Open Sets and Metric SpacesDate: 11/17/2002 at 08:38:30 From: Jan Jansen Subject: Open sets / metric spaces (M,d) is a metric space. 1) Prove Ø is open. 2) Prove M is open. Date: 11/19/2002 at 00:57:02 From: Doctor Mike Subject: Re: Open sets / metric spaces Jan, You need to look at the definition of "open" for a metric space. Also, while you are at it, look at the definition of "open ball" (or some similar term), which is the set of all elements whose distance from some particular point is less than some particular positive number. The point is called the "center" and the number is called the "radius" of the ball. In the usual Real Plane, such a "ball" is a circular disk WITHOUT its circular border. So, the definition will say something like S is open if and only if FOR EVERY x in S, there is some open ball that contains x and which is a subset of S. So you have 2 things to prove, namely, (1) For every x in Ø, there is some open ball that contains x and which is a subset of Ø. (2) For every x in M, there is some open ball that contains x and which is a subset of M. These two things are VERY easy to prove, but they are also tricky. The tricky part of (1) is the "For every x in Ø" part. Think! Just exactly how many x's ARE there in Ø anyway? None! So ask yourself what exactly is needed to prove this. The tricky part of (2) is the realization that ALL open balls, with ANY center, and ANY radius, are still sets in the metric space. Thus, each is a subset of M. So, for any particular x in M, pick some open ball with center x, and ....... It takes a while for budding mathematicians to get the hang of what is involved in constructing a proof. Just keep trying. NOTE: There are other ways of defining "Open" for a metric space, but they are all equivalent to the one described above, and the proofs will be similarly simple AND tricky. Hope this helps. - Doctor Mike, The Math Forum http://mathforum.org/dr.math/ Date: 11/19/2002 at 05:50:08 From: Jan Jansen Subject: Open sets / metric spaces Thank you for explaining it to me. It now is much easier to understand. However, I always have difficulties with proving it, although I understand why it is. Could you please PROVE 2) M is open? I understand why it is, but can you prove it? Thank you so much! Date: 11/19/2002 at 11:14:45 From: Doctor Mike Subject: Re: Open sets / metric spaces Hi again Jan, Okay. I'll show all the steps and reasons in great detail so you can see everything. Let's start with reviewing what is to be proved: (2) For every x in M, there is some open ball that contains x and which is a subset of M. Note: Whenever the statement to be proved starts out "For every x ..." the straightforward way to do the proof is to ASSUME that x represents some arbitrary element, and then proceed to show that the statement is true for THAT x. Because nothing special was assumed about x, you have proved the fact for EVERY x. See? Let x be any element in M. We must discover SOME open ball B such that x is contained in B, and B is a subset of M. We choose B to be the open ball with center x and radius one. That is, B is the set of all elements "e" of M whose distance d(x,e) from x is less than 1.0. There are two specific things to prove about B: (a) x is contained in B. One of the properties for a metric space distance function is that the distance d(x,x) is zero. Thus d(x,x) = 0.0 and 0.0 is less than 1.0, so x qualifies as an element in the open ball B. Done with this part. (b) B is a subset of M. By the way B was defined, it is the set of all elements of M that have some property. We don't care right now what that property is. The important thing to notice is the "of M" part. If "b" is in B then b is an element "of M" having some such property, but all we need from that is just that b is "an element of M." Thus we have shown that any b in B is also an element in M. This is exactly what it means to say that B is a subset of M. Done with this part. Therefore, B is an open ball that contains x and which is a subset of M. This completes the proof. We are totally done. Comments: As you look over this proof, your reactions to each of the steps may be something like "well yes, sure, that's obvious" because there is nothing deeply mysterious about any of these steps. They ARE logical and obvious once you look at them. The trick is to realize exactly what is needed, and then do it. Another comment I should make is that I could have just as well defined B to be the open ball with center x and radius 123.456 or some other strange number. The proof steps would have been exactly the same, namely, x is in B because d(x,x) is zero, which is less than 123.456, and B is a subset of M because B is defined as a set of elements of M. As you begin to grow in your understanding of the logical proof techniques commonly used in math, you will look back at things like this and say with mature understanding, "Oh yes, some of those early basic proofs were so easy they were hard!" Good luck. - Doctor Mike, The Math Forum http://mathforum.org/dr.math/ Date: 11/20/2002 at 11:15:57 From: Jan Jansen Subject: Thank you (Open sets / metric spaces) Thank you very much for your explanation! |
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