Is $\mathbb{R^2} \subset \mathbb{C}$? 
Is $\mathbb{R^2} \subset \mathbb{C}$?

I've heard that $\mathbb{R^2}$ is isomorphic to $\mathbb{C}$, but can we say that $\mathbb{R^2} \subset \mathbb{C}$?
$\mathbb{R^2}$ is defined as $$\mathbb{R^2} = \{ (x,y) | x, y \in \mathbb{R}\}$$ and correct me if I'm wrong, but we can define $\mathbb{C}$ over the reals as follows: $$\mathbb{C} = \left\{(x, iy) | \ x, y \in \mathbb{R} \  \ \text{and} \  \ i = \sqrt{-1}  \right\}$$
But I don't see how we could show $a \in \mathbb{R^2} \implies a \in \mathbb{C}$. A counterexample could be that $(1,1) \in \mathbb{R^2}$, but $(1,1) \not\in \ \mathbb{C}$, however $(1, i) \in \mathbb{C}$.
So is my conclusion that $\mathbb{R^2} \not\subset \mathbb{C}$ correct?
 A: First your definition of $\mathbb C$ is not correct notation.  It is not $(a,ib)$, it is generally writen as $a+ib$.  It is important to consider an element of $\mathbb C$ as one thing rather than as two things.  The form $a+ib$ is only one way of writing a complex number.  $\mathbb C$ is generally defined by taking the field $\mathbb R$ and adjoining an $i$ as a new number that happens to satisfy $i^2=-1$.  Think about it more as taking polynomials with real coefficients (the set $\mathbb R[x]$), such as $1+3x+5x^2-\pi x^3$ etc..., and then
plugging in $i$ for $x$ to get $1+3i+5i^2-\pi i^3$.  You can simplify that with the equation $i^2=-1$ if you want.  In this way we have that the real numbers are naturally a subset of the complex numbers, while the real numbers are not naturally a subset of $\mathbb R^2$.
${\mathbb R}^2 = \mathbb C$ naturally as a set (ignoring any structure).  $a+ib \in {\mathbb C}$ is the element $(a,b) \in \mathbb R^2$ and vice versa.  As for isomorphic, that always depends on context.  $\mathbb R^2$ is naturally a real-vector space, but $\mathbb C$ is a field, so they are different type of objects and talking about them being isomorphic does not make sense.  It is sort of like asking if a certain car has the same horsepower as a shopping cart.
However, in the case of $\mathbb C$ we could consider it a real-vector space as well (see definition of a vector space).  Then as real vector spaces, $\mathbb C$ and $\mathbb R^2$ are in fact isomorphic.
Alternatively, you could define multiplication on $\mathbb R^2$ and then prove that with this extra operation you obtain a field, and this field is isomorphic as a field to $\mathbb C$.
Those are two different statements, and in each case we take one of the objects and we somehow throw it into a different category by unnaturally either taking away or adding some of its properties.  It's like taking your car, taking out the engine, forgetting that you can drive it on the highway and using it as a shopping cart.  So a small enough car could be isomorphic to a a certain shopping cart in the category of shopping carts.  Alternatively you could add an engine to a shopping cart and drive it on the highway and then say it is isomorphic to a BMW in the category of cars.
A: Two spaces $A, B$ being isomorphic means they "have the same form" i.e. you can find a function $\Phi: A \to B$ that is a bijection. If $A$ and $B$ have some operations defined, then $\Phi$ must also preserve those operations. Say $A$ has operations $\triangle_i$ and $B$ has operations $\circ_i$. Then we need that
$\Phi(x \triangle_i y) = \Phi(x) \circ_i \Phi(y)$ for all $i$
You say $A \subset B$ if $A$ has only elements taken from $B$, if $A$ is a subset of $B$. That would mean all elements in $A$ are actually elements from $B$ as-is. $\Bbb{R}^2 \not\subset \Bbb{C}$ since elements from $\Bbb{R}^2$ are vectors or pairs or 2-tuples and elements from $\Bbb{C}$ are complex numbers. One further hint showing that $\Bbb{R}^2 \not\subset \Bbb{C}$ is that in $\Bbb{C}$ you naturally have both addition $+$ and product $*$ defined while in $\Bbb{R}^2$ you usually only use the operation $+$ (even though you could define many more operations, including a similar thing to the product of complex numbers in $\Bbb{C}$).
