# Let $S\subseteq\mathbb{R}^2$ be the set of points where $f$ is continuous - is $S$ open or closed? Is it empty?

Define function $f : \mathbb{R^2}\rightarrow \mathbb{R}$ by

$$f(x, y) = \begin{cases}1,&\text{if }xy=0\\2,&\text{otherwise}\;.\end{cases}$$

If $$S = \{(x, y): f\text{ is continuous at the point }(x,y) \}\;,$$

what can we say about $S$? Whether $S$ is open, closed, or empty set ?

I tried by picking up some points of $\mathbb{R}$ to check its inverse images. I need proper proof. I appreciate any kind of help. And my sincere thanks to you.

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What’s a simple geometric description of the set of points where $f$ is $1$? –  Brian M. Scott May 13 '12 at 20:54
@BrianM.Scott fixed points? –  srijan May 13 '12 at 20:58
No, $f$ is $1$ at points on the coordinate axes. If a point is on one of the axes, are there points very close to it where $f$ is $2$? –  Brian M. Scott May 13 '12 at 21:00
Try to draw the graph of $f$. You will see where discontinuities are. –  Norbert May 13 '12 at 21:01
@BrianM. $f$ is 2 every where in $\mathbb{R^2}$ except at coordiante axes. –  srijan May 13 '12 at 21:05

Let \begin{align*} Q_1&=\{(x,y)\in\mathbb{R}^2\mid x>0, y>0\}\\ Q_2&=\{(x,y)\in\mathbb{R}^2\mid x<0, y>0\}\\ Q_3&=\{(x,y)\in\mathbb{R}^2\mid x<0, y<0\}\\ Q_4&=\{(x,y)\in\mathbb{R}^2\mid x>0, y<0\} \end{align*} be the four open quadrants in the plane, and let \begin{align*} A_x&=\{(x,y)\in\mathbb{R}^2\mid y=0\}\\ A_y&=\{(x,y)\in\mathbb{R}^2\mid x=0\} \end{align*} be the $x$- and $y$-axes, respectively. Then $f(x,y)=2$ if and only if $$(x,y)\in Q_1\cup Q_2\cup Q_3\cup Q_4$$ and $f(x,y)=1$ if and only if $$(x,y)\in A_x\cup A_y.$$ To provide a picture, $f(x,y)=2$ in the blue region, and $f(x,y)=1$ in the red region:

$\hskip1.6in$

By definition, the function $f$ is continuous at $(a,b)\in\mathbb{R}^2$ if and only if, for any open set $V\subseteq\mathbb{R}$ containing $f(a,b)$, there is an open set $U\subseteq\mathbb{R}^2$ containing $(x,y)$ such that $f(U)\subseteq V$.

Suppose $f(a,b)=1$. The set $V=(\frac{1}{2},\frac{3}{2})$ is an open subset of $\mathbb{R}$ containing $1$. Because $f$ only takes the values $1$ and $2$, the only way to have $f(x,y)\in V$ is to have $f(x,y)=1$. Thus, the only way to have $f(U)\subseteq V$ is if $f(x,y)=1$ for every $(x,y)\in U$, or in other words, $U\subseteq A_x\cup A_y$. Are there any non-empty subsets $U\subseteq A_x\cup A_y$ that are open as subsets of $\mathbb{R}^2$? What does that tell you about whether it's possible for $f$ to be continuous at any point in $A_x\cup A_y$ (the red region)?

Now do the same analysis but for when $f(a,b)=2$. You could let $V=(\frac{3}{2},\frac{5}{2})$, for example.

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many many thanks to you –  srijan May 14 '12 at 15:55
It is clear that $f(x,y)=1$ for the $x$ and $y$ axes (i.e. $x=0$ or $y=0$). For every point $(x,y) \in f^{-1}(2)$, i.e. all the points that are not on the axes, $f$ is continuous as a constant function. So $f^{-1}(2) \subset S$. However, $f$ is not continuous on the axes (edit: there is no open ball $B$ in $\mathbb{R}^2$ such that $B \subset f^{-1}(2)$) , and since they are closed subsets of $\mathbb{R}^2$ their complement is open. One can see directly that $S= f^{-1}(2)$ is an open subset, since every point $(a,b)$ (with $a,b \ne 0$) has an open ball with radius $r = \min(a/2,b/2)$ which is fully contained in $S$.
I used the standard topology on $\mathbb{R}^2$ and $\mathbb{R}$ and the definition of continuity using real limits ($\lim_{ (x,y) \to (a,b)}f(x,y) = f(a,b)$).
Yes, of course $f(x,y)=1$ (but since $0\ne2 , 1 \ne 2$ it really doesn't change much the solution). –  Zachi Evenor May 13 '12 at 21:37