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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.

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  • $\begingroup$ That is correct, but what has this to do with probability-theory? $\endgroup$ Commented Jan 5, 2021 at 10:47
  • $\begingroup$ @JoséCarlosSantos Doesn't probability theory work with sets and their intersections as well? $\endgroup$
    – Fib
    Commented Jan 5, 2021 at 10:48
  • $\begingroup$ 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. $\endgroup$ Commented Jan 5, 2021 at 10:50
  • $\begingroup$ 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. $\endgroup$
    – Fib
    Commented Jan 5, 2021 at 10:51
  • $\begingroup$ @JoséCarlosSantos Isnt $\mathbb{R}^0 \cap \mathbb{R}^1 \neq \emptyset$? $\endgroup$
    – Eminem
    Commented Jan 5, 2021 at 10:52

2 Answers 2

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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 $m\ne n$.

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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.

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