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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.

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  • $\begingroup$ A set $A$ is said to be open in $R^2$ with the usual metric, if $\forall x\in A$ there exist $r>o$ such that $B_r(x)\subset A$. $\endgroup$ – David Sep 23 '15 at 6:27
  • $\begingroup$ So I am to assume that it CAN be shown the way I did, but for my specific example it was probably not the easiest/best way to do it? $\endgroup$ – seakay Sep 23 '15 at 6:58
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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.

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  • $\begingroup$ @seakay if you are finding the answer satisfies your needs then do not forget to accept it. that will help other not to discuss more this.. $\endgroup$ – David Sep 23 '15 at 6:49
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$U=\{(x,y)|x+y<1\}\cap \{(x,y)|x>0\}\cap \{(x,y)|y>0\}$, 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.

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$U^{c}$ 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.

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