Mathematics - 2020-01-21 (page 2 of 2)

Ted Shifrin

7:03 PM

Hi, @Lukas. Félicitations!

Lukas Heger

Merci :)

manooooh

Hello all!!

Thorgott

That inequality doesn't always work for all $n$. Maybe you want $|\int|f_n|-\int|f-f_n|-\int|f||\le2\int|f-f_n|$?

Ted Shifrin

Howdy @manooooh.

manooooh

Question: we know that $p\wedge q\to p$

slapbot

7:13 PM

Hi there! I am actually looking for a term, I have googled for like past couple of hours and I had no luck so far.

So there exists a thing called, minimum vertex cover which implies " a set of vertices such that each edge of the graph is incident to at least one vertex of the set".

I'm looking for a similar concept only slightly different to the original, it is to find a set of vertices such that the set covers all of the vertices in a graph regardless of each edge is covered or not without any duplicates"

Lukas Heger

Hey @EdwardEvans

Edward Evans

yooo man

How's it going

Ted Shifrin

Oh, there's that person with the funny name.

Lukas Heger

really well :)

and for yourself?

Edward Evans

@TedShifrin sorry who are u

Ted Shifrin

7:13 PM

I forget.

Edward Evans

rofl

@Lukas yeah well too thanks, learning a load about cyclotomic extensions has given me a massive motivation boost :P

manooooh

Can we apply this to $(p\wedge q)\vee(p\wedge r)\to (p\vee r)$?

Or the OR operator is first?

Ted Shifrin

Huh?

manooooh

I mean we have two conjunction, so apply reduction on both of them

Ted Shifrin

I don't know what that means.

manooooh

7:16 PM

But I am not sure if this is correct since the disyunction is first

Lukas Heger

@manooooh you can apply the rule you mentioned along with this one: $((a \to p) \land (b \to p)) \to ((a \lor b) \to p)$

Ted Shifrin

Can't you regroup, if you want, and say this is equivalent to $p\wedge (q\vee r)$?

I HATE hate hate symbolic stuff.

Edward Evans

it's all made up by nerds like @Alessandro

Thorgott

that's de Morgan, isn't it

Ted Shifrin

hides in the corner

slapbot

7:18 PM

Sorry if I am being annoying, but just a last message to give some visual idea.

Graph: https://i.imgur.com/L4Tr9HC.png

Solution: https://i.imgur.com/C8Fr3j2.png

So the solution isn't a minimum vertex cover since all edges aren't covered, however the set of vertices do cover every other vertex in the graph without any overlaps/duplicates.

Ted Shifrin

@slapbot: I don't think we have any graph theory experts here. I personally know almost nothing.

Alessandro Codenotti

@EdwardEvans I hate that kind of stuff too

slapbot

Ah okay no problems, thanks for the headup tho @TedShifrin :)

Ted Shifrin

I just didn't want you to think we were being rude to you.

slapbot

Gotchu!

Leaky Nun

7:21 PM

@LukasHeger disjunction is the coproduct in the category of propositions

Ted Shifrin

smacks Leaky

Lukas Heger

@LeakyNun that's a good way to think about it

manooooh

@TedShifrin yes but I started from $p\wedge (q\vee r)$

Ted Shifrin

LOL, well, I'm not a mind reader.

manooooh

So that's why I ask if $(p\wedge q)\vee(p\vee r)$. Hahah don't worry I didn't clarify that

Ted Shifrin

7:22 PM

So you're trying to ask if that implies $p\vee r$?

I think it's clearer in the shorter form, btw.

manooooh

This comes from a major problem: Prove that $A \cap (B-C) = (A \cap B) - (A \cap C)$ (set theory)

@TedShifrin exactly, or $q\vee p$ or $q\vee r$

Ted Shifrin

if $p\wedge s$ implies $p$, why are you making this so difficult?

Edward Evans

@Alessandro fair

I was just generalising

manooooh

@TedShifrin because if $x\in A\cap(B-C)$ then $x\in A\wedge x\in B\wedge x\in\overline{C}$

Ted Shifrin

Wow, you're sure making basic set theory harder. Write down the definition. $x\in A\cap (B-C)$ iff ....

manooooh

7:24 PM

Yes, then $x\in\overline{C}$ implies $x\in\overline{C}\cup\overline{A}$ and we are done

Ted Shifrin

So use associativity of $\wedge$?

whoa. Where did $\bar A$ come from?

manooooh

But a friend of mine did another thing, and I am trying to help him

@TedShifrin if $x\in C'$ then $x\in C'\vee x\in A'$ (since $p\to p\vee q$), then $x\in C'\cup A'$, ...

Edward Evans

@Lukas I'm applying to partake in the Heidelberg Laureate Forum lol, I get an automatic nomination as a DAAD scholar but still need to apply, worth a shot I guess :P

Ted Shifrin

Why introduce $A'$? I totally do not understand the purpose.

manooooh

@TedShifrin because we want to show that $x\in A\wedge x\in B\wedge x\in C'$ is equivalent to $x\in A\cap B\wedge x\in(A\cap C)'$

Ted Shifrin

7:27 PM

Huh? I'm totally lost.

Oh. Crazy

manooooh

$A \cap (B-C) = (A \cap B) - (A \cap C)$. Proof: take $x\in A \cap (B-C)$. Then $x\in A\wedge x\in B-C$

Then $x\in A\wedge x\in B\wedge x\in C'$. Then $x\in A\wedge x\in B\wedge x\in C'\cup A'$. Then $x\in A\cap B\wedge x\in(A\cap C)'$. Then $x\in (A\cap B)\cap(A\cap C)'=(A\cap B)-(A\cap C)$

Is it clear now?

Edward Evans

where are the words

Ted Shifrin

$A\cap C'$ should be $(A\cap C)'$?

And you have union where it should be intersection.

manooooh

@TedShifrin yes, fixed both, thank you

Ted Shifrin

No, you still have one wrong.

manooooh

7:31 PM

Now

Ted Shifrin

OK, now fixed.

manooooh

But that's not the question

Ted Shifrin

So to clarify, it should be about reassociating (and commuting) with $\wedge$.

Of course, you have only half the proof here. Unless every step is $\iff$.

manooooh

My question is other: From $x\in A\wedge x\in B\wedge x\in C'$ my friend did: $x\in A\wedge x\in B\wedge x\in C'\wedge(x\in A\vee x\in A')$

@TedShifrin yes I know

Ted Shifrin

Something's wrong there.

You can't write stuff like that.

$x$ cannot be an element of a statement.

manooooh

7:34 PM

But I am trying to prove (or disprove) that with this proposition: $x\in A\wedge x\in B\wedge x\in C'\wedge(x\in A\vee x\in A')$ doesn't follow $x\in(A\cap B)-(A\cap C)$

Ted Shifrin

Well, the last statement holds for all $x$.

manooooh

@TedShifrin but how can we go from that to what we want to prove? I don't see it possible

Unless this property: $(p\wedge q)\vee(r\wedge t)\to p\vee r$ holds

Ted Shifrin

Because you distribute and the $x\in A$ case is already covered, so that leaves you with

$x\in A\wedge x\in B\wedge x\in C' \wedge x\in A'$, so the empty set.

So we're just back at the original.

anakhro

@Thorgott I would hope that inequality works. Otherwise this paper has an error. :(

Ted Shifrin

Your friend is trying to regroup and say $x\in C'\wedge x\in A' \iff x\in (A'\cap C') \iff x\in (A\cup C)'$?

manooooh

7:38 PM

@TedShifrin that's a good argument. Can't we distribute otherwise to (1) avoid that absurdity, (2) try to get to what we want to prove?

Ted Shifrin

So this is no good.

I think there's the wrong conjunction in there. An $\wedge$ needs to be an $\vee$?

manooooh

@TedShifrin where?

Ted Shifrin

We ended up with $(A\cup C)'$, by my reckoning, and as you said earlier you want $(A\cap C)'$.

manooooh

Oh, yes, you are right

anakhro

@Thorgott yeah it's true I just figured out how to do it. I always forget the reverse triangle inequality.

Thorgott

7:40 PM

You mean the alternative?

anakhro

No, the original inequality I posted.

Oh wait

manooooh

@TedShifrin I don't know if my friend wants to come up with that expression, but he did: from $x\in A\wedge x\in B\wedge x\in C'\wedge(x\in A\vee x\in A')$ we distribute in this way: $x\in A\wedge x\in B\wedge((x\in C'\wedge x\in A)\vee(x\in C'\wedge x\in A'))$

anakhro

I put integrals in it

Why did I put integrals in it

It's not supposed to have integrals

I copy and pasted the wrong thing.

SORRY.

Ted Shifrin

You wanted to confuzle us, @anakhro. I looked at it and gave up.

anakhro

It's $||f_n| - |f-f_n| - |f||\leqq 2|f|$

manooooh

7:42 PM

The left part is what we want: $x\in A\wedge x\in B$ is equals to $x\in A\cap B$

Semiclassical

||x|-|y|| or some such

manooooh

Furthermore we have a conjunction between $x\in A\cap B$ and $(x\in C'\wedge x\in A)\vee(x\in C'\wedge x\in A')$ , needed since we want to show that $x\in(A\cap B)\color{red}{\wedge}x\in(A\cap C)'$

anakhro

reverse $\triangle$ inequality ya

Thorgott

oh lol

anakhro

i.e. the $\triangledown$ inequality

Ted Shifrin

7:45 PM

Well, @manooooh, the first one gives $A-C$ but the second one gives $(A\cup C)'$, which is extra.

manooooh

So we have to prove (or disprove) that $((x\in C'\wedge x\in A)\vee(x\in C'\wedge x\in A'))$ is equal to $x\in(A\cap C)'\equiv x\in A'\cup C'$

Ted Shifrin

Oh, but if $x\in A$, then $x\notin (A\cup C)'$, so we throw that away.

Thorgott

I love the triangle inequality, it's so incredibly useful

Semiclassical

my problems lately can be summed up as "how to procrastinate by thinking about the spherical triangle inequality"

Ted Shifrin

You need no extra techniques in procrastination, @Semiclassic.

Semiclassical

7:47 PM

it's an application of known methods

Ted Shifrin

@manooooh: or, reassociate and you get $(x\in A\wedge x\in A')\wedge x\in C'$. Regardless, there's no such $x$.

anakhro

I had an Iranian professor who called it "the lazy ass theorem": if you have a right angled path between a donkey and its food, the donkey will take a different path.

manooooh

@TedShifrin I didn't see that, many thanks. But we don't want to came up with that expression, we want to came up (if possible, that's why I am asking) to $x\in A'\vee x\in C'$

Thorgott

lol

Ted Shifrin

That's from the first case, @manooooh.

Your friend had a $\vee$ statement, and what you want comes from the first part.

manooooh

7:50 PM

@TedShifrin but don't put sticks in the way xD. I mean hell yes, the way we are trying to prove this seems more tricky

Ted Shifrin

Huh? I'm just saying your friend's approach works.

manooooh

@TedShifrin oh okay, so any combination gives us a contradiction

@TedShifrin ... I think you are showing me that it does not work, since there's no such $x$

Ted Shifrin

He put in a universal $x\in A \vee x\in A'$.

manooooh

Yes, which is always true

Ted Shifrin

I'm saying that you get two terms from his universal statement. The first term gives what you want. The second term gives the empty set.

manooooh

7:53 PM

@TedShifrin the first term does not give what we want. The first term is: $x\in C'\wedge x\in A$ . That is not equal to $x\in(A\cap C)'$

Ted Shifrin

No, you're stuck on your approach, not mine. This is $A-C$, which is what you want.

manooooh

@TedShifrin no I don't want $A-C$ but the red part of $(A\cap B)\color{red}{-(A\cap C)}$ trying to use the work of my friend, if possible

But I think that since you showed that any combination of distributive in this proposition: $x\in A\wedge x\in B\wedge x\in C'\wedge(x\in A\vee x\in A')$ gives us a contradiction, then we can't proceed further; we want to find other way (which we found)

Ted Shifrin

Oh, so let's see. $(A\cap B)\cap (A-C) = (A\cap B) - C = (A\cap B)-(A\cap C)$, so it's certainly fine.

No, there's no contradiction. It just eliminates the part of the disjunction that we already eliminated.

manooooh

@TedShifrin we don't have $(A\cap B)\cap (A-C)$. We have: $(A\cap B)\cap( (A-C)\cup(A\cup C)')$

Ted Shifrin

The second piece gives the empty set. For the third time.

I hate this stuff. You get lost in symbols and stop thinking.

manooooh

8:02 PM

@TedShifrin sorry I don't see why $(A\cup C)'$ is the empty set

Ted Shifrin

It's not. But when you intersect with $A$ it is.

manooooh

@TedShifrin but where do we have the intersection between $(A\cup C)'$ and $A$?

Rithaniel

It looks like we're unioning it with $A-C$, so some of $A$ might still be in that union

Ted Shifrin

We're not talking about that piece, @Rithaniel.

What does it mean for $x\in A\cap B$? It means $x\in A$. How then can it be in $(A\cup C)'$? (I've said this twice already ...)

Rithaniel

Ah, alright

anakhro

8:05 PM

Its mathematically legal to go $\lim (a_n - b_n) = \liminf (a_n - b_n) \leqq \liminf a_n$, where $a_n,b_n$ are positive sequences?

manooooh

@TedShifrin if $x\in A\cap B$ means $x\in A$ (which I agree) then we replace $x\in A\cap B$ by $x\in A$, right?

But if we replace $x\in A\cap B$ by $x\in A$ then we lost the first part!!!

I can open a question on the main site if you want

Thorgott

$\liminf$ is monotonous

Ted Shifrin

You don't replace, @manoooh. Just think about regrouping the $wedge$. I've said several times you have to reassociate if you do it this way. $x\in A\wedge x\in B\wedge x\in (A\cup C)'$.

Write this as $x\in B\wedge (x\in A\wedge x\in (A\cup C)')$.

Lunch time for Ted.

manooooh

@TedShifrin we don't have $x\in A\wedge x\in B\wedge x\in (A\cup C)'$ so I don't know why are you writing it

0

Q: Show $A \cap (B-C) = (A \cap B) - (A \cap C)$ using an alternative proof

I am trying to prove that $$A \cap (B-C) = (A \cap B) - (A \cap C)$$ holds for any sets $A,B,C$. This is equal to $$A \cap (B-C) = (A \cap B)\cap\color{red}{(A\cap C)'}.$$ I have seen a similar question: Proving : $A \cap (B-C) = (A \cap B) - (A \cap C)$ but I am trying to use another approach, ...

Ted Shifrin

8:33 PM

I give up. You need to read carefully what we've said and do everything with paper and pencil. Sometimes people in here are just too busy talking and not thinking.

anakhro

8:55 PM

A function $f\colon\mathbb R\to\mathbb R$ and an $a\in\mathbb R$: we know that we can prove continuity at $a$ using only monotone sequences (showing that any monotone $a_n\to a$ converges $\implies$ $f(a_n)\to f(a)$), but can we further reduce the number of sequences we need to check?

Ted Shifrin

If your sequence isn't arbitrary, I can always make up a function that works for the sequences you used and doesn't have a limit. Same thing for paths in multivariable.

Lukas Heger

@TedShifrin Pourquoi est-ce qu'il français tant difficile? L'italien est plus facile.

anakhro

In multivariable do you have something like the monotonic thing where it suffices to consider sequences "monotonic along a line through $a$"?

Ted Shifrin

No, @anakhro, I don't care about monotonic. I'm talking about checking on certain families of curves. Unless you allow arbitrary curves, I can always make a function that'll mess up.

Thorgott

9:15 PM

You do need arbitrary curves, as Ted says, but it suffices to check monotonic sequences through each curve (this reduces to the one-dimensional case immediately).

anakhro

Is it the obvious way to cook up examples? Just pick your favourite curve that was not checked and then force a normally continuous function to be discontinuous along this curve?

Ted Shifrin

9:27 PM

Define a function to be $0$ on the union of your curves and $1$ elsewhere.

anakhro

But I asked about linear curves?

The union of all the lines through a point in the plane is the plane.

Lukas Heger

9:43 PM

define $f(x,y)=\frac{x^2}{y^4}$ for $y \neq 0$ and $f(x,0)=0$ for all $x$. On every linear curve through the origin, $f$ is continuous, but on the parabola $x=y^2$, it isn't

Ted Shifrin

Oh, you got me. OK, if you're using all lines, define $f(x,y) = 0$ if $|y|\ge x^2$ and $1$ otherwise.

@Lukas: That's harder to generalize to all polynomial curves, say.

@Lukas: I don't think your example works. Upside down maybe?

Lukas Heger

it's continuous on $x=0$, on $y=0$ and on $y=ax$ for any $a \in \Bbb R\setminus 0$, so on all linear curves through the origin

Ted Shifrin

On $y=ax$ I have $1/a^4x^2$.

Lukas Heger

oops, right

I'm dumb

yeah, switch denominator and numerator

Ted Shifrin

Just sayin' ... :D

Lukas Heger

9:52 PM

then it should work though

Ted Shifrin

There are other famous examples, but my method generalizes to all monomials, for example.

Lukas Heger

@Ted are you sure no family works? What about all smooth curves? Every continuous curve can be approximated by smooth curves, so it might work

Ted Shifrin

Well, sure, by "family" I intended some constraints, not just "all curves."

But it is the case that if $f$ does not have a limit at $0$, then I can create a continuous curve that fails.

I suppose in @anakhro's original question, if we allowed the uncountable collection of all sequences, of course, that would suffice. My understanding was that we had a more "finite" collection.

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$Mathematics$ Mathematics