Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have the following exercise:

Prove that $$A=\{(x,y)\in \mathbb{R}^{2} \mid x >0\}$$ is a open set.

I try to solve that exercise with the help of definition, so :

To prove that $A$ is open, we show for every point $(x,y) \in A$ there exists an $r>0$ such that $D_{r}(x,y)\subset A$. Now I must know the definition for $D_{r}(x,y)$ and from the definition we find out that: $\displaystyle D_{r}(x,y)=\{(\alpha,\beta)\mid{(\alpha,\beta)-(x,y) <r}\}.$

My question is: How do I prove that there is an $r>0$ such that $D_{r}(x,y) \subset A$ ?

Thanks :)

share|cite|improve this question
up vote 2 down vote accepted


  1. Consider the point $P=(x,y)=(7,0)$, can you think of an $r$ such that $D_r(P)\subset A$?

  2. Consider the point $P=(x,y)=(7,4)$, can you think of an $r$ such that $D_r(P)\subset A$?

  3. Consider the point $P=(x,y)=(7,y_0)$ (for some $y_0$), can you think of an $r$ such that $D_r(P)\subset A$?

  4. Consider a point $P=(x,y)=(x_0,y_0)$ where $x_0>0$, can you think of an $r$ such that $D_r(P)\subset A$?

share|cite|improve this answer
Btw, it is a beautiful town Timisoara, but the Mathematical department need more colour. :) – AD. Aug 23 '12 at 19:54
Timisoara is one of the most beautiful cities in Romania (I'm born there). I know at least 2 very skilled mathematicians at Mathematical department (not personally). – user 1618033 Aug 23 '12 at 21:37

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.