Mathematical analysis Zorich

[lament path]

a) a) Let deltaX be the diagonal of the set X^2 and deltaY be the diagonal of the set Y^2. Show that if relations R1 is part of XxY and R2 is part of YxX are such that (R2 o R1 = deltaX) and (R1 o R2 = deltaY), then both of them are functional and define mutually inverse mappings

[lament path]

a) Let deltaX be the diagonal of the set X^2 and deltaY be the diagonal of the set Y^2. Show that if relations R1 is part of XxY and R2 is part of YxX are such that (R2 o R1 = deltaX) and (R1 o R2 = deltaY), then both of them are functional and define mutually inverse mappings

I need to prove that I wrote the translation.

[strange ginkgo]

by diagonal, you mean {(x, x) : x in X} right?

[lament path]

Yes

Likewise {(x1, x2) in X^2 | x1=x2}

[strange ginkgo]

By functional you mean if (x1, y), (x2, y) in R then x1=x2, right?

[strange ginkgo]

what do you have so far for a proof?

[lament path]

(x R y1) and (x R y2) => y1=y2. This means that the relation R is a function (synonymous with mapping). Or in other words, R is a functional.

x R y1 <=> (x, y1) in R.

[strange ginkgo]

oh derp. Yeah that's right. I mixed it up with surjective

Okay, so have done any of the proof yet?

[lament path]

My logic is such that as a result I get the expression (1)

In order for R1 and R2 to be mappings, I could require x=x1, y=y1. This requirement is a special case of the expression (1) indicated. In this case, this would correspond to the definition of the function for the relations R1 and R2.

But then a circular dependence arises, which I have indicated with the (?) sign.

[lament path]

This is just my logic, I don't know where this terrible cycle came from 😅

[lament path]

I use the definition from the problem and the definition of diagonal and that's it

[strange ginkgo]

My, this is a bit hard to read. I'll need a little while

[strange ginkgo]

yeah this is too confusing to read. I feel like you're just writing definitions anywhere, but I'm not seeing any real progress toward proving that R1 and R2 are functional

[lament path]

(x1 R1 y*) and (y* R2 x2) => x1=x2
(y1 R2 x') and (x' R1 y2) => y1=y2
y1, y2 in Y. x1, x2 in X.
D(R1) = X. D(R2) = X.

If x1=x2=x. y1=y2=y:
(x R1 y*) and (y* R2 x)
&
(y R2 x') and (x' R1 y)
x in X
y in Y

compression and removed the error

[gusty helm]

пипец

[lament path]

У вас есть решение?

[lament path]

please check it

[lament path]

Version 2

[sinful bobcat]

.close