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Open Sets and Metric Spaces

Date: 11/17/2002 at 08:38:30
From: Jan Jansen
Subject: Open sets / metric spaces

(M,d) is a metric space.
1) Prove  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 , there is some open ball
     that contains x and which is a subset of .
(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 " part. Think! Just exactly 
how many x's ARE there in  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!
Associated Topics:
High School Sets

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