if $f: D\subseteq \mathbb{R}^n \to R$ is continuous at $a \in D$, then there exists a $\delta > 0$ s.t $f$ is continuous on $B(a, \delta)$?

Question

Question:

Is it true that if $f: D\subseteq \mathbb{R}^n \to R$ is continuous at $a \in D$, then there exists a $\delta > 0$ s.t $f$ is continuous on $B(a, \delta)$ ?

Reasoning:

if $f$ is continuous at $a \in D$, given $\epsilon > 0$, $\exists \delta >0$ s.t $$f(B(a, \delta) ) \subseteq B(f(a), \epsilon),$$ so take $\delta_1 > 0$ in such a way that $c\in B(a, \delta)$ and $B(c, \delta_1) \subseteq B(a, \delta)$ (it is trivial to show the existence of $\delta_1$).

Now, we do know that $f (B(c, \delta_1)) \subseteq B(f(a), \epsilon)$, but how about $f (B(c, \delta_1)) \subseteq B(f(c), \epsilon_1)$ where $\epsilon > \epsilon_1 > 0$ ?

Answer 1

You cannot prove it because it's false.

Take for example the modified Dirichlet function: $$f(x)=\begin{cases}0 & \text{if } x\in\mathbb{R}\setminus\mathbb{Q},\\ \frac{1}{q} & \text{if } x=\frac{p}{q}\text{ as reduced fraction.}\end{cases} $$

This function is continuous in the irrationals and discontinuous in the rationals, so you cannot find such an open set around $\sqrt{2}$ where $f$ is continuous, as the rationals are dense in $\mathbb{R}$.

Answer 2

Here is another example, why it is false. Consider the following function

$$ g(x) = \begin{cases} 0,& x\in \mathbb{Q}, \\ x^2,& x \in \mathbb{R}\setminus \mathbb{Q}. \end{cases} $$

It is continuous in 0 (even differentiable), as we have

$$ 0\leq \left\vert \frac{g(x) - g(0)}{x-0} \right\vert \leq \left\vert \frac{x^2}{x} \right\vert = \vert x \vert \rightarrow 0, \qquad x\rightarrow 0.$$

Thus, by the squeeze theorem, we get $g'(0)=0$. However, for $x\neq 0$ the function is not continuous.

Take $x\in \mathbb{R}\setminus \mathbb{Q}$, and a sequences $(y_n)_{n\in \mathbb{N}} \subseteq\mathbb{Q}$ such that $\lim_{n\rightarrow \infty} y_n = x$. Then we have

$$ \lim_{n\rightarrow \infty} f(y_n) = \lim_{n\rightarrow \infty} 0 = 0 \neq x^2 = f(x). $$

Thus, $f$ is not continuous in $x$. Similarly one can argue for $x\in \mathbb{Q}\setminus \{0\}$.