6

I temporarily say function $f(x,y)$ (from $\mathbb{R}^2$ to $\mathbb{R}$) is coordinately continuous (shortly as c.c.) iff it is continuous everywhere regarded as an single-variable function while another coordinate is given.

I am curious about how large the set of uncontinuous (in sense of $\mathbb{R}^2$) points of a c.c. function can be .

It is not hard to see that there are c.c. functions which is not continuous at one point.

For example, $f(x,y)= 1−4(xy/(x^2+y^2))^2$ is not continuous at $(0,0)$, which can be seen from the form of polar coordinate $1 - \sin(2\theta)^2$ except $(0,0)$.

Intuitionly, I think discontinuity while maintaining c.c. property needs a well designed neighbor. Therefore I think the set of discontinuity points is not dense.

It has been proved that such function cannot be discontinuous everywhere: separately continuous functions $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ but nowhere continuous

Chennes
  • 287
  • 1
    Perhaps this answer gives a simple example of a cc function that has a discontinuity at one point. – Theoretical Economist Oct 17 '19 at 04:00
  • 2
    https://math.stackexchange.com/questions/3019208/separately-continuous-functions-f-mathbbr2-rightarrow-mathbbr-but-now This might help to find a dense discontinuity set. – Jason Gaitonde Oct 17 '19 at 04:09
  • 1
    @J.G This is interesting. The answer says that the set of discontinuities is of first category (http://mathworld.wolfram.com/FirstCategory.html). – amsmath Oct 17 '19 at 04:16
  • @amsmath yes, I read that this is a result of Baire (of course!) and has since been substantially generalized, but couldn’t think of a good argument off the top of my head so figured I’d take the easy way out and give a link! – Jason Gaitonde Oct 17 '19 at 04:21
  • 1
    @Chennes This implies that the set of continuities $C$ is dense. Namely, $C^c\subset\bigcup_n E_n$, where each $E_n$ is closed with empty interior. So, each $E_n^c$ is open and dense. Hence, $\bigcap_n E_n^c\subset C$ and $\bigcap_n E_n^c$ is dense by Baire's theorem. – amsmath Oct 17 '19 at 04:37

1 Answers1

4

The set of points of discontinuity can be everywhere dense; in fact, it can be any countable subset of the plane.

Let $D=\{(x_n,y_n):n\in\mathbb N\}$ be a countable subset of $\mathbb R^2$.

You can easily contruct a function $h:\mathbb R^2\to[0,1]$ which is discontinuous at $(0,0)$ and continuous everywhere else, and has the property that its restriction to any straight line in the plane is continuous; for example, the function $$h(x,y)=\begin{cases} \quad\ \ 0\quad\quad\quad\text{ if }\quad y\le2\pi x^2,\\ \sin(y/x^2)\quad\text{ if }\quad2\pi x^2\lt y\lt3\pi x^2,\\ \quad\ \ 0\quad\quad\quad\text{ if }\quad y\ge3\pi x^2.\\ \end{cases}$$

For each $n\in\mathbb N$ the function $h_n(x,y)=h(x-x_n,y-y_n)$ is discontinuous at $(x_n,y_n)$ and continuous everywhere else, and is continuous on every straight line.

Since the series converges uniformly, the function $$f(x,y)=\sum_{n=1}^\infty\frac{h_n(x,y)}{2^n}$$ is discontinuous at each point of $D$ and continuous everywhere else, and is continuous on every straight line in the plane. In particular, $f(x,y)$ is continuous in $x$ for each fixed value of $y$, and continuous in $y$ for each fixed value of $x$.

bof
  • 82,298
  • Thank you for your nice construction! I am thinking whether there is some medium property between dense property and interior area to investigate. – Chennes Oct 18 '19 at 14:30