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Let $\{f_n:A\to\mathbb{R}\}_{n\ge 1}$ is a sequence of functions that converges pointwise to some function $f:A\to\mathbb{R}$ and let $A_n:=\{x\in A:|f_n(x)-f(x)|\ge\alpha\}$ for some fixed constant $\alpha$

(Some other hypotheses on $f_n$ and $f$ may be added)

I'm trying to prove the following:

for every $\varepsilon\gt0$ there exists $n\in\mathbb{N}$ and a finite closed interval covering $\{U_1,U_2,\cdots,U_k\}$ of $A_n$ such that $\sum_{i=1}^{k}\mu(U_i)\lt\varepsilon$

I don't even know if this is true (please do not post a proof/counterexample yet) but the first strategy is to negate the statement conclude that $f_n$ does not converge pointwise. So, how is it negated?

There exists some $\varepsilon\gt0$ such that for every $n\in\mathbb{N}$ and every finite closed interval covering $\{U_1,U_2,\cdots,U_n\}$ we have $\sum_{i=1}^{k}\mu(U_i)\ge\varepsilon$

My question is that, after the [for every $\varepsilon\gt0$] there is [exists $n\in\mathbb{N}$ AND exists finite closed...] but negating the AND doesn't make sense: [exists $\varepsilon>0$] such that [for every $n\in\mathbb{N}$ OR for every finite closed...] we have [$\sum_{i=1}^{k}\mu(U_i)\ge\varepsilon$]. What is the "finitely closed..." if the $n$ wasn't previously found??

Thanks

NewBornMATH
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augustoperez
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1 Answers1

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The words "finitely closed" assume some $n$, indeed, and this is natural language, after all-

Perhaps, in the negated sentence, it is better "for every $n\in\Bbb N$ there is no set with $n$ elements such that ..."

ajotatxe
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