# proof for open set

• Jan 30th 2010, 04:26 AM
alexandros
proof for open set
Given the set S={(x,y): |x|<1 , |y|<1} prove that is open in $R^2$.

Using the definition of the open set we have:

S is open in $R^2$ iff for all $(x_{1},y_{1})\in S$ ,there exists an ε>0 such that ,for all $(x_{2},y_{2})$:

$(x_{2},y_{2})\in B((x_{1},y_{1})\epsilon)\Longrightarrow(x_{2},y_{2 })\in S$.

.......................................OR......... ..................................................

if $|x_{1}|<1,|y_{1}|<1$ ,there exists an ε>0 such that:

$\sqrt{(x_{1}-x_{2})^2 + (y_{1}-y_{2})^2}<\epsilon\Longrightarrow |x_{2}|<1,|y_{2}|<1$

What is the epsilon ,so that the implication can be satisfied ??

That is as far i can go.

Any suggestions??
• Jan 30th 2010, 04:38 AM
Dinkydoe
Choose a point $x:=(x_0,y_0)\in S$

Then the maximum $\epsilon > 0$ such that $B^o(x, \epsilon)\subset S$ is:

$\epsilon_{\max} = \min\left\{|1-x_0|, |1-y_0|\right\}$.

But, if you want to show the given square $S$ is open, you need to show that $\epsilon_{\max}> 0$ for all $(x_0,y_0)\in S$
• Jan 30th 2010, 09:10 AM
alexandros
Quote:

Originally Posted by Dinkydoe
Choose a point $x:=(x_0,y_0)\in S$

Then the maximum $\epsilon > 0$ such that $B^o(x, \epsilon)\subset S$ is:

$\epsilon_{\max} = \min\left\{|1-x_0|, |1-y_0|\right\}$.

But, if you want to show the given square $S$ is open, you need to show that $\epsilon_{\max}> 0$ for all $(x_0,y_0)\in S$

So we have:

$\sqrt{(x_{1}-x_{2})^2 + (y_{1}-y_{2})^2}<\epsilon=$min{ $(1-|x_{1}|),(1-|y_{1}|)$} $\Longrightarrow |x_{2}|<1,|y_{2}|<1$.

But how do we get that : $|x_{2}|<1,|y_{2}|<1$ now??
• Jan 30th 2010, 09:45 AM
Plato
Quote:

Originally Posted by alexandros
So we have:

$\sqrt{(x_{1}-x_{2})^2 + (y_{1}-y_{2})^2}<\epsilon=$min{ $(1-|x_{1}|),(1-|y_{1}|)$} $\Longrightarrow |x_{2}|<1,|y_{2}|<1$.

But how do we get that : $|x_{2}|<1,|y_{2}|<1$ now??

If $\left(x_1,y_1\right)\in \mathcal{B}\left(x_0,y_0\right); \varepsilon)$ then $\left| {x_1 } \right| \leqslant \left| {x_1 - x_0 } \right| + \left| {x_0 } \right| < \left( {1 - \left| {x_0 } \right|} \right) + \left| {x_0 } \right| = 1$.
Likewise $\left| {y_1 } \right| < 1$

That proves that $\mathcal{B}\left((x_0,y_0); \varepsilon\right) \subseteq \mathcal{S}$.