map $x+y \le 1, x,y >0$ to $R^2$? Is there a bijective continuous function which can map $x+y \leq 1, x,y >0$ to $R^2$? 
I appreciate any idea and comment.
 A: This is going to be quite similar in spirit to what stewbasic suggested in a comment, with the intermediate steps made more explicit.
Let's first look at what happens if we approach a point $a\in \overline{D}\setminus D$. Here, I write $D=\{(x,y): x,y>0, x+y\le 1\}$, and the closure is taken in $\mathbb R^2$.
I claim that if we had a homeomorphism $\varphi: D\to\mathbb R^2$, then $|\varphi(x_n)|\to\infty$ for any sequence $x_n\in D$ with $x_n\to a$ for such an $a$. This follows because otherwise $\varphi(x_n)\to y\in\mathbb R^2$ on a subsequence, but this contradicts the fact that also $y=\varphi(b)$ for some $b\in D$, and then a whole neighborhood of $y$ is obtained as the image of a neighborhood of $b$, so $\varphi$ would not be injective.
So we obtain an extended continuous surjective map $\varphi: \overline{D}\to S^2$, by mapping that part of the boundary that wasn't in $D$ to start with to $\infty$. Now the boundary of our triangle $\overline{D}$ gets mapped to a Jordan curve $C$ on $S^2$, and we can now take two points $a,b\in S^2$ (one from the interior and one from the exterior region of $C$) that can not be joined on $S^2$ without crossing $C$.
However, they can clearly be joined in this fashion on $D$, so we have obtained a contradiction.
Remark: This doesn't quite address the original question (which I hadn't read carefully enough), it shows that there is no homeomorphism. The continuity of $\varphi^{-1}$ is used in the third paragraph.
A: If you make the boundary $x+y \lt 1$ as you say in a comment, you can make such a bijection because the set is open. First take your favorite bijection $(0,1) \leftrightarrow \Bbb R$.  Mine is $z \leftrightarrow \tan (\pi (z-\frac 12))$.  Now given a point in your triangle, biject it to the open unit square by $(x,y) \leftrightarrow (\frac x{1-y}, \frac y{1-x})$, now apply the stretch to $\Bbb R$ independently in each axis.
