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I want to prove that a function from $\mathbb{R}$ to $\mathbb{R}$ satisfying $f(0)=0$ and $f(1/n) \to 1$ as $n \to \infty$ is discontinuous at $0$ straight from the $\epsilon, \delta$- and the $\epsilon,N$-definition. The difficulty I see is how to relate discrete information to something about continuity. The way I think of is to create a sequence $x_n$ with $x_n \to 0$ so that $f(x_n)$ does not converge to $f(0)=0$. Define the sequence $x_n=1/n$, then $x_n \to 0 $ which means that for $n \geq N$ for some $N$, $x_n \in (-\delta,\delta)$ for a positive $\delta$. Since $f(1/n) \to 1$, for some $N_2$, when $n\geq N_2$, $f(x_n) \in 1-\epsilon, 1+\epsilon)$. So when $n\geq \max(N,N_2)$, $x_n \in (-\delta,\delta)$ and $f(x_n) \in 1-\epsilon, 1+\epsilon)$. Taking $\epsilon=1/3$, $f(x_n)\in (2/3,4/3)$ which does not belong to $(-1/3,1/3)$. Hence it cannot be continuous. However, I used a theorem that states roughly continuity at $a$ is equivalent to that every sequence $x_n$ with $x_n \to a$ must satisfy $f(x_n) \to f(a)$. Is there a neater way?

Arnaldo
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user30523
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1 Answers1

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The definition of continuity says "for each $\epsilon>0$, there exists a $\delta>0$ such that...". Thus, to disprove continuity, you have to show "There is an $\epsilon>0$ for which there is no $\delta>0$ such that..."

In this case, it's enough to choose $\epsilon = \frac12$. What we want to prove, more specifically, is that

For any $\delta>0$, there is a $y\in(-\delta, \delta)$ such that $|f(0)-f(y)|\geq \epsilon = \frac12$

Can you see how to finish this now?

Arthur
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  • Shouldn't it be $x\in (-\delta,\delta)$ such that $|f(x)-f(0)| \geq \frac12$? – user30523 Mar 08 '17 at 17:50
  • @user30523 Yes, of course. You're right. – Arthur Mar 08 '17 at 18:04
  • What about this? Let $\epsilon=1/2$. Since $x_n=1/n \to 0$, for any $\delta>0$ there is $x_n \in (-\delta,\delta)$, (for $n \geq N$, $N\in \mathbb{N}$). Then since $f(x_n) \to 1$, for $x=x_n$, $|f(x)-1|<1/2$ so $|f(x)-f(0)|\geq 1/2$. – user30523 Mar 08 '17 at 19:09
  • That's the idea. However, you should expand a little more. For instance, you should say specifically that $N$ is such that $f(x_n)>1/2$ for all $n>N$, and such an $N$ exists by definition of convergence of a sequence. Then you pick an $n$ such that both $n>N$ and $n>1/\delta$, which gives your conclusion. – Arthur Mar 08 '17 at 19:18