Which two of the following space are homeomorphic to each other?
\begin{align}
X_1&=\{(x,y)\in \mathbb{R}^2:xy=0\}\\
X_2&=\{(x,y)\in \mathbb{R}^2:xy=1\}\\
X_3&=\{(x,y)\in \mathbb{R}^2: x+y\geq 0 \text{ and } xy=0\}\\
X_4&=\{(x,y)\in \mathbb{R}^2: x+y\geq 0\ \text{ and } xy=1\}
\end{align}
I want to use the contentedness property.
Efforts:
$X_1$ and $X_4$ are not homeomorphic as $X_1$ is connected and nd $X_2$ is disconnected(two connected component)
By same logic $X_1$ and $X_4$ are not homeomorphic.
If I remove the origin from $X_1$, there are 4 components, but if we remove the origin from $X_3$, there are three connected component. So $X_1$ and $X_3$ are not homeomorphic.
Am I going in the right direction?
How to proceed after this.
Best Answer
Actually, $X_4$ is connected (it's a branch of a hyperbola) and therefore your argument fails. But if you remove $(0,0)$ from $X_1$, then what remains has four connected components. This proved that $X_1$ and $X_4$ are not homeomorphic.
On the other hand, $X_3$ and $X_4$ are homeomorphic. Just consider the map$$\begin{array}{ccc}X_4&\longrightarrow&X_3\\(x,y)&\mapsto&\begin{cases}(0,y-x)&\text{ if }y\geqslant x\\(x-y,0)&\text{ otherwise.}\end{cases}\end{array}$$