Factor spaces

[sour tundra]

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[short cradle]

How would <= look like?

Or am I overcomplicating this?

[slim sedge]

<= can be achieved with double inclusion maybe

If w = v + u, then w = v + u - v' + v' = v' + u'

Where u' = u - v' + v

Same reasoning for the reverse inclusion

[slim sedge]

I would say it's required

[short cradle]

After all, we can treat this like an equation though, no?

-v' on both sides

[slim sedge]

it's not an equation

[short cradle]

$v - v' + U = U$

[mossy quest]

Kepe

[slim sedge]

it's a set equality

[short cradle]

Yeah it's not but I mean

We could prove that we can treat it like an equation

No?

[mystic hill]

the claim is

[slim sedge]

In a sense yes but it isn't the most straightforward way to prove that two sets are equal

[mystic hill]

this holds if and only if v-v' in U

[slim sedge]

two sets are equal when they contain the same elements

which is exactly what double inclusion proves

[short cradle]

Yes

[slim sedge]

this is the most straightforward way to prove that those two sets are equal

[short cradle]

That's the claim

[mystic hill]

on the one hand, if the set equality holds, then v = v' + u for some u in U

in particular u = v-v'

[slim sedge]

while it is very tempting to do set operations here as you suggest, there is little guarantee that they are logically sound

unless you have explicitly proven said properties for set operations

[short cradle]

Are you referring to -v'-ing both sides or what aL said

[slim sedge]

the former

[mystic hill]

no

[short cradle]

ok

[short cradle]

Yes

[slim sedge]

I think the OP has already proven =>

[mystic hill]

the + in "v+U" doesnt have the same meaning as within V itself

so it need not have the same properties

Suppose v-v' in U. Take v+u in v+U. Then v+u = v' + (v-v' +u) in v'+U

the other inclusion is symmetrical

@short cradle

[short cradle]

Hmm

[short cradle]

How

v + u = v + (v - v')

[mystic hill]

look again

[short cradle]

Suppose that $v - v' \in U$ for $v, v' \in V$. Then also $v' - v \in U$. We need to show that $v + U = v' + U$. \begin{align*} v + U &= {w \in V \mid w = v + u \text{ for a $u \in U$}} \ &= {w \in V \mid w = v + (v' - v + u') \text{ for a $u' \in U$}} \ &= {w \in V \mid w = v' + u' \text{ for a $u' \in U$}} \ &= v' + U\end{align*}

[mossy quest]

Kepe

[short cradle]

Does this look fine?

[short cradle]

This would be <= now

[slim sedge]

Line 2 and line 3 are very fishy

I think line 2 contains a mistake and line 3 is not super transparent, about whether it is really the same set or not

Again, that's why a reasoning by double inclusion is much more straightforward

[short cradle]

Line 3 is just cancelling the v

I agree line 2 is a bit fishy but maybe we can save it?

[slim sedge]

the reason why it is not easier is that you still need to prove that your "simplification" doesn't exactly say whether you still have the same set

it's very difficult to carry set equality over several lines because you always need to justify whether they really are the same set or not

or due to a certain algebraic operation, one is a subset of the other

[short cradle]

Justification for line 2: $v' - v \in U$ is not necessarily $= u$. But we can find some $u' \in U$ so that $v' - v + u' = u$.

[mossy quest]

Kepe

[slim sedge]

here, it happens to be the same set for every line, but you never justified why it's the same set (and the effort you would spend on this justification is far more than the effort spent on just double inclusion from the very beginning)

you get what i'm trying to say?

[short cradle]

But I'm just simplifying the condition, they're equivalent

It's not like I'm changing something

[slim sedge]

this "equivalent" part is exactly what needs to be justified every line

[short cradle]

What's the difference to [{(x, y)^{\tr} \in \mathbb R^2 \mid x + y = 1 +1} = {(x, y)^{\tr} \in \mathbb R^2 \mid x + y = 2}?]

[mossy quest]

Kepe

[short cradle]

There, it's just a simplification too

[short cradle]

Do I really need to justify here why they're equivalent?

[slim sedge]

Here it is trivial, but from line 1 to line 2 it's not trivial

[short cradle]

Oh, I thought you meant line 2 to 3

[short cradle]

Yeah I agree line 1 to 2 is a bit fishy

But can't it be justified by this

[slim sedge]

It's not a very rigorous justification

[short cradle]

Or would it still not be clear

[slim sedge]

Again, remember that two sets are equal WHEN they contain the same elements

this is the definition you can't really escape

so, you must prove that the set from line 1 and the set from line 2 contain the same elements, in order to justify the step

[short cradle]

Ok, so it's better to argue with inclusions, thank you!

[slim sedge]

Yeah, I've been trying to argue in favor of that for a while 🤣

You're welcome in any case

[short cradle]

Yeah haha I was just trying to save my proof by arguing that what I'm doing is just a simplification of the condition, the condition remains the same

But yeah, to prove that we probably need to either again do inclusions

Or do my argument over again more rigorously

[slim sedge]

It's not worth the effort. A double inclusion is very clean and straightforward

especially when the reasoning is symmetrical here

[short cradle]

What does that mean?

[slim sedge]

when you prove that LHS is a subset of RHS

the same method is used to prove that RHS is a subset of LHS

[short cradle]

Just in reverse

[slim sedge]

yeah in a sense

[short cradle]

Thank you

[raven crown]

@short cradle has given 1 rep to @slim sedge

[short cradle]

+close