Which two of the following space are homeomorphic to each other?

Question

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.

Answer 1

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$$\begin{array}{ccc}X_4&\longrightarrow&X_3\\(x,y)&\mapsto&\begin{cases}(0,y-x)&\text{ if }y\geqslant x\\(x-y,0)&\text{ otherwise.}\end{cases}\end{array}$$

Answer 2

Yes you are right. In fact the only two homeomorphic sets are $X_3$ and $X_4$ since $X_4$ can be interpreted as a warped version of $X_3$. $X_3$, $X_4$ and $X_1$ are connected and $X_2$ is not. Therefore $X_2$ is not homeomorphic to any of the others. Also removing origin from $X_1$ leaves us with $4$ distinct parts in $X_1$ while leaving us $2$ distinct parts in $X_3$ and $X_4$. Therefore the only homeomorphic sets are $X_3$ and $X_4$.