Show that the set $A = \{(x,y) \in \Bbb R^2 \mid x>0, y>0 \}$ is open in set in $\Bbb R^2$ with the Euclidean metric. 
Show that the set $A = \{(x,y) \in  \Bbb R^2 \mid x>0, y>0 \}$ is open in set in $\Bbb R^2$ with the Euclidean metric.

Pick $(x,y) \in A$. We need to show that there's an $\varepsilon >0$ for which $B((x,y), \varepsilon) \subset A.$ Let $\varepsilon = \min\{x,y\}$ and pick some $(x',y') \in B((x,y), \varepsilon).$ If we can show that $x'$ and $y'$ are positive we're done right? From $B((x,y),\varepsilon)$ we have that $$\|(x,y)-(x'-y')\| = \sqrt{(x'-x)^2+(y'-y)^2} < \varepsilon$$.
How to proceed from here? I'm not sure how can I show the positivity of $x'$ and $y'$.
 A: Consider the open square $S=(x-\epsilon,x+\epsilon)\times(y-\epsilon,y+\epsilon)$. Show that

*

*$S\subseteq A$

*$B((x,y),\epsilon)\subseteq S$
A: $x-x’\le|x’-x|\le\sqrt{(x'-x)^2+(y'-y)^2}< \varepsilon\le x$
implies that
$x-x’<x$
implies that
$x’>0\;.$
Analogously you can prove that $\;y’>0\;.$
A: Alternate proof:  If you accept that the product topology on $\Bbb R^2$ agrees with the Euclidean metric topology, then we have $A=(0,\infty)\times(0,\infty)$ is open.
A: The function $f(x,y) = \min(x,y)$ is continuous and $A = f^{-1}((0,\infty))$.
A: Let $(a,b) \in A$.  That means $a > 0$ and $b > 0$.  Try to draw a circle around $(a,b)$ that doesn't intersect either the $x$ axis or $y$ axis.  Suppose the circle has radius $r$.  Then the point $(a-r, b)$ will be in the circle so we need $a-r > 0$ (or in other word $r < a$) to avoid hitting the $y$ axis.  And the point $(a,b-r)$ will be in the circle so we need $b-r > 0$ (or $r < b$) to avoid hitting the $y$ axis.  So let $r = \min(a,b)$.  Will such a circle always avoid hitting the $x$ or $y$ axis?
Let's think it out.  let's show if $(x,y) \not \in A$ (i.e. if either $x < 0$ or $y < 0$) then $d((x,y),(a,b))$ must be larger than $r$.  Thus that circle is indeed small enough to avoid both axises.
Suppose $x,y \not \in A$.  the either $x < 0$ so $d((x,y), (a,b)) = \sqrt {(x-a)^2 + (y-b)^2} \ge \sqrt {(x-a)^2 }= |a-x| = a + |x| > a > r$.  the $(x,y)$ is not in the open ball around $(a,b)$ with radius $r$.   Or $y < 0$ and by the same argument $(x,y)$ in not in the open ball.
So the open ball of radius $r$ centered and $(a,b)$ is entirely within $A$.
So $(a,b)$ is an interior point of $A$ and $A$ is open.
