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Show that the subset of $\mathbb{R}^2$ given by $\{(x_1, x_2)\in \mathbb{R}^2:x_1>x_2\}$ is open.

Can anyone give me some help, or a proof for it. The book I'm using defines a subset $S$ of $E$ as open if for each $p\in S$, $S$ contains some open ball of center $p$.

Any input is appreciated.

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    Sorry, $E^2$ is the Euclidean plane with the standard distance function. I see that is true, I just can't find a way to show it.2012-11-23

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Let's call this subset $S$, and let $(x_1,x_2) \in S$. Suppose that $x_1 - x_2 = r > 0$. Can see why the open ball of radius $r/10$ around $(x_1,x_2)$ should remain in $S$?

Suppose that there is a point $(y_1,y_2)$ in this ball. The triangle inequality tells us

$ r = |x_1 -x_2| \leq |x_1 - y_1| + |y_1 - y_2| + |y_2 - x_2|$

We know that $|x_1 - y_1|$ and $|y_2 - x_2|$ can be at most $r/10$, so their sum is at most $r/5$. This means that $|y_1 - y_2|$ is at least $4r/5 > 0$, and so $(y_1,y_2) \in S$.

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    Yes, thank you.2012-11-23
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Define $f:E^2\to\mathbb R$ by $f(x_1,x_2)=x_1-x_2$. Your set is then $f^{-1} (0,\infty)$ which is open because $(0,\infty)$ is open and $f$ is continuous.