Subset of a Cartesian product?

Question

Is it true the following

$ R \subset R^2$ ?

(If yes, I would like to see a rigorous proof.)

Answer 1

An isomorphic COPY of $\mathbb R$ is, but the set itself is not, as the elements of $\mathbb R ^2$ are ordered pairs, and the elements of $\mathbb R$ are numbers

Answer 2

Formally it's wrong, $R$ can't be a subset, since the elements are no tuples $(a,b)$, but you can identify $R$ in $R^2$ by $\pi:R \to R^2, \pi(r) := (r, 0)$. If you take $\mathbb{C} $ instead of $R^2$, you don't need that identification formally.

Answer 3

It all depends on what $R$ and $R^2$ should be? Think about set-theory: how can you prove that $r \neq (r,r')$ for any Element of $r$?

If $f:A \rightarrow B$ is injective, you can always remove $f(A)$ from $B$ an construct a new set $B_1$ as a disjoint union $$ B_1 := B\setminus f(A) \cup A $$ Now $A \subseteq B_1$ and sometimes you can transport any structure (operations, topologies,...) to from $B$ to $B_1$ and get something isomorphic.