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Prove that if the point $x_0$ in $D$ is a limit point, then a function $f:D\rightarrow\mathbb{R}$ is continuous at $x_0$ iff $\lim\limits_{x\rightarrow x_0}f(x)=f(x_0)$


$\Longrightarrow$ Since the point $x_0\in D$ is a limit point and the function $f:D\rightarrow\mathbb{R}$ is continuous at $x_0$, then by the definition of limit point, we can have $\lim\limits_{x\rightarrow x_0}f(x)=f(x_0)$.

$\Longleftarrow$ Since $\lim\limits_{x\rightarrow x_0}f(x)=f(x_0)$, this tells us $x_0$ is a limit point in $D$. Then there must exists a sequence $\{x_n\}$ converges to $x_0$ where for all $n$, $x_n\in D-\{x_0\}$, we can have $\lim\limits_{n\rightarrow\infty}f(x_n)=f(x_0)=\lim\limits_{x\rightarrow x_0}f(x)$; the function $f:D\rightarrow\mathbb{R}$ is continuous at $x_0$.


Can someone tell me where I did incorrect? Thanks in advanced.

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  • $\begingroup$ Is the limit point property of $x_0$ independent of the equivalence? $\endgroup$ – mvw Nov 6 '15 at 3:45
  • $\begingroup$ @mvw yes, it is. $\endgroup$ – Simple Nov 6 '15 at 3:47
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    $\begingroup$ What definition of "continuous at $x_0$" are you using? The equivalence you have given here is almost always given as the definition of continuous at $x_0$? $\endgroup$ – charlestoncrabb Nov 6 '15 at 3:58
  • $\begingroup$ @charlestoncrabb I am using the sequential continuity $\endgroup$ – Simple Nov 6 '15 at 4:01
  • $\begingroup$ The $\Rightarrow $ is very likely incorrect. If you are using the sequential definition of continuity, you need to show that $\lim f(x_n) = f(x_0)$ for all sequence $x_n\to x_0$ implies $\lim_{x \to x_0} f(x) = f(x_0)$. $\endgroup$ – user99914 Nov 6 '15 at 4:10
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From the comment we know that the OP is using the sequential definition of continunity: $f$ is continuous at $x_0 \in D$ if $$\lim_{n\to \infty} f(x_n) = f(x_0)$$ for all sequences $\{x_n\}$ in $D$ converging to $x_0$. Now

$$\tag{1} \lim_{x\to x_0} f(x) = f(x_0)$$

is by definition:

For all $\epsilon >0$, there is $\delta >0$ so that if $$|f(x) - f(x_0)| <\epsilon$$ whenever $x\in D$ and $|x-x_0| <\delta$.

Now we show $(\Rightarrow)$. Assume that $(1)$ does not hold. Then:

There is $\epsilon_0 >0$, so that all for $\delta >0$, $$|f(x) - f(x_0)| \ge \epsilon_0$$ for some $x\in D$ and $|x-x_0| <\delta$.

In particular, setting $\delta = 1/n$, there is a sequence $\{x_n\}$ in $D$ so that $|x_n - x_0|<1/n$ and $|f(x_n) - f(x_0)| \ge \epsilon_0$. So $x_n \to x_0$, but $\{f(x_n)\}$ does not converge to $f(x_0)$. So $f$ is not continuous at $x_0$. By contrapositive, we have shown the $(\Rightarrow)$ part.

For the $(\Leftarrow)$ part, it is somehow easier. Given $(1)$, we want to show that $f$ is continuous at $x_0$. So let $\{x_n\}$ be any sequence in $D$ converging to $x$. Let $\epsilon >0$. By $(1)$, there is $\delta >0$ so that $$\tag{2} |f(x) - f(x_0)|<\epsilon$$ whenever $x\in D$ and $|x-x_0|<\delta$ $(3)$. As the sequence $\{x_n\}$ converges to $x_0$, there is $N \in \mathbb N$ so that $|x_n - x_0| <\delta$ whenever $n\ge N$. Combining $(2)$ and $(3)$, we have $$|f(x_n) - f(x_0)|<\epsilon$$ whenver $n\ge N$. As $\epsilon >0$ is arbitrary, we have $$\lim_{n\to \infty} f(x_n) = f(x).$$ As a result, $f$ is continuous at $x_0$.

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