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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 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.

(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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