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Real Analysis | Continuity, connected sets, and the IVT.

we're continuing to look at the

interaction of continuous functions with

the topology of the real line

and in this video we're going to look at

connected sets

let's go ahead and recall what a

connected set is so a set

a of the real numbers is said to be

disconnected

if there exists e and f also subsets of

real numbers

such that a is the union of e and f

and in fact it's the disjoint union of e

and f

plus some extra stuff and that extra

stuff is that

e intersected with the closure of f is

empty

and the closure of e intersected with

f is also empty so

in other words e contains

no limit points of f

and vice versa in other words

f contains no limit points of e just

recall that the closure of a set is the

set

union its limit points and then next we

say that a is connected if it is

not disconnected and i also want to

recall the intermediate value theorem

from a calculus one type class

and that's because the big theorem that

we're going to prove in this video

will imply as a corollary this

intermediate value theorem

so the intermediate value theorem says

that if you have a continuous function

on a

closed interval that closed interval

we'll call it a to b

and the value of the function at the

endpoints

is not the same in other words f of a is

not equal to f

b then for every y naught

between f a and f b there is

an x naught on the open interval a to b

such that f of x naught equals y naught

so notice what we've got going on here

is y naught is like our value which is

intermediate to the value on the

endpoints

and this says that we are able to

achieve that intermediate value

somewhere on the open interval okay

great

now we're going to prove this following

theorem and we'll show how that

implies this as a corollary at the end

so the theorem that we want to prove

says that if we have a continuous

function

on a and so this is continuous at every

point

in a and a is a subset of the real

numbers

and then we have a subset of a which

we'll call

b and that's a connected subset

then f of b is also connected so in

other words the image of a connected set

is connected

okay and so we want to prove that using

this definition of uh

connected sets so let's suppose that

we've got something that looks like a

disconnection of f

of b and show that in fact it isn't one

so limit points from one set will be in

the other set or vice versa

so what we're going to do is suppose

that we can take

f of b and rewrite it as e

union f with e

intersect f equals the empty set

so notice you can do this for any set of

real numbers

just like pick a point in the set and

take everything above that point and

everything less than or equal to that

point

that's most definitely going to be this

kind of setup you can always write a set

of real numbers

as a disjoint union but

you may not be able to do it in a way

that the closures don't overlap like

this so let's go ahead and point out

what we want to show

so we want to show that f of b is

connected in other words we want to show

that

e closure intersect f is

non-empty or

e intersect f closure is non-empty

in other words we don't have a

disconnection and so that means we have

a connected set

okay so let's get to it so we want to

get back to the set b because we know

something about that set b

we know that that set is connected so

let's do that using inverse images

so let's set c equal to the inverse

image of

e but we don't want the entire inverse

image of b we want the inverse image of

b which is inside

of e which is inside of b so we'll just

intersect that with b

and that's because e may contain some

stuff that also comes from

outside of b just if f is not a one to

one function or something

and then similarly we're going to set d

equal to

the inverse image of f intersected with

b

now what we want to claim

is that b is equal to c

union d so let's see how we can do that

so let's suppose that x

is in b now we want to show that x

is either in c or d but now notice the

fact that x is in b

that tells us that f of x is inside of

f of b the image of b but we know

the image of b is equal to e union

f but by the definition of union that

means that

f x is in e or

f x is in f

but that means that x is in f

inverse e the pre-image of e

or x is in f inverse of f

the pre-image of f but the fact that we

have it's already in

b and then it's either here or here

that tells us that it's in c or

d because we have it's in b and it's an

f

inverse of e but that and statement

gives us an intersection

or it's in b and it's in the inverse

image of

f but again that and statement gives us

an intersection which is this

set d here so here we have x is in

c or x is in d but that tells us that x

is in c union d

so we finished the proof of this

claim great and so now let's go ahead

and maybe clean up the proof of this

claim

and then we'll move on to the next step

so we just constructed two sets

c and d related to our function

f whose union is b which we have assumed

to be a connected set

so the fact that b is connected that

tells us

that c intersected with d bar in other

words the closure of d

is non-empty or c bar

intersected with d in other words the

closure of c intersected with d

is not empty so let's maybe write that

down so

either c bar intersect d

is non-empty or c

intersect d bar is non-empty

so without loss of generality let's

maybe

assume that c bar intersect

d is non-empty it's going to be

exactly the same if we have the other

case so it's safe to do this

so notice what that means is that d

contains a

limit point of

c remember c bar is equal to c

and then all of its limit points but

notice that that tells us

there exists some sequence we'll call it

x

n n goes from 1 to infinity totally

contained in

c such that the limit as n goes to

infinity of x sub n

equals x is inside of

d and then notice

since this is a limit of a sequence of

guys that are inside of c that also

tells us that x

is inside of c bar so we proved that a

long time ago that if you have a limit

point of a set

then you can construct a sequence

that converges to that limit point okay

great

and then now we haven't used the fact

that f is continuous so we better do

that

so since f is continuous

we have the limit

as n goes to infinity of f of x

sub n equals f of x

so this is the sequential version of

continuity

but now what i want to notice is that

this

sequence defined by the function

and our original sequence so in other

words the sequence of numbers

f of x sub n that's going to be

inside of e

because it was inside of the original c

and notice that puts it inside of e

after evaluating it at the function

but now since this is a sequence of

numbers

in e that tells us that the limit of

this sequence of numbers

is in the closure of e we'll call that e

bar

and then also since

x is in d from our

earlier argument we have f

evaluated at x is inside of f

because of our construction of d up here

so let's see what we have

we have f of x is in e bar

and f of x is in f

which tells us that f of x

is inside of e closure intersected with

f which must be a non-empty set if it

has an

element so we tried for a disconnection

of our image of b f of b

and we ended up finding that such a

disconnection was impossible

so in other words the image of a

connected set is connected

okay so let's maybe get rid of this

proof and then we'll look at how this

implies the intermediate value theorem

now we're ready to use this general

theorem in order to prove the

intermediate value theorem

so let's recall that the intermediate

value theorem says that if we have a

continuous function

on a closed interval a b and

f of a is not equal to f of b then

for every y naught between f of a and f

b

there is an x naught on the open

interval a b

such that f of x naught equals y naught

okay so let's get to the proof

so let's first notice that

this closed interval a b is connected

and it's compact so recall that compact

in the real numbers means

closed and bounded so that's most

definitely closed because it's a closed

interval

and it's bounded because it is of finite

length

so notice that that tells us that the

image we'll call that f of a b

like this is also connected and compact

so also

connected and compact

so it's connected because of this

previous theorem that we just got done

proving

and it's compact from a theorem in a

previous video which says that the image

of a compact set is compact

now if we've got a connected and a

compact set of real numbers or i should

say

a set of real numbers that is both

connected and compact

that implies that it's a closed interval

so that tells us

that f of a b

equals c d

for sum

c and d and r so again it's a closed

interval

another thing that we want to notice is

that we also know that

f of a is an element of c d

and f of b is also an element of

c d so why is that well that's because f

a is most definitely in the image of f

of the interval a b and similarly for f

of b

now the next thing that we're going to

do is assume

that f of a is less than f of b

so it may seem like we're losing

something here but we're really not

and we can do that by replacing

f with the function negative f as needed

so notice f and negative f are

essentially the same function

except what happens is you're going to

change the ordering here

good and then the next thing that we

want to notice is that the open interval

f of a

to f of b is totally contained in this

closed interval

c to d great and now

next we want to take a y naught in the

open interval

f of a to f of b which like i said is a

subset of this closed interval c

to d which is equal to the image

of f of a b and notice the fact that

it's in the

image of f of a b that tells us that

there exists an

x naught in the closed interval a b

such that f of x naught equals y

naught so that's the definition of being

in the image of that set

a b great and then also

we want to notice that x

naught is not equal to a or b

because we have y

naught is not equal to f of a or f b

so let's say y naught is not equal to f

of a

or f of b by our construction via this

open interval right here and that's

exactly what we needed to finish

in order to prove this intermediate

value theorem from our more general

theorem

and that's a good place to stop