Continuity of Functions

Definition.

Let be a non-empty subset of and let a function. Suppose . Then is said to be continuous at if for every $ε$ , such that

ε ε ε

Example 1.

some non-empty subset of fix
we can show is continuous function with the above definition

Example 2 (Identity Function).

Suppose is a non empty subset of

Consider any
now,

take $ε$ then for

ε

so is continuous at all

Example 3.

let
consider defined by

Consider
we need to estimate

Take $ε$ Then for

ε ε

Example 4.

let
show that is continuous

Fix

if then
Take $ε$
Then for then $ε$

complete the exercise

Theorem.

Let A be a non-empty subset of and let be function suppose Then is a continuous at iff for every sequence in , converges to . converges to

Proof.
Suppose is continuous at . Suppose is a sequence in converges to .
for $ε$ such that

ε

As converges to , there exist such that

Hence for , and hence from ^089c2b (4.1)
so converges to .

Suppose is not continuous at . Then there exist $ε$ such that for no

ε

holds
so for take then
such that

ε

As do not converge to

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#24-sep
Suppose and does that mean converge?
No, let and

Definition (continuity at a point

ε

version).

Let A be a non empty subset of let be a function .
Fix then is said to be continuous at if for every $ε$ such that

ε

Definition.

Let be a non-empty subset of . Then is said to be isolated in if there exist such that

Corollary.

suppose is an isolated point in . Then any function is continuous at

Definition (Continuity of function).

Let be a non-empty and let be a function. Then is said to be continuous if it is continuous at every

Theorem.

Let A be a non-empty subset of Let and be functions. Fix . Suppose and are continuous at .
then is defined by
is also continuous at c

Proof.
for $ε$ , there exist such that

ε

similarly there exist such that

ε

Take then
so for this value of both ^ba9cbb (4.2) and ^1eabe6 (4.3) are true

ε

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#26-sep

Theorem.

Let A be a non-empty subset of and let suppose are functions continuous at c

For defined by m is continuous at
defined by is continuous at c
if then defined by is continuous at c

Proof.
(using sequential version of continuity.)

Take we want to show that is continuous at .
Suppose is a sequence in A converging to we wan tho show that converges to

Take we want to show that is continuous at .
suppose is a sequence in A converging to we wan tho show that converges to
Take we want to show that is continuous at .
Suppose is a sequence in A converging to we wan tho show that converges to and for all in then

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Example.

Suppose then the polynomial defined by is continuous

If you can't prove it leave ISI

Theorem.

Let and let . Suppose is continuous at . Suppose define by then is continuous at c

Proof.
for $ε$ there exist such that

ε

consider then and

ε

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Remark.

is known as the restriction of to and is denoted by

Corollary.

If is a non-empty subset of and is a polynomial then is continuous
suppose is another polynomial, with
then defined by
is continuous

Theorem.

Let A be non empty subset of
suppose and be functions
Assume suppose for some is continuous at is continuous at
then defined by
is continuous at

Proof.
suppose is a sequence in A converging to .
We want to show that converges to

as is continuous is also continuous
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#1-oct

Definition.

Let A be a non-empty subset of and let be a function then is said to be bounded above if there exist
similarly is said to be bounded below if there exist such that
it is said to be bounded if there exist such that
let be bounded above then is called the supreimum of
it is said to be a maximum if there exist such that

suppose is bounded below
is called the infimum of
it is said to be a minimum if the infimum is attained, that is there exist . .

Theorem.

Let with let be a continuous function then is bounded
moreover attains its supremum and infimum in

Proof.
suppose is not bounded
then for any natural number , is not a upper bound
this means that there exist such that

as is bounded by bolzano-Weierstrass theorem bolzano-Weierstrass theorem it has a convergent sequence
Take then
then by continuity of is convergent in particular it is bounded but we have

so it cant be bounded this is a contradiction

take
For is not upper bound for so there exist such that

once again by bolzano-Weierstrass theorem bolzano-Weierstrass theorem has a convergent sequence say converging to some
thus
as tends to by squeeze theorem is convergent and converges for as is continuous at
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#3-oct

Theorem (Location of roots ).

Let with . Let be a continuous function. Suppose then there exist a can also be called as 'root'

Proof. 1
We will use nested intervals property of
take and and

consider
if we are done
else if take
else if take
take

Note that and
repeat the process consider
if we are done
else if take
else if take
we continue so on
if it terminates we have a

If it doesn't terminated we have sequence of nested interval

by nested interval property
the length of
thus the is a singleton say
then

Claim.

For every every
By continuous of hence
similarly, for every
by continuity of at
so,
Hence

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Proof. 2

consider

is non empty as
is bounded by
by the limit upper bound property has a supremum
Take

Claim.

for consider .
As is not a supremum of there exist
such that

So
Hence
in particular
Now consider
note for any
let

as
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