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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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 $x_0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.