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Let $(f_n)_{n=1}^{\infty}$ and $f $ be real valued functions defined on $\mathbb{R}$. For $ε > 0$ and $\forall m ∈ \mathbb{N}$ define $E_m (ε) = \{x ∈ \mathbb{R}\, |\, |f_m (x) − f(x)| ≥ ε\}$. Let $S = \{x ∈ \mathbb{R} \,|\, {f_n (x)},n\in \mathbb{N}$, does not converge to $f(x)\}.$ Express $S$ in terms of the sets $E_m (ε)$, $m∈\mathbb{N}$, $ε>0$ (using the set theoretic operations of unions and intersections).

I think the answer is $\bigcup_{ε>0}\bigcap_{n=1}^{\infty}\bigcup_{m=n}^{\infty}E_m(ε)$, but am unable to derive it. Any help. Thanks beforehand.

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    @user71352 thanks, why did you not write an answer directly instead of posting 3 comments? And the union and intersection must be interchanged in your comment. – 2017-01-03
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    It appears I have made an error. I will remove my comments. – 2017-01-04

1 Answers 1

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Your answer is correct.

Take the def'n of convergence and observe from it that $(f_n(x))_n$ does NOT converge to $f(x)$ iff for some $e>0$ the set $\{m:|f_m(x)-f(x)|>e\}$ is infinite.

For ease of notation let $F_n(e)=\cup_{m\geq n}E_m(e).$

(1). If $(f_n(x))_n$ does not converge to $f(x)$. For some $e >0$ we have: For every $n$ there exists $m\geq n$ such that $x\in E_m(e).$ Hence for some $e>0$ we have: $x\in F_n(e)$ for all $n.$

So $x\in \cap_{n=1}^{\infty}F_n(e)$ for some $e>0$. Therefore $$x\in \cup_{e >0}\cap_{n=1}^{\infty} F_n(e).$$

(2). On the other hand if $x\in \cup_{e>0}\cap_{n=1}^{\infty}F_n(e)$. For some $e>0$ we have: $x\in F_n(e)=\cup_{m\geq n}E_m(e)$ for every $n.$

So for some $e >0$ we have: For every $n$ there exists $m\geq n$ such that $x\in E_m(e).$ From this, and the def'n of $E_m(e)$, we have, for some $e >0$ : For every $n$ there exists $m\geq n$ with $|f_m(x)-f(x)|>e.$

So for some $e >0 $ the set $\{m:|f_m(x)-f(x)|>e \}$ is infinite. Therefore $ (f_n(x))_n$ does not converge to $x.$

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    Thank you for correcting me. I shall delete my comments from the comment section. – 2017-01-04