# Related to Compact and Connected set

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..

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

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.

• Thank u so much. I just remembered that I've studied this criteria of connectedness in metric space.. But on the spot i suddenly forgot it. Thanks again😇 – user684646 Jun 25 at 8:52

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.

• Thanks a lot😇 I'll try to show that A and B are open sets. – user684646 Jun 25 at 8:53
• @user684646 You're welcome! I've been assuming that you are using the Euclidean Topology, which implies the following definition of "open": a set is open if every point in the set can be "wrapped up" in some ball (ball=circle in the case of $\mathbb{R}^2$) which is also strictly contained within the set. You may want to look up the precise definition but this is the direction I would recommend looking at. – karlahrnndz Jun 26 at 3:27

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.