Can anyone explain to me why [math]\mathbb{R}^2 \nsubseteq \mathbb{R}^3[/math]

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>>9092947
>>>/reddit/

R2 has elements of the form (x,y) with x,y in R.

R3 has elements of the form (x,y,z).

how is (x,y) gonna be contained in (x,y,z)?

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>>9092943

A plane in 2-space contains elements of the form (x,y), whilst one in 3-space contains elements, say, of the form (x,y,0). The difference is subtle but significant. In the latter you can look up from your spot on the plane, in the former there simply is no up.

>>9092969
But complex numbers are defined as elements of the form (x,y) with x,y in R yet the real numbers are a subset of the complex numbers.

>>9092976
https://math.stackexchange.com/questions/1553537/why-mathbbr-is-a-subset-of-mathbbc

>>9092943
There are infinitely many strictly distinct planes in R^3. Which of them would you consider to be R^2?

>>9093028

Not very clear answer. There is nothing wrong with having multiple copies of R^2.

The point is that there exists a natural "embedding" of R^2 space into R^3. Notice that R^2 is not the same thing as plane (even though they are isomorphic) since R^2 consists of ordered pairs of real numbers and plane is a strictly geometric concept.

Anyway, this embedding takes element (a,b) from R^2 and maps it to (a,b,c) of R^3, where "c" is arbitrary real number. Now, if you want to keep structure of field given on R^2, you would have to choose that c=0 so you can keep neutral element for addition.

The problem is in terminology. Since you have this embedding, you "think of" R^2 as subset, but in general, it is not.
On the other hand, you can "define" R^2, to be set of all (a,b,0) and then you would get that it really is a subset of R^3.

>>9093075
Thank you, a very lucid explanation.

>>9093075
Of course we can just take {(a,b,0): a,b in R} (plus operations) and call it R^2, just like we call {(a,0): a in R} the set of real numbers. But I meant to answer OP's claim:

>Surely any plane is contained within the 3D space.

If you think in terms of euclidean geometry, any plane, no matter the orientation, would be a subset of the 3D space. You cannot, however, deduce R^2 is a subset of R^3 from that unless you define what (class of) plane(s) R^2 represents.

>>9093092

Yes. That is why i made distinction between R^2 and geometric plane.

Again, problem rises from the fact that when you write R^2={(a,b)} you take your base to be "x-axis" and "y-axis". If you could somehow (and I'm not really into geometry) switch, or take all possible pairs of linear independent vectors to be your base, you would have it as subset of R^3.

Strictly speaking: no you cannot.
According to set theory those sets are completely different and disjoint simply due to the fact that R3 has a third element. However you can map every R2 to R3 with (x, y, 0) and so R3 can have all the functionality of R2, but they are still different sets. Bottom line is that there are multiple ways to represent the same thing.

>>9092943
Because their elements are different. That said, you can construct an isomorphism between the plane and a plane in 3D space and then through that isomorphism say that it is contained. The most natural choice would be to choose the plane z=0.

>>9093154
Not just natural but necessary if you want R^2 structure kept. If you take z=1, then adding (a,b,1) and (c,d,1) won't belong to the same plane (with standard addition)

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>>9092943
You're starting from a place that is not descriptive enough. Yes R2 can describe 2D space and R3 can describe 3D space. But R2 cannot describe anything in 3D space--you need 3 components!. You could prove a relationship between all R3 planes and R2 itself, but cannot just use a single subset symbol. They just don't automatically overlap.

>>9093192
You could also take x=0. Or x+y = 0
As long as it goes through the origin.