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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?

Snaw
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Paul
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    Why do you need to point out that the sets are potentially empty? I don’t see the problem here. – Mark Saving Dec 13 '21 at 17:07
  • Because otherwise, the reader may think that property C exists for very high x. But that is not necessarily true – Paul Dec 13 '21 at 20:45
  • That clarification seems completely unnecessary in this case. If they were more complicated sets, I think your “potentially empty” phrasing would be the best way to say it. – iosce Dec 14 '21 at 09:50
  • I now added more info on the property that will hold. I would feel betrayed as a reader if I would believe that the sequence A>B, $A\leq B$, A>B is always there. Although for some "external reasons" $A\leq B$ always. – Paul Dec 15 '21 at 12:24

1 Answers1

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There exists $0\leq x_1\leq x_2\leq \infty$ such that:

  1. If $x\in[0,x_1)$ then $A>B$.
  2. If $x\in[x_1, x_2)$ then $A\leq B$.
  3. 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.

Snaw
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  • Hi Snaw, thanks for that answer and I will grant you the 50 units. So the notational device of extending the real numbers allows us to say $x_2=\infty$, which then implies that case c) is empty, right? – Paul Dec 16 '21 at 15:56
  • @Paul Yes, exactly. The notation $[\infty, \infty)$ is still uncommon though and if you keep the formulation as in the first part then you might want to add a comment that in case $x_2=\infty$ we consider the third set empty. I would go with the simplified formulation "$A\leq B$ when $x\in[x_1,x_2)$, and $A>B$ otherwise" which does not require us to talk of $[x_2,\infty)$ explicitly in the first place, so the problem does not arise. – Snaw Dec 16 '21 at 18:10