Sample Assignment - Set Theory Assignment

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  1. Recall that if is a metric space and, then denotes the set of all limits points of.

  1. a) Suppose that and are subsets of a metric space having. Using the open ball definition of limit points, show that.

Solution: Given . Also, -------------(1)

Suppose is open relative to .To each limit point , there is positive number such that the condition imply that .

Let be the set of all such that and define

Then is an open subset of . Since, for all , it is clear that

. By our choice of , we have for every so that

. Thus, ---------------(2)

Suppose is open relative to.To each limit point, there is positive number such that the condition imply that.

Let be the set of all such that and define

Then is an open subset of. Since, for all, it is clear that

. By our choice of, we have for every so that

. Thus, ---------------(3)

From equation (1), (2) and (3)

proved

1. b) Show that if are two subsets of a metric space, then.

Solution: Given

To prove:

We can say that as well as

Now, let be an arbitrary element that belong to

i.e. . Then there exists an such that . Since ,

we have . i.e.

Now, it follows that

as well as

On the other hand, let then and . Therefore, there exist

and such that and . Let .

Clearly, and .

i.e. . Hence, proved.

c) Find two subsets with

Solution: Given

To prove:

Let be a set of real numbers. Also, are subset of .

Indeed, if and

then

. Here, it is a closed interval.

Now, since

and then

, it is an open interval.

Therefore we can say that

Let us take another example and try to solve the given problem.

and , then closed interval.

Again,

and , then = (p), it is open interval.

Therefore, we can say that

  1. We are in a metric space and is a point of . Show that the set is a closed set.

Solution: To prove this we need to show is open set in .

Suppose is closed in . We need to show that is open in .

If then and it is open. So, we may suppose that

. Let be a point in . Since, is closed and,

cannot be a limit point of .

So, there exists an such that . Thus for each point

Of is contained in an open set contained in

This means is open.

Now, Let , Then . If

Then

Moreover, .

Indeed, if then

Thus, i.e.

The following fundamental properties of closed sets are analogues of the properties of open .

Sets.

3. Suppose that is a collection of open intervals in with the special property that if are not equal, then. For each there is a point because is a dense subset of.

3. a) Show that the function is 1-1.

Solution: Theorem: In order to prove that certain function, say f(x) as one to one we

need to show that if =, then.

Given, and

(From the question)

Now, let

And (from question)

Therefore, for every there exists a unique value of

So, is

  1. b) Show that the function is not onto.

Solution: Theorem: A function is said to be onto when its image equals its range, i.e.

Using this theorem, we can write, and then there exists some value for which there is no pri-image ofin. Also, onto function means “a function from to is called onto if for all in there is an in such that. All elements in are used.



Hence, from the figure we can say that is not onto.

3. c) Using or otherwise, show that the collection must be either finite or

countably infinite.

Solution: From 3.a), we can say that is one – one.

i.e., there exists a for every.

Therefore, must be finite and countably infinite.



  1. Suppose andare bounded, nonempty subsets of . We know that each has a least upper bound, say and . Let . Prove that

Solution: and

---------------------(1)

Let be be .

If a is the if b is the , then

---------------(2)

----------------(3)

Also, (from eq(1))---------------(4)

Putting the value of a and b from equation (2) and (3) in equation (4), we get

Hence, it is proved.