Open Balls and Open SetsDate: 10/04/98 at 10:26:50 From: Robert Subject: Open Balls/ Open Sets I am having trouble understanding the concepts of open balls and open sets. I know the definitions for both: Open Ball: An open ball B in a set E centered at a point v, with radius r > 0, is the set of all x in E such that |x - v| < r. Open Set: We define a set U to be open if for each point x in U there exists an open ball B centered at x contained in U. I was wondering if you could give me some examples for these definitions, in order for me to understand them better. Can you also explain how every open ball is an open set? Thank you so much for your help. - Robert Date: 10/08/98 at 19:47:16 From: Doctor Mike Subject: Re: Open Balls/ Open Sets Hello, For an open ball, another word you could think of here would be "sphere." It is the same thing. Inside a spherical ball is every point whose distance from the center of the sphere is less than the radius of the sphere. The surface of the sphere has points whose distance from the center is equal to the radius of the sphere. The open ball is the inside part. For examples, think of any enclosed volume: a square box, a rectangular box, an egg, a cylinder with top and bottom, a balloon of any shape. What these have in common is one or more surfaces that enclose some space. If you just consider what is on the "inside" (or interior) then that would be an example of an open set in space. The reason: By the exact definition of "open," around any point in such an interior you can fit a complete Ball which is also completely in the interior. Granted, if the point you pick is just a fraction of a nanometer from the surface, the ball is going to be very small, but you can do it. If you understand why every open ball is an open set, you have the concept. How can we prove that this ball qualifies to be "open"? We just use what the definition says. Let x be some point (any point) inside this spherical ball. Keep in mind that it is not actually on the outside surface of it, but actually inside it. If D is the distance from the point x to the center of the spherical ball, and if R is the radius of the spherical ball, then it must be true that D < R . [The distance from the center must be less than the radius. Right?] Now, how far is the point x from the surface of the spherical ball? Answer: (R-D). So, if you place a ball of radius (R-D)/2 at the point x, it will fit inside the ball of radius R with room to spare. Done. (We just finished doing all we had to do to qualify by the definition.) If this is not immediately clear to you, think about it a few more times (and re-read the definition of "open set" again) and I think pretty soon it will begin to make sense. Here's one more thing to think about. In the definition of open, it depended in a major way on balls of a certain radius. What else has a radius? Circles, of course. All that we did above for open sets and open spheres in space could be done for open sets and open "circular disks" in a plane or on some other kind of flat surface. In advanced math the concept of distance (and so the concept of radius) is generalized to define something called a Metric Space. An abstract form of "open ball" can be defined in such an environment, too. I hope this helps. - Doctor Mike, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/