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We have the Lemma

If $\alpha_i \uparrow \alpha$, $\beta_i \uparrow \beta$, then $\alpha_i\beta_i \uparrow \alpha\beta$.

And I have to show

The analogue of this Lemma for decreasing directed families is false.

I thought that repeating the proof for the lemma now with "> If $\alpha_i \downarrow \alpha$, $\beta_i \downarrow \beta$, then $\alpha_i\beta_i \downarrow \alpha\beta$." would fail at some point, but I cannot see it. I also tried to draw an example, but I can't find any on the reals, maybe this is because $\mathbb{R}$ is totally ordered?

user2820579
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1 Answers1

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I don't know where this stuff comes from, so what I write might be garbage.

The proof of the analogue lemma should fail where the original proof uses the "directed to the right" property. Now for a counterexample (the easiest I could come up with):

Let $I = \{x,y,z\}$ with $x \leq z, y \leq z$. Then if I'm not mistaken, I has this "directed to the right" property (and is partially ordered). Now let $\alpha_x = 1, \alpha_y = 2, \alpha_z = 2, \beta_x = 2, \beta_y = 1, \beta_z = 2$. Then $\alpha_i \downarrow 1, \beta_i \downarrow 1$ but $\alpha_i \beta_i \downarrow 2$.

Kameldieb
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  • Probably it's ok. To be honest, I was stuck trying to understand what the exercise wanted to be proven. I actually thought it was $\alpha_i\geq \alpha_j$ for all $i\leq j$, but I was always essentially ending with the previous Lemma ($\alpha_i\leq \alpha_j$). This is from the book of Sterling Berberian, Measure and Integration, AMS. – user2820579 Aug 18 '22 at 12:40