I thought it meant you need to find a solid which did not have a boundary - which is not possible. What you need to do is to find a solid & exclude it's boundary - the new solid does have a new smaller boundary, but that boundary is not the boundary of your "Open set" - so that's OK.

And I do understand the diff between < & <=. I just misunderstood the remaining part of the original definition.

FYI, I found the defn on Wikipedia - I was reading a book on Computer Graphics. The book defined open set in a more mathematical way which was not easy to understand - hence I checked wikipedia.

 

[Stephen Tashi]

FYI, I found the defn on Wikipedia

What you quoted is part of the intuitive discussion in the wikipedia article; it isn't from the mathematical definitions given in the article. The intuitive discussion is only to help your intuition. It doesn't provide a mathematical definition.

 

[SW VandeCarr]

Ok - I got it now.

I guess I misunderstood the original defn. I thought it meant you need to find a solid which did not have a boundary - which is not possible. What you need to do is to find a solid & exclude it's boundary - the new solid does have a new smaller boundary, but that boundary is not the boundary of your "Open set" - so that's OK.

You still don't seem to get it. What is the lower boundary of the open interval (0,1)? For any number greater than zero, there is a smaller number greater than zero. For a solid the same principle holds. The neighborhood of every point in an open ball contains only interior points. I think I stated this clearly in post 2. This is equivalent to the neighborhood of any point on the open real number interval (0,1) not containing 0 or 1.

 

[phiby]

You still don't seem to get it. What is the lower boundary of the open interval (0,1)? For any number greater than zero, there is a smaller number greater than zero. For a solid the same principle holds. The neighborhood of every point in an open ball contains only interior points. I think I stated this clearly in post 2. This is equivalent to the neighborhood of any point on the open real number interval (0,1) not containing 0 or 1.

I got this fully - I just am not able to put it into proper words like you guys.

What you quoted is part of the intuitive discussion in the wikipedia article; it isn't from the mathematical definitions given in the article. The intuitive discussion is only to help your intuition. It doesn't provide a mathematical definition.

The mathematical defn isn't important to me. I am just trying to understand the general concept as a base for Graphics.