Prove that a set is open 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. 
 A: Remember One can do in this way also,

I am going to use these two aspects.
$1.$ If $f$ is contiuous and $A$ is open, then $f^{-1} (A)$ is also open.
$2.$ Finite intersection of open sets is again open.

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.
A: $U=\{(x,y)|x+y<1\}\cap \{(x,y)|x>0\}\cap \{(x,y)|y>0\}$, because finite intersection of open sets is open, it's enough to prove each individual set is open.
Let $f(x,y)=x+y$, $g(x,y)=x$, $h(x,y)=y$ be functions from $R^2\rightarrow R$, they are clearly continuous. We have $U=f^{-1}(-\infty, 1)\cap g^{-1}(0,\infty)\cap h^{-1}(0,\infty)$. Because inverse image of an open set under continuous function is open, the original set is open.
A: $U^{c}$ i.e. compliment of $U$ which is same as $$\{ (x,y)\in\mathbb{R}^{2} |x+y\geq1,x>0,y>0 \}\cup \{ (x,y)\in\mathbb{R}^{2}|x>,y>0\}^{c}$$  is  a closed set as it contains all of its limit points. 
