Theorem: Expression for Set of Points at which Sequence of Functions does not Converge to Given Function

Let $X$ be a set. 
Let $f : X \to \R$ be a real function.
For each $n \in \N$, let $f_n : X \to \R$ be a real function.
Then we have: 
:$\ds \set {x \in X : \sequence {\map {f_n} x}_{n \in \N} \text { does not converge to } \map f x} = \bigcup_{k \mathop = 1}^\infty \bigcap_{N \mathop = 1}^\infty \bigcup_{n \mathop = N}^\infty \set {x \in X : \size {\map {f_n} x - \map f x} \ge \frac 1 k}$

Proof:
It helps to rewrite:
{{begin-eqn}}
{{eqn	| l = \bigcup_{k \mathop = 1}^\infty \bigcap_{N \mathop = 1}^\infty \bigcup_{n \mathop = N}^\infty \set {x \in X : \size {\map {f_n} x - \map f x} \ge \frac 1 k}
	| r = \bigcup_{k \mathop = 1}^\infty \bigcap_{N \mathop = 1}^\infty \set {x \in X : \size {\map {f_n} x - \map f x} \ge \frac 1 k \text { for some } k \ge n}
}}
{{eqn	| r = \bigcup_{k \mathop = 1}^\infty \set {x \in X : \text {for each } N \in \N \text { there exists } n \ge N \text { such that } \size {\map {f_n} x - \map f x} \ge \frac 1 k}
}}
{{eqn	| r = \set {x \in X : \text {there exists some } k \in \N \text { such that for each } N \in \N \text { there exists } n \ge N \text { such that } \size {\map {f_n} x - \map f x} \ge \frac 1 k}
}}
{{end-eqn}}
We have that $\sequence {\map {f_n} x}_{n \in \N}$ does not converge to $\map f x$ {{iff}}:
:there exists a real number $\epsilon > 0$ such that for each $N \in \N$ there exists $n \ge N$ such that $\size {\map {f_n} x - \map f x} \ge \epsilon$
Clearly then, setting $\epsilon = 1/k$, if $x \in X$ is such that: 
:there exists $k \in \N$ such that for each $N \in \N$ there exists $n \ge N$ such that $\size {\map {f_n} x - \map f x} \ge \dfrac 1 k$ 
we have that: 
:$\sequence {\map {f_n} x}_{n \in \N}$ does not converge.
So:
:$\ds \set {x \in X : \text {there exists some } k \in \N \text { such that for each } N \in \N \text { there exists } n \ge N \text { such that } \size {\map {f_n} x - \map f x} \ge \frac 1 k} \subseteq \set {x \in X : \sequence {\map {f_n} x}_{n \in \N} \text { does not converge to } \map f x}$
Now suppose that $x \in X$ is such that: 
:there exists a real number $\epsilon > 0$ such that for each $N \in \N$ there exists $n \ge N$ such that $\size {\map {f_n} x - \map f x} \ge \epsilon$
Pick $k \in \N$ sufficiently large so that: 
:$\epsilon \ge \dfrac 1 k$ 
Then we have that: 
:there exists $k \in \N$ such that for each $N \in \N$ there exists $n \ge N$ such that $\size {\map {f_n} x - \map f x} \ge \dfrac 1 k$.
So:
:$\ds \set {x \in X : \sequence {\map {f_n} x}_{n \in \N} \text { does not converge to } \map f x} \subseteq \set {x \in X : \text {there exists some } k \in \N \text { such that for each } N \in \N \text { there exists } n \ge N \text { such that } \size {\map {f_n} x - \map f x} \ge \frac 1 k}$
Then from the definition of set equality:
:$\ds \set {x \in X : \sequence {\map {f_n} x}_{n \in \N} \text { does not converge to } \map f x} = \set {x \in X : \text {there exists some } k \in \N \text { such that for each } N \in \N \text { there exists } n \ge N \text { such that } \size {\map {f_n} x - \map f x} \ge \frac 1 k}$
That is:
:$\ds \set {x \in X : \sequence {\map {f_n} x}_{n \in \N} \text { does not converge to } \map f x} = \bigcup_{k \mathop = 1}^\infty \bigcap_{N \mathop = 1}^\infty \bigcup_{n \mathop = N}^\infty \set {x \in X : \size {\map {f_n} x - \map f x} \ge \frac 1 k}$
{{qed}}
Category:Convergence
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