What am I missing about isomorphisms?

[frog_measure No.8484978]

I wrote part of the question in the picture.
http://mathworld.wolfram.com/Isomorphism.html
Does axiomatic set theory take this into account (i.e. isomorphic sets being somewhat equal)?
[math]\mathbb{R}\not \subset \mathbb{C}[/math].
This is fucked up.

[Anonymous No.8484987]

>>8484978
Surely -1 is a real number

[Anonymous No.8485002]

>>8484978
Define C as the algebraic closure of Q and R as the subset of C
[eqn] \mathbb{R} = \{ x\in \mathbb{Q} : x^2 = |x| ^2 \} [/eqn]
or something

[Anonymous No.8485004]

>>8484987
But [math] (-1,0) \in \mathbb{R} \times \mathbb{R} [/math] isn't

[Anonymous No.8485007]

>>8484978
Isomorphism of what? R is not isomorphic to C in any category with some algebraic structure (Vector Spaces, Rings, Fields, etc.). Isomorphisms of sets are just bijections and do not carry any structure other than the vague set theoretic structure.

[Anonymous No.8485014]

>>8484978
Yeah, set theory is a pretty awkward foundational language. Usually, people will get around this by saying something like "[math] \mathbb{R}\subset\mathbb{C}[/math] where we identify [math] \mathbb{R}[/math] with the subalgebra of [math] \mathbb{C}[/math] given by [math] \{(x,0):x\in\mathbb{R}\}[/math]".

Here's a similar situation for you to ponder: in many introductory real analysis textbooks, a real number is defined as a Dedekind cut, that is to say a particular kind of subset of [math]\mathbb{Q}[/math]. The ordering is then defined so that a<b iff [math] b\subset a[/math]. One consequence of this is that [math] \pi\in 2[/math].

[Anonymous No.8485017]

>>8485007
>Isomorphism of what?
Yeah, I meant isomorphism from [a subset of] [math]\mathbb{C}[/math] to [math]\mathbb{R}[/math].
The function is the trivial, usual one in the picture.

[Anonymous No.8485018]

>>8485017
Thats not what I meant. Isomorphism w.r.t what structure? (i.e. in what category?)

[Anonymous No.8485020]

>>8485002
Does this allow for [math]\mathbb{R}\subset \mathbb{C}[/math]?
Is there any reference for this?
My references are analysis and topology textbooks.

[Anonymous No.8485024]

>>8485007
>R is not isomorphic to C in any category with some algebraic structure (Vector Spaces

>not specifying the ground field
>not realizing they are isomorphic as [math]\mathbb{Q}[/math] vector spaces

W E W

[Anonymous No.8485031]

>>8485018
>Shilov.

[Anonymous No.8485043]

>>8485014
>Usually, people will get around this by saying something like ...
That's like the only approach I have encountered.
I was wondering whether this was merely a simplification.
So, this is the state of things in 2016?
>One consequence of this is that π∈2.
Rudin uses this approach in his book but I found the construction from Cauchy sequences too intuitive to bother with understanding cuts.
I will certainly have a look at it in the future.

[Anonymous No.8485059]

>>8485043
As far as I know, it's the only way to be completely correct if you insist on working with sets. Another thing you could say is that [math] \mathbb{R}\hookrightarrow \mathbb{C}[/math], meaning there exists an injective function from R into C (and you can even specify it to be the "obvious" one). This has the advantage that it doesn't depend on the particulars of how you choose to define R or C (a different but equivalent definition of R will also change the definition of the function, but it doesn't change the fact that such a function exists), and in practice, it's pretty much "as good" as having an actual inclusion of sets.

[Anonymous No.8485082]

>>8485059
Thank you for the answers.
>As far as I know, it's the only way to be completely correct if you insist on working with sets.
What is the alternative to sets?

[Anonymous No.8485089]

>>8485020
Yeah, if you define R to be a subset of C, then R is a subset of C.
also I messed up the definition
[math] \mathbb{R} = \{ x\in C : x^2 = |x|^2 \} [/math]
And C should probably be the analytic completion of the algebraic completion of Q.
> Is there any reference for this?
I kinda came up with it on the spot.
You just have to define C however you want, and define R to be some subset with the properties you want.

[Anonymous No.8485100]

>>8485082
One alternative is to work in categories. You can say that "R embeds in C in the category of fields", and this is nice because it's still true if you replace R with something that is isomorphic to it in this category (so you don't have to worry about the details of the construction as long as then end result is same). There are also topoi, and, more recently, Homotopy Type Theory, but I don't know anything about these.

[Anonymous No.8485125]

>>8484978
They have different dimensions, so no.

[Anonymous No.8485128]

>>8485089
>Yeah, if you define R to be a subset of C, then R is a subset of C.
Thank you for the clarification.
I misread your post when I replied.
Defining [math]\mathbb{C}[/math] from [math]\mathbb{Q}[/math] seems nice and I'll look into it.
>You just have to define C however you want, and define R to be some subset with the properties you want.
What I mean is that I don't have a formal education in algebra but I guess any book would be OK.

[Anonymous No.8485135]

>>8485100
OK.
>>8485125
Yeah.
I guess that given the *usual* [math]\mathbb{R}[/math] there is no [math]\mathbb{C}[/math] with the properties I'm asking for.
Unless:
>>8485089
>>8485100.

[Anonymous No.8485229]

>>8485002
>Define C as the algebraic closure of Q
>18 posts
>no one point out this mistake

[Anonymous No.8485257]

over the multiplicative groups, rings, and fields of R and C:

f(-1) = f(i * i) = f(i) * f(i)

f(-1) = -f(1) = -1

f(i)^2 = -1 has no solutions in the real numbers, therefore they are not isomorphic

[Anonymous No.8485265]

>>8485229
Pretty sure I corrected my mistake in>>8485089
Should be the algebraic closure of the analytic closure right?

[Anonymous No.8485277]

>>8485265
>algebraic closure of the analytic closure
yes this is right since analytic closure of Q is R and algebraic closure of R is C

in >>8485089 you say analytic completion of algebraic completion which would give you the analytic completion of [math] \bar \mathbb{Q} [/math]... I'm not actually sure what this is but I have a feeling it's not all of C

[Anonymous No.8485288]

>>8485277
test
[math] \bar{ \mathbb{Q} } [/math]

[Anonymous No.8485296]

>>8485288
>>8485277
i take it back
http://math.stackexchange.com/questions/68231/why-does-mathbfq-lie-dense-in-mathbfr-and-overline-mathbfq-lie