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Prove that the set $ U={\{(x,y) \in \mathbb R^2|y>0}\} $ is an open subset of $ \mathbb R^2 $.

John Taylor's Definition for being open: If $U$ is an open subset of $\mathbb R^d$ we will say that $U$ is open if, for each point $x \in U$,there is an open ball centered at $x$ which is contained in $U$.

attempt at proof: Take $ U$ as defined. We need to show $U$ is open. Let $(x,y) \in U$ be an arbitrary point. Suppose $B_r(x,y)$ is an open ball centered at $(x,y)$. We need to show that $U$ is open amounts to showing that $B_r(x,y)$ is contained in $U$.

proof idea: I was wondering if I should take $U$ and treat it as an open ball and then show that $B_r(x,y)$ is just another ball contained in $U$. I'm also thinking about using the triangle inequality to show that $||(x,y)-(x_0,y_0)||<r$.

Note: $B_r(x,y) = \{(x,y) \in \mathbb R^2:||(x,y)-(x_0,y_0)||<r \}$

Any hints or advise to help me get in the right direction would be helpful.

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  • $\begingroup$ $\forall (x,y) \in U$, take $r = \frac{|y|}{2}$. Show $B_r((x,y)) \subset U$ $\endgroup$ – Good Morning Captain Jan 23 '18 at 17:05
  • $\begingroup$ One possible way is to use the topological definition of open sets in combination with the continuous function $f: U \rightarrow (0,\infty), (x,y) \mapsto f(x,y) := y$: The preimages of open sets (in the range of $f$) under the continuous function $f$ are open sets (in the domain of $f$). Clearly, $(0, \infty)$ is an open subset of $\mathbb{R}$, and $f$ is obviously continuous. Hence, $f^{-1}((0, \infty)) = U$ is open by definition. $\endgroup$ – ComplexFlo Jan 23 '18 at 18:26
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You don't need to show complex relations. All you need to do is, for every $(x,y)$ in $U$, to find a $r$ so that $B_r(x,y)$ is a subset of $U$.

Hint : pick a $r$ such that $r < y$.

Hint 2 : A way to show that $(x_1,y_1) \in B_r(x,y)$ is in $U$ : show that happens if and only if $(x, y_1)$ is in U and $B_r(x,y_1)$. Then using the triangle inequality on $(x_1, y_1)$, $(x, y_1)$ and $(x,y)$ you can easily see that $\mathrm{dist}((x,y),(x,y_1)) \leq 2r$. If you chose $r < \frac y2$, like $r = \frac y4$, you're done.

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As stated in my comment, $\forall (x,y) \in U$, take $r = \frac{|y|}{2}$. Show $B_r((x,y)) \subset U$.

For $(x', y') \in B_r((x,y))$,

Then $|y'| > |y| - \frac{|y|}{2} > 0 \implies (x',y') \in U$

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