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.
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.