Sample Assignment  Set Theory Assignment
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Recall that if
is a metric space
and, then
denotes the set of
all limits points of.

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

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 11.
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

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 priimage 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.

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.