You and yours doing ok?

although i much much doubt in one part of one sentence written there

@CaptainAmerica16 Didn't I send you the exercises and stuff?

@Ante: Here's a cool exercise (from Spivak's Calculus). Any function with the intermediate vslue property is continuous provided it takes on each value finitely many times.

so, to prove is IVT+f^{-1}(c) is finite for every c => continuous, right?

very nice exercise

IVP, not T

Yes. Do you see why that finiteness condition is required?

suppose it has IVP, if it takes some value an infinite number of times, let that value be d, then f^{-1}(d) is infinite, since it is infinite it has accumulation point, say c, so there is a sequence c_n, converging to c such that f(c_n)=f(c) for every n, and also, for every m,n in N it is true that f(c_m)=f(c_n),

but then for every m,n in N the function takes on all values between f(c_m)=d and f(c_n)=d, so it is constant between those points, so the constant is the only continuous function that has IVP and takes some value more than finitely many times?

whoops, what did i do there?

is this weird argument?

i can make it more rigorous, if needed

@TedShifrin Yeah, I never went through all of them.

Meaning I started, but didn't finish

Sorta like you, @Captain 😈

;-;

@Ante Your logic is flawed.

where?

If the function values are the same, you know nothing about the function in between.

i know, since it has IVP, as assumed

That was back when I was a child.

Huh?

There are lots of continuous functions that take the same value infinitely often.

Now, I make To-Do lists.

The point is to have a discontinuous one with the IVP.

And lists of your lists, @Captain?

Actually, yes. :/

I make school lists and then each task has sub-tasks

Wait, I don't think that's what you meant.

This could be an infinitely nested loop.

The IVP tells you that the range contains an interval, not that it's equal to that interval, so you can't conclude that it would be constant

Considering I rarely finish what I have planned for each day. Tasks forever carry over to tomorrow.

it is piecewise-constant, seems to be at least

nah

Write down IVP carefully. You're misunderstanding it, as Thorgott said.

could be, but something is proven there

although i do not know what exactly

shrug

this is written on Encyclopedia of mathematics:

*(the Darboux property can be characterized in terms of images but not in terms of pre-images). *

is this really true, and why?

If two points are in the image, so is the interval with those endpoints. The comparable statement about preimages is false, but it seems to be what you keep asserting.

hm

yes

i am thinking that also something with pre-images can be done in the sense of characterization

No.