Cool!

Still like "wiskunde" the best.

@BrianM.Scott Honorable Sir.

@PeterTamaroff Jawohl, mein Herr.

@BrianM.Scott Wie gehts dir?

@PeterTamaroff At the moment I'm seriously considering taking a nap!

@PeterTamaroff I prefer that you address me as "Grandmaître". Do you think you can do that? Good! Thanks!

@Peter: Psst! Try Grandmère; if you speak quickly, the difference is hardly noticeable!

Oh no!

@BrianM.Scott Well, naps are nice.

@BrianM.Scott But can I ask you a quick question? =D

Go ahead.

@BrianM.Scott Today I had some free time so I went ot my uni's libray. I took a good analisys book (Rey Pastor) and it had the proof that every closed interval on $\bf R$ is "compact".

He used the bisection method.

I've also said that some hours ago, Peter. Didn't look so quick after all 8-)).

But I couldn't get the last part.

@JonasTeuwen Whut?

How do you define compact?

@PeterTamaroff That being?

@BrianM.Scott Well, the statement was the following

If to each point $x$ of $[a,b]$ we assing a nbhd $N_x$ of it, there exists a finite number of these such that every poiont of the interval $[a,b]$ has one of these as a nbhd.

The proof is by contradiction: Suppose no finite amount suffices to "cover" al points of $[a,b]$. Then there must be a half where these "rogues" are, call it$[a_1,b_1]$.

Huh?

(We choose the left side if both halfs have "uncovered" points.

What if you just set $\epsilon = (b - a) N^{-1}$ and actually just take $[0, 1]$?

In this half there, a half must have also uncovered points, call it $[a_2,b_2]$

Then divide the interval in $N$ pieces, and take $N_x$ with radius $\frac32 \epsilon$?

Keep on and construct a sequence of fitted intervals and apply the lemma of iftted intervals.

What...?

there is a $\xi$ in every $[a_n,b_n]$

Your statement probably is missing some "unique" or so.

But the $N_\xi$ that we assigned by hypothesis must contain all the $[a_n,b_n]$ after some $n$.

@PeterTamaroff It’s not clear what you mean by rogues.

We had assumed we couldn't cover tehse points with any finite amount of nbhds, but we have covered them with one, thus our original assumption msut have been wrong.

@BrianM.Scott Sorry, I mean the points that are not covered by the finite cover.

@PeterTamaroff But what finite ‘cover’? (It’s not actually a cover.)

I do not fancy the contradiction part here.

@BrianM.Scott The theorem states that if to each $x\in[a,b]$ we assign a nbhd $N(x)$ then there is a finite amount of points $x_1,\dots,x_n$ such taht $\bigcup N(x_i)$ covers $[a,b]$-

I know. But you’re trying to prove that this is possible, so you don’t have any finite collection at the moment.

Pick a cover of $[0, 1]$. Can be easily refined to countable one, right ($\mathbb Q \cap [0,1]$ is dense in $[0, 1]$)?

@BrianM.Scott Yes. The proof is as follows: For the sake of conttadiction, we assume that there are uncovered points. Thus, there must be uncovered points in either one or the other halve of $[a,b]$, or both.

Also, you can just apply a continuous mapping so you talk about $[0, 1]$. (linear even). No need to carry $a$ and $b$ around.

Uncovered by what?

Son of a cow, I'll help somebody else.

@BrianM.Scott We assume no finite number of $N$s can cover the interval.

That is, no finite cover exists.-

Thus any finite number of $N$s msut not cover some poiunts.

I think peter means that suppose, it is not true. so suppose there is an infinite cover (no finite cover), then either $[0,1/2]$ would have the infinite elements of the finite cover or $[1/2,0]$ will have the infinite elements of the finite cover. Go on applying bisection to the resultant interval, finally you get a single point by finite intersection property and hence contradiction.

@JayeshBadwaik No no, we assume no finite number of $N(x)$ can cover the interval

@PeterTamaroff Agreed. But that doesn’t make sense of the rest of the argument as you’ve written it. You want to say that some points are uncovered, but you never say by what. The fact is that if you give me any finite set of points, I can find a finite subcollection of your original cover that does cover those points.

@BrianM.Scott You have a "direct". proof?

I am trying to prove Fermat's point.

@PeterTamaroff I have several proofs. I’m trying to figure out what argument it is that you’re trying to describe.

@hhh I am sorry to inform that you have been scooped by Andrew Wiles. Sorry.

@BrianM.Scott Can you read Spanish?

@PeterTamaroff Not really, but in this context I might be able to figure it out.

I cannot understand where I get the 120 degree condition.

@BrianM.Scott It is "Spanish" strictly, from Spain, so maybe it is easier.

"Entorno" means nbhd, so $e$ is short for "entorno".

"If to each point of a closed interval $[a,b]$ we assign a nbhd of it, there is a finite number of these, such that each point of the interval has one of these nbhds as a nbhd of itself."

"Suppose, for the sake of a contradiction, that for each finite set of such nbhds $n$ there are uncovered points $x$; bisecting $[a,b]$, in one of its halves $[a_1,b_1]$ there must be uncovered points $x$ (if both, choose left)"

@PeterTamaroff That’s what I thought it said. Frankly, the argument doesn’t make sense. I’ll have to think a bit to see whether I can turn it into something sensible.

@BrianM.Scott OK. Thanks anyways.

@JayeshBadwaik :D.

@JasperLoy i have thunk about it.

@JasperLoy :-) Its a more proper answer than the other one which got upvotes. It does not answer the OP's doubt.

@JasperLoy An OP has a reputation below ten (I think it is) can’t upvote; he can only accept.

@BrianM.Scott no, he has rep 219.

What you are seeing must be the monthly/daily figure (that is the one which comes up in the search I guess)

@JasperLoy =)

I have no clue how much rep I have.

@JonasTeuwen 6184

Thank you. That does not sound much.

But if I would be able to exchange it for euros I would be really happy.

@JonasTeuwen :-)

6184 years?

I prefer not.

That's where the critical branch got chosen?

Scary stuff!

This is a nice time-travel movie.

Poeh! Time travel? Do that every day. Tell me something I don't know.

@JonasTeuwen No, this is a little different in that it is not sci-fi

Neither is my time travel.

@JasperLoy You can do that too!

(simple question about vector calculus)

Then $x$ is hell. Or a black hole if you prefer those.

(Means you die. Quick! Run!)

I don't like yoru tone, sorry stupid q but I find this improtant detail.

Actually I think the mistake is in the first statement, it is a trivial zero vector, scalar namely there.

ofc I need to just check that the unit vector sum is not a zero vector, have to do that to marginal, sorry noise.

Wooow

I gave the answer, read between the lines. I don't like your tone either, this is a chat room not the Q&A site math.SE. If you get division by zero for some cases, consider them separately.

@JayeshBadwaik Will milk make me sound like Louis Armstrong?

@JonasTeuwen No. Whisky is always smoother.

That makes you sound like Tom Waits. Do you call that smooth?

Need some Chocolate Jesus'.

Hell Yeah.

Running now!

:D :D

"The operation is a go!"

Waltzin mathilda....!!!?!

Be happy :D

@JonasTeuwen Roberto?

Yes! Sir!

@hhh Also, you might be an asshole yourself and that makes others asshole. Feedback etc etc. Coffee from assholes for assholes.

@JonasTeuwen That is right :P ..that is why it is funny :)

Here the zero -point had anyway fascinating interpratation: the subgradient is an unit ball, solved :D

Good!

@JasperLoy I will not call him anything.

Funky!