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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 :)

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1 Answer 1

up vote 2 down vote accepted

Hint:

  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 $D_r(P)\subset A$?

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1  
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). –  Chris's sis Aug 23 '12 at 21:37

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