Related to Compact and Connected set

Question

Is the following set $S$ compact subset of $\mathbb{R}^2$?

$$S= \{(x,y) \in \mathbb{R}^2 | xy<0 \}$$

Is the set $S$ connected subset of $\mathbb{R}^2$?

I've done the compact part by looking the set in $xy$ plane.. Clearly it's not bounded and hence not compact. But how to approach connectedness?? Help me figure it out..

Answer 1

The considered set is the union of the second and the fourth quadrants. These are quite “clearly” disjoint, so the set is not connected.

Answer 2

It is not connected, because the image of the continuous function $f\colon S\longrightarrow\mathbb R$ defined by $f(x,y)=x$ is $\mathbb R\setminus\{0\}$, which is not connected.

Answer 3

It is the union of the disjoint open sets $\{(x,y): x<0, y>0\}$ and $\{(x,y): x>0, y<0\}$. So by definition of connectedness the set is disconnected.

Answer 4

A set $S$ is not connected if it is equal to the union of two or more disjoint non-empty open sets.

1) In your case, it is easy to show that $S=A \cup B$, where: $$A\equiv \{(x,y)\in \mathbb{R}^2|x<0\text{ and } y>0\},$$ $$B\equiv \{(x,y)\in \mathbb{R}^2|x>0\text{ and } y<0\}.$$

2) Most importantly, $S=A\cup B$ where $A$ and $B$ are disjoint, non-empty, and open.

• It is easy to see that $A$ and $B$ are non-empty and disjoint.
• Depending on how rigorous you need to be, it may take a tiny bit of work to show that they are open.

3) Therefore, $S$ is not a connected set.