Continuity Problem

[uncut fractal]

Help me prove this

We have to prove that for all a, lim y-->a g(y)=g(a)=lim y-->a f(y) i.e., lim y-->a [g(y)-f(y)]=0

Which amounts to saying a-del<y<a+del and y ≠ a implies f(y)-eps<lim x-->y f(x)<f(y)+eps

But what if graph of f looks like this

In this case f(y) can't be made arbitrarily close to l

[carmine rain]

Fix a in the domain. Let eps > 0. There exists delta such that for every x we have
0< |x-a| < delta = > |f(x) - g(a)| < eps/2

Take b such that 0 < |a-b| < delta. Then there exists z_b != b such that 0 < |z_b - a| < delta and |f(z_b) - g(b)| < eps/2. So from the above we get |f(z_b) - g(a)| < eps/2.

Now conclude with a standard triangle inequality trick

@uncut fractal

[uncut fractal]

Didn't get it

[carmine rain]

i don't expect you to in the first 2 minutes

think about it and fill in the details

it's not without reason the exercise is marked as *

[uncut fractal]

How do we know that such z_b exists?

[carmine rain]

well, what do you know about f?

for every element y

[uncut fractal]

Well, nothing

Just that every limit of it exists

[carmine rain]

that sounds like something to me

[carmine rain]

if we didn't have this there are obvious counter examples to the initial claim

so you had better use this assumption in your proof

[uncut fractal]

Ofc I am trying to use this, but idk how to get past this problem

[uncut fractal]

.

[carmine rain]

that is what I adressed

you pick a point that is close enough to the removable discontinuity

[uncut fractal]

Maybe I am being stupid, but can you explain how? How is this true when f(y) jumps

[carmine rain]

that cant happen

all discontinuities are removable

[uncut fractal]

Isn't it possible to define f(y)>>lim x-->y f(x) for some arbitrary y

[carmine rain]

what does >> mean?

[uncut fractal]

Very large, informally

[carmine rain]

then y is not removable

[uncut fractal]

Why

[carmine rain]

consult the definition please

[uncut fractal]

A point of discontinuity a is removable if lim x-->a f(x) exists but not equal to f(a)

Isn't it

[carmine rain]

yes, but your function is defined as the limit

so what you're probably getting at are one point jumps

[uncut fractal]

Yes, perhaps

[carmine rain]

we don't care about these, the claim is that if you define g the way you do you obtain a continuous function

the claim isn't that g(y) = f(y) at a removable discontinuity

[uncut fractal]

This was my initial question

I think it is clear

@carmine rain Is the paraphrasing correct

[carmine rain]

yes

you exclude the jump

you're saying that if you're close enough to a, but not at a, then the function values are close to the limit

what you want to show is that for |x-y| small enough around some fixed point a in the domain you have

$$|\lim _{t\to x}f(t) - \lim _{t\to y}f(t)| < \varepsilon$$

[marble swallow]

aL

[carmine rain]

all this is is just an unentanglement of the above

[uncut fractal]

You still didn’t justify existence of z_b

[carmine rain]

I'll leave that to you

[uncut fractal]

Gn

[carmine rain]

nini

[uncut fractal]

Suppose lim y-->a f(y)=l, we would like to have lim x-->y f(x)=l as well.
First observe that for all eps there is a del for which a-del<y<a+del implies |f(y)-l|<eps, if it were not the case the limit wouldn't exist. In other words, f(y) behaves nicely on the interval (a-del,a+del)
Thus for lim x-->y f(x) for all eps we can choose del'=min(y-a+del,y+a-del), this choice ensure that x is also in (a-del, a+del), hence |f(x)-l|<eps. Which shows that lim x-->y f(x)=l

@carmine rain check this out

[carmine rain]

let f(x) = x and redefine f(0) = + inf. What you're saying is if I take a != 0, then lim _{t to 0} f(t) = lim _{t to a}f(t) = a

that will not work

"Suppose lim y-->a f(y)=l, we would like to have lim x-->y f(x)=l as wel"

if |a-y| is small, then you would like the two limits be close to each other

use the simple example I gave and think about how to pick delta such that so and so..

[uncut fractal]

This?

It is only simple to you, unfortunately

[carmine rain]

simple because it contains only one discontinuity

the point is, do not ever invoke f(a) at a discontinuity point a

[carmine rain]

keep practicing and it'll be simple to you too

[uncut fractal]

What should I practice again? It has already been three days I am stuck on this

[carmine rain]

definition of limit for starters

one of your questions was about "why does such z_b" exist

it follows from the definition of limit

or alternatively, if no such point exists, then the discontinuity is not removable, which stands in contradiction with the assumption

[uncut fractal]

I showed that lim t-->a f(t)=lim x-->t f(x)=l for all a-del<t<a+del and a ≠ t, is it really the same as what you said?

[carmine rain]

your notation is overloaded, in one case t is changing, in the other case it's fixed

you should also be more precise. Let a be a discontinuity of f. You want to pick two points x,y around a such that |lim _{t to x} f(t) - lim _{t to y} f(t)| < eps

[uncut fractal]

First fix an eps, then there is del>0 such that if a-del<y<a+del then |f(y)-l|<eps; y is the dummy variable here, so we'll use y' to indicate a fixed number in (a-del,a+del)-{a}, so |f(y')-l|<eps. For this eps and y' if we consider all x in (y'-del', y'+del')-{y'}, where del'=min(y'-a+del,a+del-y'), but then x is also in (a-del, a+del)-{a} so |f(x)-l|<eps

[carmine rain]

l = lim f(t) as t to a, correct?

why is this number positive?

[uncut fractal]

Let me correct it plz

[carmine rain]

i gtg work anyway, buenos dias

[uncut fractal]

y' ∈ (a-del, a+del)

[uncut fractal]

@carmine rain I have proved that for every individual y in (a-del,a+del)-{a} f(y)-eps<lim x-->y f(x)<f(y)+eps holds, should I accept it as a valid proof?

[carmine rain]

that is correct provided your delta is small enough

because every discontinuity is removable the existence of such delta is also guaranteed

earlier you proposed this

but that can't possibly be true

[carmine rain]

add a triangle inequality argument to conclude continuity of g

[uncut fractal]

We have |f(y)-lim y-->a f(y)|<eps/2 and also |lim x-->y f(x)-f(y)|<eps/2, thus triangle ineq yields |lim x-->y f(x)-lim y-->a f(y)|<eps i.e., |g(y)-g(a)|<eps

[uncut fractal]

But for every eps>0 I have chosen a del'>0 and satisfied the limit condition, still why?

[carmine rain]

if a < y, then f(x) = x is a counterexample, can even be left continuous

[uncut fractal]

By f(x)=x do you mean the identity function?

[untold ingot]

@uncut fractal

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[untold ingot]

@carmine rain The user still needs help with this help request.

[uncut fractal]

OK, so your claim is that if a<y, then f(x)=x is a counterexample, which mean the argument doesn't work

[marble swallow]

ASR is Mathematicoid

ASR is Mathematicoid

ASR is Mathematicoid

[uncut fractal]

@carmine rain What is the dubious point

[carmine rain]

if a<y then y-a > 0, in particular |a-y| >= (y-a)/2

this is false, I don't need to see the rest

[uncut fractal]

It should be \forall y,(0<|y-a|<...) btw

[carmine rain]

in that case you choose delta = epsilon and that's it

there's nothing else to prove

[uncut fractal]

But if this is false that would mean lim y-->a y ≠ a

Since the limit condition won't hold

[carmine rain]

I'll remind you that what you claimed was this

$$a<y \overset{?}\Rightarrow \lim _{t\to a}f(t) = \lim _{t\to y}f(t)$$

[marble swallow]

aL

[carmine rain]

not true

[uncut fractal]

It should be lim t-->a f(t)=lim y-->t f(y)

[carmine rain]

overloaded notation

t is variable and t is fixed, you can't do that

[uncut fractal]

Ok, you accepted that my argument was correct, so how can I express it in notationally correct way

[carmine rain]

which argument?

[uncut fractal]

.

[carmine rain]

same question stands, I didn't say it was correct

[uncut fractal]

Ig we're both at an impasse

[carmine rain]

this statement is true

but for some reason that's not what you are expressing with your later claim

[uncut fractal]

Ok is there a correct notation for writing it

[carmine rain]

1. Let a be a removable discontinuity. Then there exists delta small enough such that (as you say)

for all y in (a-del, a+del), y != a implies |g(y) - f(y)| < eps

[uncut fractal]

But I have corrected this point later, it should be del'=min(y-(a-del),(a+del)-y)

[carmine rain]

how does this even work

you have fixed y in (a-del, a+del) and then you're defining a new del' depending on y?

what are you even proving anymore

[uncut fractal]

First I would be interested to know what's wrong with the argument then think about what can I deduce from it
For some reason I cannot get this outta my mind

[carmine rain]

what is the claim that this proof is for?

i can't understand what is going on

[uncut fractal]

What's unusual about this, I am deciding which number f(x) approaches as x gets closer to y

[carmine rain]

[uncut fractal]

🤦‍♂️

[carmine rain]

answer this question please

[carmine rain]

you are not deciding, we already have that this number is g(y) because by definition g(y) = lim f(t) as (t -> y)

[uncut fractal]

For every eps>0 there is del such that for all y 0<|y-a|<del implies |f(y)-l|<eps, then for every y satisfying this, f approaches l near y

[carmine rain]

you can't decide that it's approaching something else

[uncut fractal]

Gtg, coming back after 5 min

[carmine rain]

yes and l = g(a), specifically

this is in no way a justification for this to be true

I think you are confused at the quantifier level

what is true is this

$$\lim _ {y\to a}(\lim _{x\to y}f(x)) = g(a)$$

[marble swallow]

aL

[uncut fractal]

Is this the correct notation I was asking for

[carmine rain]

I have no way of knowing

I still don't know what it is you want to prove

[uncut fractal]

I don't know, either

[carmine rain]

that is unfortunate

luckily we know what needs to be proved

and you have done that already, whether by accident or not

[marble swallow]

ASR is Mathematicoid

ASR is Mathematicoid

ASR is Mathematicoid

[carmine rain]

you already have information about what it approaches

[uncut fractal]

Basing on that information we have to prove this

[uncut fractal]

It should be delta' actually, no?

[uncut fractal]

We have used delta once here

[carmine rain]

that's fine, you can distinguish between them if you need them both in the same argument

[uncut fractal]

Now we have to do something with lim x-->y f(x)

[uncut fractal]

In this case, we can let $\delta'=\min(y-a+\delta, a+\delta -y)$

[marble swallow]

ASR is Mathematicoid

[carmine rain]

careful

this delta ' is valid for only one fixed value of y

[uncut fractal]

It can't be helped because we have to involve lim x-->y f(x) in the proof

[carmine rain]

your y is changing

oh it can be helped

[uncut fractal]

How?

[carmine rain]

I told you at the very start, it is enough to find a point z "close" enough to a such that

|f(z) - g(a)| < eps/2 and |g(y)-f(z)| < eps/2 for all y in some (a-del, a+del)

you have effectively done this already

[uncut fractal]

What's that z

[carmine rain]

that is what I meant by this reply

[uncut fractal]

This is correct, right?

[carmine rain]

you are making the same mistake with the overloaded notation

don't write it like this

$$\lim _{y\to x}f(y)\qquad \lim _{x\to y}f(x)$$

[marble swallow]

aL

[carmine rain]

in the same sentence

your y is fixed, it doesn't change

gtg work anyway, you're basically there, just be careful about not overloading your notation and always think about the quantifiers in your statements

[uncut fractal]

Which amounts to saying that
It is enough to find a point z such that

|f(z)-lim x-->a f(x)|<eps/2 and |lim x-->y f(x)-f(z)|<eps/2 for all y in some (a-del, a+del)
But if we have to deal with all y in (a-del, a+del) then y must change, which is the problem you're addressing with my argument

[uncut fractal]

@carmine rain