Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
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
2  
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. –  no identity 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

2 Answers 2

up vote 0 down vote accepted

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:

$\hskip1.6in$enter image description here

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.

share|improve this answer
    
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)$).

share|improve this answer
    
you meant f(x,y)=1 for the x,y axes? –  user1412 May 13 '12 at 21:08
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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