# Topology of a metric space

• Feb 22nd 2009, 10:57 AM
Julie22
Topology of a metric space
Hi I have started a new subject this week called Topology, and it a bit harder than the stuff that I am used too, but anyway

We have metric space called $(M, d)$ where two $m_1, m_2 \in M$ and $m_1 \neq m_2$. Then I need to show that there exists open sets $S_1, S_2 \in \mathcal{T}_d$

which statifies that 1) $m_1 \in S_1$ 2) $m_2 \in S_2$ and 3) $S_1 \cap S_2 = \emptyset$

Firstly I know that the definition of open set or ball is as follows

$\{\forall x \in U \exists r > 0 : B_r(x) \subseteq U\}$ In other words: A set U is considered open if there for every element in the set is the center of an open ball of the set.

I also know from the metric subject field that if x,y are point in the set T and if they a $x \neq y$ then d(x,y) > 0.

So the way to show 1) and possible 2) isn't that to claim that for set $S_i$ to be open I must show that every point m_j on that set will be the center of an open ball?

I can see all the definitions that I need in my head I just need some assistance to connect them :(

Cheers
Julie
• Feb 22nd 2009, 11:09 AM
Plato
Quote:

Originally Posted by Julie22
We have metric space called $(M, d)$ where two $m_1, m_2 \in M$ and $m_1 \neq m_2$. Then I need to show that there exists open sets $S_1, \S_2 \in \mathcal{T}_d$ which statifies that 1) $m_1 \in S_1$ 2) $m_2 \in S_2$ and 3) $S_1 \cap S_2 = \emptyset$

Firstly balls themselves are open sets.
Then let $\delta = \frac{{d\left( {m_1 ,m_2 } \right)}}
{4},\;S_1 = B_{\delta}\left( {m_1 } \right)\;\& \,S_2 = B_{\delta}\left( {m_2 } \right)
$

Using the triangle inequality is easy to show the third part.
• Feb 22nd 2009, 11:27 AM
Julie22
Quote:

Originally Posted by Plato
Firstly balls themselves are open sets.
Then let $\delta = \frac{{d\left( {m_1 ,m_2 } \right)}}
{4},\;S_1 = B_{\delta}\left( {m_1 } \right)\;\& \,S_2 = B_{\delta}\left( {m_2 } \right)
$

Using the triangle inequality is easy to show the third part.

Okay then if to prove that the balls are open if $d(m_1,m_2)> 0 \Rightarrow \|m_1 - m_2 \| < \epsilon$ then there is a $\delta > 0$ does this then imply that $0 < \|m_1 - m_2 \|> \delta$
Have I used the definition correctly?

Cheers
Julie
• Feb 22nd 2009, 11:33 AM
Plato
No that is not correct.
Look into the textbook. One of the first theorems ought to be that "Balls are open sets."
• Feb 22nd 2009, 12:11 PM
Julie22
Quote:

Originally Posted by Plato
No that is not correct.
Look into the textbook. One of the first theorems ought to be that "Balls are open sets."

For you convience my textbook is Apostol "Mathematical Analysis second edition".

On page 49 there is not exactly a theorem but it states

Let a be a point in $\mathbb{R}^n$ and let r be a possitive number then th set of all points x in $\mathbb{R}^n$such that

$\|x-a| < r$ is called an open n-ball of radius and center a, and I et that epsilon and delta are some sort measures which are used to determen what happens at a.

There is a definition on the same page which states that every interior point a of a set S can b surrounded by an "n-ball" $B(a) \subset S$ and any set containing a ball with cente a is refered to as a neighbourhood of a.

So basicly my task is to show there exist a neighbourhood around m1 which only holds m1 and not m2? (Headbang)
• Feb 22nd 2009, 12:38 PM
Plato
Look at the two open sets I suggested: $\delta = \frac{{d\left( {m_1 ,m_2 } \right)}}{4},\;S_1 = B_{\delta}\left( {m_1 } \right)\;\& \,S_2 = B_{\delta}\left( {m_2 } \right)$

What if $z \in S_1 \cap S_2$? Well then we get a contradiction:
$d\left( {m_1 ,m_2 } \right) \leqslant d\left( {m_1 ,z} \right) + d\left( {z,m_2 } \right) < 2\delta < \frac{{d\left( {m_1 ,m_2 } \right)}}
{2}$
.
So they are disjoint.
• Feb 22nd 2009, 01:02 PM
Julie22
Quote:

Originally Posted by Plato
Look at the two open sets I suggested: $\delta = \frac{{d\left( {m_1 ,m_2 } \right)}}{4},\;S_1 = B_{\delta}\left( {m_1 } \right)\;\& \,S_2 = B_{\delta}\left( {m_2 } \right)$

What if $z \in S_1 \cap S_2$? Well then we get a contradiction:
$d\left( {m_1 ,m_2 } \right) \leqslant d\left( {m_1 ,z} \right) + d\left( {z,m_2 } \right) < 2\delta < \frac{{d\left( {m_1 ,m_2 } \right)}}
{2}$
.
So they are disjoint.

This may be stupid question but I know the axioms of metric
there is one which states d(x,y) > 0 -> $x \neq y$ and you choose a small delta and bit larger epsilon and the bigger delta gets the smaller epsilon until it reaches m1 and thusly m1 does not contain m2?

I am not trying to sound stupid or anything but its just then I get these generalized assignments I crash and burn, but then there is assignment with numbers then everything is fine.
I don't suppose you could explain to me in a example on how is much proceed? then everything works much better :)
• Feb 23rd 2009, 10:48 AM
HallsofIvy
Quote:

Originally Posted by Julie22
This may be stupid question but I know the axioms of metric
there is one which states d(x,y) > 0 -> $x \neq y$ and you choose a small delta and bit larger epsilon and the bigger delta gets the smaller epsilon until it reaches m1 and thusly m1 does not contain m2?

I am not trying to sound stupid or anything but its just then I get these generalized assignments I crash and burn, but then there is assignment with numbers then everything is fine.
I don't suppose you could explain to me in a example on how is much proceed? then everything works much better :)

Well, its not "stupid" but it is confused. In the first place, while responses above have mentioned " $\delta$, no one has said anything about an " $\epsilon$ so I don't know where you got that. The point is to show a neighborhood about each point that is disjoint from the other. Making $\delta$ small should do that.

Also "m1 does not contain m2" makes no sense. m1 and m2 are points, not sets.