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.

(1) Is $U=\{(x,y)\in \mathbb{R^2} : x^2+y^2 \neq 1\}$ open in $\mathbb{R^2}$?

(2) How can i show $(a,b)\times(c,d)$ is open in $\mathbb{R^2}$?

(3)Is $S=\{(x,y)\in \mathbb{R^2}:xy\neq 0\}$ open in $\mathbb{R^2}$?

In all these examples i draw the figure in (1) it is interior part of circle and exterior part of circle so i can draw any ball with some radius it is also containing in the set.

Similar argument for (2) which is rectangle region in $\mathbb {R^2}$ and in the (3) is unbounded region without containing X and Y axis. It is very clear from the geometry but Anyone help me how can i prove all these three sets are open using definition of open set or triangle inequality? If it is possible then give me some hints. Thanks in advance.

share|improve this question
Hint: If f is continuous then the preimage of any open set is open. –  vanna Aug 28 '12 at 13:29
@Siddhant What is your definition of an open set? –  process91 Aug 28 '12 at 19:08
(X,d) be a metric space and let G$\subset$X for all a $\in G$ there exists corresponding r such that S(a,r)$\subset$G then G is called open set. –  Siddhant Trivedi Aug 29 '12 at 2:50
@vanna how can i find preimage without function? –  Siddhant Trivedi Sep 1 '12 at 4:48
@SiddhantTrivedi : $$U = \{(x,y)\in\mathbb{R}^2, \ f(x,y) \in \mathbb{R}\backslash \{1\}\}$$ where $f(x,y) = x^2 + y^2$ is continuous. Then $U = f^{-1}(\mathbb{R}\backslash\{1\})$. As $\{1\}$ is closed, its complement is open. –  vanna Sep 1 '12 at 15:40
show 2 more comments

4 Answers

up vote 3 down vote accepted

They are all open. Let's do Question $(3)$, quite formally. The ideas for the others are the same.

Our region is the plane with the union of the two axes removed. Let $(a,b)$ be in our region, and let $\epsilon=\min(|a|, |b|)$. Then $\epsilon$ is positive.

Now consider the open disk $D$ with centre $(a,b)$ and radius $\epsilon$, or, if you want some wiggle room, radius $\epsilon/2$. I claim that all of $D$ is in our region. That will show that there is an open "ball" with center $(a,b)$ that lies entirely in our region. That is one of the definitions of open set in a metric space, I hope the official one you are using in your course.

We need to show that there is no point in the union of the two axes which is in $D$. Note that $\sqrt{(x-a)^2+(y-b)^2}\ge \max(|x-a|,|y-b|)$. If $\sqrt{(x-a)^2+(y-b)^2}\lt \epsilon$, then $\max(|x-a|, |y-b|)\lt \epsilon$.

But by the choice of $\epsilon$, we have $|0-a|\ge \epsilon$, and $|0-b|\ge \epsilon$. It follows that no point in $D$ can lie on either of the axes.

Remark: There is much less to this argument than meets the eye! Take for example $(a,b)$ in the first quadrant, but not on an axis. Let $\epsilon$ be the smaller of $a$ and $b$. Then the (open) disk with centre $(a,b)$ and radius $\epsilon$ cannot meet an axis.

share|improve this answer
how can i express this mathematically i know it is very clear from the geometry. –  Siddhant Trivedi Aug 28 '12 at 14:00
It is done in the post above, using the language of inequalities. The key inequality is $\sqrt{(x-a)^2|(y-b)^2}\ge \max(|x-a|,|y-b|)$ ($\ge \min(|x-a|,|y-b|$ would be enough.) –  André Nicolas Aug 28 '12 at 14:05
add comment

For (1) and (3) it might be easier to prove the complements are closed. You can prove that the unit circle is closed by showing that every sequence on the unit circle has limit on the unit circle. And similarly for (3) were you can prove that all limit points are on the lines $x=0$ or $y=0$.

share|improve this answer
add comment

(2) is open by definition, in some sense. It is the product of open intervals and if you pick a point in this product, you can easily find a small radius such that the ball with this radius lies inside $(a,b) \times (c,d)$. It is a matter of elementary geometry. For (1) and (2), consider their complements, and prove that they are closed. The complement of $S$ is the union of the two axes, while the complement of $U$ is a circle. But a direct approach is also easy: let's see in case (3).

Pick $(x_0,y_0) \in S$, so that $x_0 \neq 0$ and $y_0 \neq 0$. Let $R=\min\{|x_0|,|y_0|\}$ and fix $0<r<R$. The circle $(x-x_0)^2+(y-y_0)^2 < r$ lies in $S$, since it cannot touch the axes.

share|improve this answer
add comment

Of course one can work with good old $\varepsilon$ to show openness, though a few simple theorems would make life easier (preimages of open sets under continuous functions are open etc., but if you don't "see" that these sets are open, you're probably not sure about the functions in (1) and (3) being continuous either.

(1) Assume $(x_0,y_0)\in U$. Then $x_0^2+y_0^2\ne 1$, i.e. $d((x_0,y_0),(0,0))\ne 1$. We want to find $\epsilon>0$ such that $d((x_0,y_0),(x,y))<\epsilon$ implies $(x,y)\in U$. By the triangle inequality, $|d((x_0,y_0),(0,0))-d((x,y),(0,0))|\le d((x_0,y_0),(x,y))$, hence letting $\epsilon = |d((x_0,y_0),(0,0))-1|>0$ does the trick: $d((x_0,y_0),(x,y))<\epsilon$ implies that $|d((x_0,y_0),(0,0))-d((x,y),(0,0))|<|d((x_0,y_0),(0,0))-1|$, hence $d((x,y),(0,0))$ cannot equal $1$. Therefore $(x,y)\in U$.

(2) If $(x_0,y_0)\in U$, then $a<x_0<b$ and $c<y_0<d$. Then $\epsilon:=\min\{x_0-a, b-x_0, y_0-c, d-y_0\}$ is $>0$. Then $d((x_0,y_0),(x,y))<\epsilon$ implies e.g. $|x-x_0|<\epsilon$ and $|y-y_0|<\epsilon$, hence $x>x_0-\epsilon\ge x_0-(x_0-a)=a$ and similarly $x<b$, $y>c$, $y<d$. Therefore $(x,y)\in U$.

(3) If $(x_0,y_0)\in U$ then $x_0\ne0$ and $y_0\ne 0$. Therefore $\epsilon:=\min\{|x_0|,|y_0|\}$ is $>0$. Then $d((x_0,y_0),(x,y))<\epsilon$ implies $|x-x_0|<\epsilon\le|x_0|$, hence $x$ cannot be $0$. Similarly we conclude $y\ne0$. It follows that $xy\ne0$, i.e. $(x,y)\in U$.

share|improve this answer
add comment

Your Answer


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.