i guess one way would be that [0,1]^2 is the square, (0,1)^2 is its interior, and [0,1]^2 \ (0,1)^2 is its boundary

Is that point (sequence) an element of $\ell^1$?

No.

Then we're done.

Wasn't that what I showed?

No, you said it wasn't a convergent sequence.

Which means that it isn't a point of $\ell^1$?

Well, indirectly, I guess that's right, because we proved that anything in $\ell^1$ must converge to $0$. I was looking for something precise.

Okay.

The main problem is the latter part of the problem.

So what does it mean to have a separating hyperplane?

I actually don't see why this is true.

That there's a hyperplane $H=\{x\in E:f(x)=\alpha\}$ such that $f(x)\leq\alpha$ for all $x\in A$ and $f(x)\geq\alpha$ for all $x\in B$ if $H$ separates $A$ and $B$.

What are you asserting about the $f$ there?

@Semiclassical is the plan for this question to deduce which identity can be found first to quickly reduce it to 2 characters and then it becomes clearer?

You also shouldn't have $\le$ and $\ge$, should you?

I dunno; I can't read the mind of the person who wrote it. But that's how I approached it.

@Semiclassic: I did it by immediately showing A couldn't be a liar.

But you derailed me.

If we have strict inequalities Brezis say that it's strictly separating.

shrug

Oh, @Oskar, so points of both $A$ and $B$ can be in the hyperplane? Doesn't feel very separate to me. But OK.

We can consider the strict case. It indeed makes more sense.

I dunno what's correct in this case.

I used to do this problem with only knight and knave, so trying to accept the fact that that there can be a spy doing both ...

Let's assume that it's strict.

@OskarTegby are u talking about the puzzle?

Anyway, did you tell me what we know about $f$, @Oskar?

No.

That's the question, @Ted.

I don't know. A hint, maybe?

It's not just a random function. Hyperplanes are linear.

Oh! Of course.

And, in fact, don't you need a continuous linear function?

Of course.

Is it topology time?

So I think the point is that any linear function that vanishes on $Y$ must take both signs at points of $X-c$.

@TedShifrin Brezis doesn't require it to be continuous in general (but he proves that the hyperplane is closed iff the functional is continuous and always talks about separation by closed hyperplanes of course)

Is this the correct logic?

@Alessandro: You confuzled me.

Eh, I think it's easiest to start with "who can be the knave"

So does this problem intend a closed hyperplane? :P

It does

BTW, @Alessandro, feel free to help out here. You've thought about all this stuff quite recently.

Problem: Let $\{a_n\}$ be some real sequence, and let $s_n := \sum_{k=1}^n a_n$. If $s_{2n}$ and $s_{2n+1}$ converge to the same number, $s_n$ converges. Proof: Given $\epsilon > 0$, we can choose $N \in \Bbb{N}$ such that $|s_{2n}-s| < \epsilon$ and $|s_{2n+1}-s| < \epsilon$ for all $n \ge N$. We want to show that $|s_k - s| < \epsilon$ for all $k \ge N$....Here's where I run into trouble...If $k$ is even, then $k=2n$ for some $n \in \Bbb{N}$; but that doesn't necessarily mean $n \ge N$...

Why would it vanish on $Y$? Why would it take both signs at $X-c$?

How do I get the indices to 'match'?

Did someone say convergence?

@Semiclassical oh wow ur starting point seems clearer, do u start with either finding the knave or knight first?

like for a general case?

@TedShifrin I'm not really sure how to approach this separation thing apart from having a feeling they can't be separated

Boy, I'm all about convergence, but let's stay focused here.

I started with the knight. But in that case I found that I could only eliminate Kenyu as the knight

LOL @Alessandro: Well, great!

So I moved on to the knave.

Answer: "I having a feel that they're separate."

@user193319: Stop using the letter $N$ everywhere.

oh so your first attempt didn't locate the Knight, just eliminated a possibility which doesn't do good?

Right.

That's still useful, but it wasn't definitive.

@Ted: Care to elaborate on the vanishes on $Y$ and have both signs in $X-c$ statement?

right ...

So then I asked who the knave could be, and that criterion worked nicely.

@TedShifrin I don't understand. That's the index I am using to show that $s_n$ converges to $s$; the one I get from the fact that $s_{2n}$ and $s_{2n+1}$ converge to $s$.

It doesn't help, though, @Oskar. It's just intuition. Consider the function $f(z) = z_{2k}-z_{2k-1}/2^k$ (for some fixed $k$).

duh, just take the maximum

I think in terms of strategy it's best to leave the spy to last, since they can be either true or false

Not sure whether knave v. knight is a better starting point in general.

Don't use it everywhere, @user193319. So you know you win for $s_k$ when $k\ge 2N$ and when $k\ge 2N-1$?

So $2N$ will be the index I need to show that $s_n$ converges to $s$?

That would indeed be zero on $Y$.

You need to write sentences that make sense, @user193319. You need some $K$ so that $|s_k-s|<\epsilon$ when $k\ge K$.

Yes, and my initial choice was $K = N$. But you are suggesting that I take $K = 2N$?

Both signs on $X-c$. Hm...

I'm suggesting you take any $K$ that actually works, @user193319.

$2N$ seems to work.

So does $2N-1$. So does $2N+739$.

@PrashinJeevaganth as a general point, I think it's handy to make a 3-by-3 grid to keep track of what possibilities you've eliminated/confirmed e.g.

We only know that $z_{2n}=-\frac{1}{2^n}$.

But can't you choose $z_{2k-1}$ to make that function positive? and choose it to make it negative?

Yeah! Exactly, I was about to write that.

What does this mean then?

If $f$ is the hyperplane function.

Intuitively, it means $Z$ is on both sides of that hyperplane containing $Y$, so it certainly didn't separate.

It doesn't separate because...

Yeah!

But I don't know what to do with an arbitrary $f$.

o..o

Note that $f$ could involve an infinite summation.

@mercio: You said something about $\overline{X+Y}=E$ and $X+Y\neq E$ with some dual to $\ell^1$ as the hyperplane.

and I said $X$ and $Y$ were subspaces

(and I did not say that the hyperplane was a dual to $\ell^1$)

Now, we only showed that there exist a hyperplane that doesn't separate them. That wasn't the question. Right? It was to show that there was no hyperplane that separates at all.

but if $f$ is a linear functional, what kind of sets can $f(X)$ or $f(Y)$ be ?

@Semiclassical noted, but I will hope I always get a definitive result on the first try if not its utterly demoralising

Not sure.

 _______|_Aiken_|_Dueet_|_Kenyu_|
 Knight |       |       |       |
 -------------------------------|
 Knave  |       |       |       |
 -------------------------------|
 Spy    |       |       |       |
 --------------------------------

there we go

can they be any subset of $\Bbb R$ ?

(or $\Bbb C$ if you are working over $\Bbb C$)

You fill that with X's (to indicate an option that isn't allowed) and check marks (to indicate one that's confirmed)

There's one check per column and per row

Are we going from sequence space to $\Bbb{R}$ now?

That's overkill in the last two problems, since you could deduce it just based on one row (e.g. knight first in the first problem). but it's a good way to organize your info in such deduction problems

What's an advisable way to approach this problem?

:46926699

I guess they can?

Given that "none of the above" is an option, I see no alternative to just going statement by statement

@mercio: So you seem to be arguing that we need to choose $f$ to vanish on $Y$, like, for example, in my example.

I guess so (and on $X$)

I haven't had time to read on all of what you have said

So $f(Y)=0$ but $f(X)\neq0$?

@OskarTegby in this message what kind of function is $f$

A continuous linear functional.

which is a linear function into ... ?

$\Bbb{R}$?

yes

Are $X$ and $Y$ disjoint, @Oskar?

Yes.

no

I said $X$, not $Z$.

Right.

They intersect in $0$.

Mea culpa.

Of course.

But that's it, right?

And you proved the closure $X+Y$ is everything.

Weren't we talking about separation of $Y$ and $Z$?

Yes, but @mercio suggests you think about $f(Y)$ and $f(X)$.

Okay.

if there is going to be an $f$ that separates $Y$ and $Z$.

<--- shuts up now

Sorry, what?

@Semiclassical Am I right to say that A is the inverse error, but I can't find a proper contradiction reason to say that B and C is wrong

its more like no such information exists

well if $f$ separates $Y$ and $Z ( = X-c) $ then there is a $b$ such that $f(X) - f(c) > b > f(Y)$

so I am suggesting you focus on what actually can $f(X)$ and $f(Y)$ be

for example take Ted's choice of $f$ from earlier

That maps $Y$ to zero, and $X$ to whatever.

So, just something that does that?

@PrashinJeevaganth well, if you can't conclude that B is true, then B isn't going to be a valid conclusion

so in that particular case, $f(Y) = \{0\}$ ?

and $f(X)$ is what exactly ?

Yes.

Anything that isn't zero?

Suppose I flip a coin. May I conclude that it comes up heads?

If it can be either positive or negative, then it doesn't separate.

are you sure there isn't

And if I can't conclude that, may I conclude that it does not come up heads?

you keep wanting to jump to conclusions

Okay.

are you sure there isn't an $x \in X$ such that $f(x) = 0$ where $f$ is Ted's example from earlier ?

@Semiclassical that was what I am about to ask, somehow when proving theorems when u contradict u often the one side and u r forced to deem that side as correct. For this you can stand on the fence and not make a remark rather than say a statement is true or false ...

No.

I might have phrased my question in a way that makes yes/no answers impossible to interpret

Well, what you often have in proofs are things like "A implies B, B is false, therefore A is false"

I think that there are $x\in X$ that can make it zero.

One way to understand that is that "If A, then B" is logically equivalent to "If not B, then not A"

yes I know about that contraposition relation

okay so then $f(X)$ is $\Bbb R$, right ?

Yeah.

On the other hand, if the premise A in "If A, then B" is false, then you can't conclude anything about B

so you have an exmaple where $f(X)$ and $f(Y)$ are among $\{0\}$ and $\Bbb R$

Yes.

can you find a linear functional $f$ with $f(X) = \{1;2;3\}$ ?

B could be false or true; its logical status is irrelevant to the truth of the syllogism "A implies B"

Uhh... I'm sure it's easy.

or is that impossible ?

Probably.

One way to look at is this problem: Given the premises that have been stated, which (if any) of the statements must be true? You're not concerned about whether they must be false, only whether they could be false.

Why wouldn't it be possible?

what do you know about linear functions ?

$f(x+y)=f(x)+f(y)$ and $f(ax)=af(x)$

If a statement could be false, then it isn't a valid conclusion and you can move on.

so what happens with $f(ax)$ if $f(x) = 1$ ?

$f(ax)=a$

is that always in $\{1;2;3\}$ ?

No.

do you have an example ?

Just $a=4$. Haha. Right?

yep

@Semiclassical so an argument can feel correct, but is not actually correct. Whereas in proofs we are really defining the truth value of the statement as a whole?

so if $f$ is a linear function such that $f(X) = \{1;2;3\}$

well you have a problem because you can pick $x \in X$ such that $f(x)=1$ and then $f(4x) = 4$ is not in $\{1;2;3\}$

Yeah!

in fact I would have liked you to say that $f(X)$ can only be a subvector space of $\Bbb R$

Okay.

@PrashinJeevaganth something like that. you don't want to confuse "a scenario that sounds plausible" with "a scenario that is logically required"

uuuh

what ?

ah

well it felt important to show him that $f(X)$ couldn't be just whatever he wanted

oh, blah

that there had to be some restrictions

Importantly, that $f(X)$ maps to $\Bbb{R}$. Right?

@Semiclassical what's the reason why C is not valid? Seems like a long chain

We have that $f(Y)$ maps to $\{0\}$ and $f(X)$ to $\Bbb{R}$. We don't have a separating hyperplane because...

because we never said that

so suppose we have proved that $f(X)$ and $f(Y)$ can only be $\{0\}$ or $\Bbb R$

what happens to $f(X) - f(c) > b > f(Y)$ if $f(Y)$ is $\Bbb R$ or if $f(X)$ is $\Bbb R$ ?

is it possible ?

I don't remember the context of those quotes and are they not talking about a single $f$ ?

Yeah. I just meant as an analogy.

@PrashinJeevaganth I think the simple point is: What about weird things which aren't reindeer?

what happens to $f(X) - f(c) > b > f(Y)$ if $f(Y)$ is $\Bbb R$ or if $f(X)$ is $\Bbb R$ ?

It isn't true.

okay so if $f$ is a linear functional that separates $X-c$ and $Y$

Any child loves santa, and therefore any child loves any reindeer. In particular, that child loves Rudolph, who has a red nose (and isn't a clown) and therefore is weird.

So any child loves at least one weird thing.

what must $f(X)$ and $f(Y)$ be ?

But it doesn't follow that said child loves every weird thing.

There's nothing stopping a child from not loving a weird frog, for instance.

ah, context

$f(X)>b+f(c)$ and $f(Y)<b$

but they have two possibilities each, $\{0\}$ and $\Bbb R$

they can't be anything else

so ?

Impossible?

o..o''

why ?

$f(c)>b>0$

NOT JUST IMPOSSIBLE , IT IS ORIMLIGT !

what's impossible about that ?

Nothing?

so ?

what must $f(X)$ and $f(Y)$ be ? (impossible is not an answer to what they have to be)

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