Set definition problem
I want to write in a Proposition:
"There exists $x_1$ and $x_2\geq x_1$ so that (a) If $x\in [0,x_1)$, then $A>B$. (b) If $x\in [x_1,x_2)$, then $A\leq B$. (c) If $x\in [x_2,\infty)$, then $A>B$."
The problem is that each of these sets can be empty based on some other conditions (which are very ugly to write down and I want to avoid that). For the first two making them empty is easy, but for the third I cannot just say $x_2=\infty$ imo. Right?
We now wanted to write: "There exist potentially empty sets $[0,x_1)$, $[x_1,x_2)$, and $[x_2,\infty)$ so that (a) If $x\in [0,x_1)$, then $A> B$. (b) If $x\in [x_1,x_2)$, then $A\leq B$. (c) If $x\in [x_2,\infty)$, then $A> B$."
But also this seems to not be entirely correct. Right?
Do you have an elegant solution? Or can the third interval be naturally empty?
Solution 1:
There exists $0\leq x_1\leq x_2\leq \infty$ such that:
- If $x\in[0,x_1)$ then $A>B$.
- If $x\in[x_1, x_2)$ then $A\leq B$.
- If $x\in[x_2, \infty)$ then $A>B$.
This is enough to make it mathematically correct; if you want to really make the reader pay attention to the fact these sets could be empty (though it is already obvious as is), you can add:
Notice that depending on $x_1$ and $x_2$, one or more of the sets appearing in $(1)-(3)$ above may be empty.
This uses the well known notational device of the extended real number line.
Explanation
We don't have to explicitly rule out the case of empty sets. For instance, if $x_1=x_2$ then the set appearing in $(2)$ is empty, but the statement "if $x\in[x_1,x_2)$ then $A\leq B$" is still correct. Since the set is empty then $x\in[x_1,x_2)$ is false and so the statement is vacuously true.
Simplification
Keep in mind that if indeed your three conclusions are $A>B$, $A\leq B$, $A>B$ as you have written, then you are saying that it is always the case that $A>B$ except when $x\in [x_1, x_2)$. If so, then you can make the wording slightly simplified:
There exists $0\leq x_1 \leq x_2 \leq \infty$ such that $A\leq B$ when $x\in [x_1, x_2)$, and $A>B$ otherwise.