I think this is right but hoping someone could verify that I am right, tell me a better way! I am trying to learn how to show that sets are open and closed.
So I have a set that I wish to show is open
$$
U = \{(x,y) \in R^{2}_{x} | x+y < 1, x > 0, y >0 \}
$$
So I need to show that
$$
B_{r}(a) = \{x \in U | d(x,a) < r\}
$$
(not sure if $a \in U$ or $a\in R^2_+$) could someone clarify?
if for all $x \in U$, $B_{r}(a) \in U $ then we can conclude that U is open
So my first step was to let $ x = (x_1,y_1) \in U, a = (x_2, y_2) \in U$ and $r = 1 – \min(x_1,x_2) – \min(y_1,y_2)$
Then I said since:
$\sqrt{ (x_1 – x_2)^2 + (y_1 – y_2)^2 } < |x_1 – x_2| + |y_1 – y_2| = \max(x_1,x_2) – \min(x_1,x_2) + \max(y_1,y_2) – \min(y_1,y_2) $
Then we can say:
$\max(x_1,x_2) – \min(x_1,x_2) + \max(y_1,y_2) – \min(y_1,y_2) < 1 – \min(x_1,x_2) – \min(y_1,y_2)$
which implies:
$\max(x_1,x_2) + \max(y_1,y_2) < 1 $
And since this last condition is true we can conclude that $B_r(a) \in U$ and therefore the set is open.
Best Answer
Remember One can do in this way also,
Define a function $f: \mathbb R^2\to\mathbb R^2$ by $ f(x,y)=x+y$ which is a countinuous function.
So, inverse of open set is an open set.
$f^{-1}((-\infty, 1))=\{(x,y)\in \mathbb R^2: x+y<1\}\tag{1}$
Now, define another function $P_x:\mathbb R^2\to \mathbb R^2$ by $P_x(x,y)=x$ Similarly, $P_y$.
So, $P_x^{-1}((-\infty,0))=\{(x,y)\in \mathbb R^2: x<0\}\tag{2}$
and $P_y^{-1}((-\infty,0))=\{(x,y)\in \mathbb R^2: y<0\}\tag{3}$
All the sets $(1),(2),(3)$ are open, so intersection is also open, since the finite intersection of open sets is also open.