# The topology of the plane (continued)

## Correction

I just fixed a rather major typo in the last class. The definition of "exterior point" should have read

**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**

*.*

and not

**Definition.** *A
point *(a,b) in **S** a subset **R**^{^2}* is an
exterior point of *

**S**

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

*that does not intersect*

**S**

*.*

I am led to conclude that either no one read it, no one noticed, or people noticed but didn't bother to comment. I hope its that last one, but in the future speak up people! Thanks :-)

## A quick review

So far the main points we have learned are:

- That subsets of the plane that are the interior of a disc are known as neighborhoods.
- That a point is interior to a set
**A**if it has a neighborhood in**A**and exterior to**A**if it has a neighborhood not in**A**. - That a set
**A**is defined to be open if every point in**A**has a neighborhood in**A**and that**A**is defined to be closed if every point not in**A**is exterior to**A**.

## Answers to questions posed in the last class

I am continuing to give proofs as rough sketches, but if anyone wants to see the details I would be happy to provide them.

**Q:** How can we give a point in **B** (a closed disk) so that it has no neighborhood in **B**?

**A:** Any point on the boundary of the disc will do. By the way, this proves that **B** is not open (remember that this is not equivalent to proving that it is closed!)

**Q:** Why can't **B** be expressed as the union of neighborhoods?

**A:** Suppose that we could express **B** as a union of neighborhoods. Then every point in **B** must be contained in at least one neighborhood. However we have already shown that this is not the case. We can easily prove the stronger result that a non open set can never be expressed as the union of open sets. Suppose we could. Then every point in it is in some open set. Therefore it is in some neighborhood. Ah ha! Its that same contradiction, because our original set, being non-open, must have had at least one point with no neighborhood in the set.

**Q:** Why is it sufficient to say that there is a disc around some point in order to garuntee it has a neighborhood, when the definition of neighborhood says that the disc must be *centered* around the point?

**A:** Suppose the point **(p _{_1},p_{_2})** is contained in a neighborhood of the point

**(c**with radius

_{_1},c_{_2})**r**. Then the neighborhood of

**(p**with radius

_{_1},p_{_2})**r - sqrt((p**is contained in the neighborhood of

_{_1}- c_{_1})^{^2}+ (p_{_2}- c_{_2})^{^2})**(c**. So it turns out that our definition of neighborhoods was much more specific than we needed them to be. We will see that there are many many ways of defining neighborhoods, some of which will work just as we expect, and others that will make put a whole new structure on the plane....

_{_1},c_{_2})**Q:** What subset of the plane besides the empty set is both open and closed?

**A:** The plane itself. Clearly every point of it has a neighborhood in it since every point has a neighborhood. Furthermore, there are no points not in it (it has an empty complement) so every point in its compliment is exterior to it!

**Q:** Can you give a subset of the plane that is neither open or closed?

**A:** Of course you can! As I said, most sets are of this form. For example, take a closed disc, and remove a single point from its boundary. The set we are left with has a point in its complement that is not exterior (namely the point we removed) and it has points which are not interior (any of the other points on the boundary). Therefore it is neither open nor closed.

## Some more definitions

**Definition.** *A point ***(x,y)*** is a ***limited point*** of a set ***A*** if every neighborhood of ***(x,y)*** contains some point of ***A**

**Definition.** *A point ***(x,y)*** is an ***isolated point*** of a set ***A*** if it is a limit point of ***A*** and there is a neighborhood of ***(x,y)*** such that its intersection with ***A*** is ***(x,y)***.*

**Definition.** *A limit point of a set ***A*** is a frontier point of *

**A**

*if it is not an interior point of*

**A**

*.*

The set of frontier points of a set is of course its boundary. Notice that both the open and closed disc we referred to in the last lesson have the exact same boundary, but that only the closed disc contains its boundary. The boundary of the open disc is contained in the disc's complement. This is generally true of open and closed sets. A closed set will always contain its boundary, and an open set never will.

**Definition.** *The union of a set and its boundary is its closure.*

As we would expect given its name, the closure of any set is closed.

I know that wasn't much, especially after I missed so many weeks, but alas it is all I have time for. I leave you with a result you may wish to prove: *the closure of a set is the smallest closed set containing it.*