Vtyjkyuk

Prove or disprove: $$A=(x,y) in BbbR^2: x+y geq 0, xy=0 $$ is homeomorphic to $$B=(x,y) in BbbR^2: x+y geq 0, xy=1 $$

$x+y geq 0$ represents the region above by the line $y=-x$ and together $xy=0$(the axes)

But $B$ represents the same region above by $y=-x$ together with the curve $xy=1$ .But $xy=1$ where $x>0,y>0$ is already covered in $x+y geq 0$ so the remaining portion(the other asymptote) is in the third quadrant

Therefore $B$ is not connected whereas $A$ is connected, so they are not homeomorphic

Am I right?

• $A$ consists of the positive $x$-axis, positive $y$-axis, and the origin.




• $B$ consists of points on the first quadrant that falls on the line $y= frac1x$.

Guide:

To construct a bijection continuous map from $B$ to $A$,

$$g(x,y) = begincases (0, y-x)&, y ge x \ (x-y,0)& x > yendcases$$

Try to construct the corresponding inverse function.

Edit: To find the inverse function for $g$,

Consider the point $(0,y')$, we want to find $(x,y)$ such that $g(x,y)=(0,y')$ where $xy=1$. Hence we have $y-x=y'$ and $xy=1$ which implies $$1-x^2=xy'.$$

We can then solve $x$ using the quadratic formula to obtain the inverse and then find the corresponding $y$.

Your proposed solution is more elegant.

$x+y ge 0, xy = 0$ doesn't mean $x + y ge 0$ or $xy = 0$ it means $x + y ge 0$ and $xy = 0$. The set $A$ is the two positive parts of the $x$ and $y$ axes (an 'L' shape).

Similarly, the set $B$ is the portion of the hyperbola $xy = 1$ contained in the set $x + y ge 0$. Namely, the portion in the first quadrant.