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I would really appreciate a proof for the Discontinuity Criterion Theorem for functions. It is stated as such...

Let $A$ be a subset of $\mathbb{R}$, let $f: A \to \mathbb{R}$ and let $c \in A$. Then $f$ is discontinuous at $c$ iff there exists a sequence $x_n \in\mathbb{R}$ such that $x_n$ converges to $c$ but the sequence $f(x_n)$ does not converge to $f(c)$.

Thank you!

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  • $\begingroup$ You should add your definition of continuity at a point. It could be exactly the contrapositive of this criterion, but I suppose that it is not. $\endgroup$ – Carsten S Dec 4 '17 at 12:19
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$(\leftarrow )$ Since $f(x_n)$ doesnot converge to $f(c)$, there exist a neighborhood $U$ of $f(c)$ such that $\{f(x_n)\} \cap U$ is finite. Now $x_n \rightarrow c$ so if we take any neighborhood $V$ of $c$, then for some $N \in \mathbb{N}$ we have $x_n \in V$ for all $n \ge N$. That is $f(V) \nsubseteq U$ for any such $V$ because $\{f(x_n)\} \cap f(V)$ is infinite. So, f is not continuous at $x=c$.

$(\rightarrow)$ Since $f$ is discontinuous at c, we have a neighborhood $V$ of $f(c)$ such that $f^{-1}(V)$ is not a neighborhood of $c$ but $c \in f^{-1}(V)$. So $c$ is a boundary point of $f^{-1}(V)$ which means there exist a sequence $\{x_n\} \subset f^{-1}(V^c)$ such that $x_n \rightarrow c$ but $\{f(x_n)\}$ cannot converge to $f(c)$ as $\{f(x_n)\} \cap V = \emptyset$ and $f(c) \in interior(V)$.

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A function is discontinuos at a point if and only if $\lim\limits_{x\to c}f(x)\neq f(c)$. I will use this definition to prove your theorem.

  • Suppose $f$ is discontinuos at $c$. We will prove there exist a sequence $x_n$ which converges to $c$ but for which $\lim\limits_{n\to\infty}f(x_n)=f(c)$ is false. Since our function is discontinuos, $\exists \epsilon>0$ $\forall\delta>0$ $\exists x\in X$ such that $|x-c|<\delta$ but $|f(x)-f(c)|\geq\epsilon$. We pick up $\delta_n=\frac{1}{n}$, thus obtaining , for each $\delta_n$ a $x_n$, defining a sequence which converges to $c$, but for which $\lim\limits_{n\to\infty} f(x_n)=f(c)$ is false.
  • Suppose now that $f$ is continuos at $c$ (we are going to prove the last implication by a contrapositive). Then $\forall\epsilon>0$ $\exists \delta>0$ such that $|x-c|<\delta$ implies $|f(x)-f(c)|<\epsilon$. Pick up an arbitrary sequence $x_n$ which converges to $c$. For this $\delta$ there exists $N :\forall n\geq N$ $|x_n-c|<\delta$. This proves that $\lim\limits_{n\to\infty}f(x_n)=f(c)$, as we desired.
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