# 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.

• That is correct, but what has this to do with probability-theory? Commented Jan 5, 2021 at 10:47
• @JoséCarlosSantos Doesn't probability theory work with sets and their intersections as well?
– Fib
Commented Jan 5, 2021 at 10:48
• Sure, as most branches of Mathematics do. If that's all that you are interested in, then I suggest that you tag your question as elementary-set-theory. Commented Jan 5, 2021 at 10:50
• I added the tag. I understand my question is general, but this is the area I came across the question thus I added it as the tag.
– Fib
Commented Jan 5, 2021 at 10:51
• @JoséCarlosSantos Isnt $\mathbb{R}^0 \cap \mathbb{R}^1 \neq \emptyset$? Commented Jan 5, 2021 at 10:52

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 $$m\ne n$$.
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).