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# The topology of the plane

In this section we will we shall take a look at the topology of the
plane, **R**^{^2} (see the guide to the notation used in this
class). . It should be understood that many of the results and
definitions given here are very specific. We will later encounter them
in more general forms. This may lead to some confusion, but I am
taking that risk as I think that there is much to be gained by
starting with more limited but familiar results. When we do come to
the more general definitions and theorems it is my hope that they will
seem familiar and their purpose will often be clear by analogy.

Note that I am denoting elements of the plane by an ordered pair of
real numbers, (x,y). I chose to do this to emphasize the familiar
Euclidean and Cartesian structure to the plane, however the reader
should be aware that we are concerned with the *point* (x,y)
and not either of the components individually. In most texts a single
letter would be given to each point: **x**=(x,y). I will introduce
and make use of this notation in later classes.

Consider these two sets

**A**={(x,y) in
**R**^{^2} | x^{^2}+y^{^2} < 1
}

and

**B**={(x,y) in
**R**^{^2} | x^{^2}+y^{^2} <= 1
}.

Geometrically these two subsets of the plane are
basically the same. They are both discs of radius 1. They are clearly
different sets though since **B** contains a circle of radius 1,
while **A** does not. This distinction becomes crucial when we look
at the sets topologically. As we will see that *there is no
continuous function that will send A to B.* We will
prove this by showing that

*, and that*

**A**is an open subset of the plane while**B**is a closed subset of the plane*there are no one to one continuous functions between open and closed sets*.

Open and closed sets will be defined in terms
of sets known as *neighborhoods*.

## Neighborhoods in the plane **Definition.** *A set of the form *{(x,y)
in **R**^{^2} | (x-a)^{^2}+(y-b)^{^2} <
r^{^2}}* for some fixed *(a,b) in
**R**^{^2}* and *r in **Z*** is known as a
***neighborhood of ****(a,b)***, or more simply just a
***neighborhood**.

Neighborhoods allow help us
understand the structure the plane and its subsets. For example every
element of of the set A described above has a neighborhood in
**A**, but this is not true for the set **B**. (Can you see why
this is?) Similarly **A** can be expressed as the union of
neighborhoods (this is trivial), while **B** can not be expressed as
the union of neighborhoods, even if we are allowed infinitely many of
them. We will prove this, but first some more definitions:

**Definition.** *A point *(a,b) in **S**
a subset of </i>**R**^{^2}* is an ***interior point**
of **S*** if there is a neighborhood of *(a,b)* in
***S***.*

In normal english, a point is
interior to a set if you can draw a circle around that point without
leaving the set.

**Definition.** *A
point *(a,b) in **R**^{^2}* is an
***exterior point** of **S*** if there a neighborhood of
*(a,b)* that does not intersect
***S***.*

As with interior points there is a
very easy and natural interpretation to this. A point is exterior to a
set only if you draw a circle around it without entering the set.

(In the two paraphrases above I said that you must be able to draw
a circle around the point, but from the definitions you would think
the circle would have to be *centered at* the point. Why are
the paraphrases still correct?)

## Open and closed sets

**Definition.**

*A set of the form*{(x,y) in

**R**

^{^2}| (x-a)

^{^2}+(y-b)

^{^2}< r

^{^2}}

*for some fixed*(a,b) in

**R**

^{^2}

*and*r in

**Z**

*is known as a*

**neighborhood of****(a,b)**

*, or more simply just a*.

**neighborhood****A**, but this is not true for the set

**B**. (Can you see why this is?) Similarly

**A**can be expressed as the union of neighborhoods (this is trivial), while

**B**can not be expressed as the union of neighborhoods, even if we are allowed infinitely many of them. We will prove this, but first some more definitions:

**Definition.**

*A point*(a,b) in

**S**a subset of </i>

**R**

^{^2}

*is an*

**interior point**of**S**

*if there is a neighborhood of*(a,b)

*in*

**S**

*.*

**Definition.**

*A point*(a,b) in

**R**

^{^2}

*is an*

**exterior point**of**S**

*if there a neighborhood of*(a,b)

*that does not intersect*

**S**

*.*

*centered at*the point. Why are the paraphrases still correct?)

We can now finally give the definitions we have been waiting for.

**Definition.** *A set ***S*** will be
called an open set if all of its points are
interior.*

**Definition.** *A
set ***S*** will be called a closed set if all the
points not in it are exterior to it.*

Note that a set may be both open and closed. This is easiest to see if we examine the empty set, the set with no elements. Every point is exterior to it, since we can draw a circle around every point with out hitting a single one of its elements! So the empty set is clearly closed. But every one of its elements is also interior to it, because when there are no elements to satisfy a condition then they all clearly satisfy it. Every one of them!

Although it doesn't have the same paradoxical appearance, it is also easy to forget that a set may be neither open or closed. In fact if we arbitrarily select a set this is what we should except. It just means that we have at least one point in that set that is not interior and at least one point its complement (the complement of a set is everything not in it) which is not exterior to it.

Can you think of another set that is both open and closed? How about a set which is neither?

## Some results

Let's go back to our two
sets **A** and **B**. You have probably already figured out
that **A** is an open set and **B** is a closed set. Now lets
prove it. (Yes, these proofs are very informal. If anyone would like
to see them done out properly I would be happy to oblige at a later
time.)

Take any point in **A**, say (x,y). Note
that the neighborhood of (x,y) with radius (1-sqrt(x^{y^2}))/2 is
entirely inside **A**. Therefore (x,y) is an interior point. Since
(x,y) was chosen arbitrarily **A** is an open set. qed

Take any point not in **B**, say (x,y). Note that the
neighborhood of (x,y) with radius
(sqrt(x^{^2}+y^{^2})-1)/2 is entirely outside of
**B**. This means (x,y) is exterior, and since (x,y) could be any
point not in **B** we see that every point in **B**'s complement
is exterior to it. Therefore **B** is closed. qed

A proof
much like the ones above could be used to prove the following
important result: *every neighborhood is an open set*.

Well, I had planned to do more today, but that is all I have time
for, so next week we will continue our examination of the topology of
the plane. We will learn more about open and closed sets, (and I will
prove the **B** is not the union of open sets as promised) and I
will formally introduce the notion of topological equivalence.

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*--Ben*