# the set of discontinuity points of function $f(x,y)=x\sin\frac{1}{y}$ is not closed

In exercise says that I have to show that the set of discontinuity points of function

$$f(x,y)=\begin{cases} x\sin\frac{1}{y}, & \text{if y \ne 0} \\ 0, & \text{if y = 0} \end{cases}$$

is not closed.

Is this function have other points of discontinuity except $0$? If not, then why one point is not closed set?

• Your question makes no sense, because $0$ does not even belong to the domain of $f$. – José Carlos Santos Sep 23 '17 at 15:58
• The set y = 0 is not a single-point set. – Xindaris Sep 23 '17 at 15:59

The function $f$ is obviously continuous on $\mathbb{R}^2 \setminus (\mathbb{R} \times \{0\})$.

For $x \ne 0$ let's show that $f$ is discontinuous at $(x,0)$:

Consider the sequence $y_n = \frac{1}{\frac{\pi}{2}+ 2n\pi}$. We see that $y_n \to 0$ and:

$$\lim_{n\to\infty} f(x,y_n) = \lim_{n\to\infty} x \sin \left(\frac{\pi}{2} + 2n\pi\right)= x$$

However, for $z_n = \frac{1}{\frac{3\pi}{2} + 2n\pi}$ we have $z_n \to 0$ and

$$\lim_{n\to\infty} f(x,z_n) = \lim_{n\to\infty} x \sin \left(\frac{3\pi}{2} + 2n\pi\right)= -x$$

Thus, since $x \ne 0$ we see that the limit $\lim_{(x,y)\to (x,0)} f(x,y)$ does not even exist.

Now, let's check that $f$ is continuous at $(0,0)$:

$$\lim_{(x,y)\to (0,0)} |f(x,y)| = \lim_{(x,y)\to (0,0)} |x|\left|\sin\frac{1}{y}\right| \le \lim_{(x,y)\to (0,0)} |x| = 0$$

Hence $\lim_{(x,y)\to (0,0)} f(x,y) = 0 = f(0,0)$.

Therefore, the set of discontinuities of $f$ is $D = (\mathbb{R} \setminus \{0\}) \times \{0\}$, which is not closed since $0 \notin D$.

Hint:

1. Remember that this is a function of two variables.
2. Is the function continuous or discontinuous in $(0,0)$?
3. What about all other points $(x, y)$ in which $y=0$? What does this mean for the exact set of discontinuity points?