Is it true that $\mathbb{R}^m \cap \mathbb{R}^n = \emptyset$ where $m \neq n$ Is it true that $\mathbb{R}^m \cap \mathbb{R}^n = \emptyset$ where $m \neq n$ ?
I would say that it is true because the elements in each set are different entities. This means that one set cannot be a subset of the other. An example would be $\mathbb{R} \cap \mathbb{R}^2$. You cannot say that $1 \in \mathbb{R}^2$ for example.
 A: Well, these sets are disjoint.
To see this, define the ordered pair
$$(x,y) = \{\{x\},\{x,y\}\}$$
and extend this definition to $n$-tuples
$$(x_1,\ldots,x_n) = ((x_1,\ldots,x_{n-1}),x_n).$$
So $n$-tuples are sets and we have clear notion of equality of two sets:
$$A= B :\Longleftrightarrow \forall x[x\in A\Leftrightarrow x\in B].$$
Equality of two $n$-tuples is then given by
$$(x_1,\ldots,x_n)=(y_1,\ldots,y_n)\Leftrightarrow \forall i[x_i=y_i].$$
Taking this into accout you see that the above sets are disjoint if $m\ne n$.
A: From rigorous point of set theory, the answer is yes. However, taking your example, let's assume the mapping $f:\mathbb R\to\mathbb R^2$ such that $x\mapsto (x,0)$ for example. Clearly, $f$ is bijective. Hence, it's safe to say that the image of $f$, namely $\mathbb R\times\{0\}$, represents the domain of $f$ i.e. $\mathbb R\sim\mathbb R\times\{0\}\subset\mathbb R^2$.
Therefore, $\mathbb R$ can be seen, in this way, as a subset of $\mathbb R^2$ and the intersection $\mathbb R\cap\mathbb R^2$ is not empty (disregarding difference in structures of elements in both sets).
I hope that example answers your question since generalizing that is relatively easy. Also, may I recommend reading about disjoint unions as a way to view unions of sets that do intersect. A lot of similar logic is used.
