# Thread: (path) connected, open, closed

1. ## (path) connected, open, closed

Consider the set $\mathbb{R}^2$\( $\mathbb{R}$ x {0})

Can I say that the sets

U:= {(x,y) in $\mathbb{R}^2$\( $\mathbb{R}$ x {0}):y>0}

V:= {(x,y) in $\mathbb{R}^2$\( $\mathbb{R}$ x {0}):y<0}

form a separation of $\mathbb{R}^2$\( $\mathbb{R}$ x {0}), meaning that it is not (path) connected?

Also, I'm not sure whether the set is open, closed, clopen...any ideas?

2. Originally Posted by TXGirl
Consider the set $\mathbb{R}^2$\( $\mathbb{R}$ x {0})

Can I say that the sets

U:= {(x,y) in $\mathbb{R}^2$\( $\mathbb{R}$ x {0}):y>0}

V:= {(x,y) in $\mathbb{R}^2$\( $\mathbb{R}$ x {0}):y<0}

form a separation of $\mathbb{R}^2$\( $\mathbb{R}$ x {0}), meaning that it is not (path) connected?

Also, I'm not sure whether the set is open, closed, clopen...any ideas?
In order for $U,V$ to be a disconnection of $\mathbb{R}^2 \setminus \mathbb{R}\times \{ 0 \}$ we require that $U\cup V = \mathbb{R}^2\setminus \mathbb{R}\times \{ 0 \}$ and $U\cap V = \emptyset$ with $U,V$ open.
The sets are indeed disjoint and do unionize to give the whole set. It thus only remains to show that they are open.
Can you see why they are open sets?

3. Originally Posted by ThePerfectHacker
In order for $U,V$ to be a disconnection of $\mathbb{R}^2 \setminus \mathbb{R}\times \{ 0 \}$ we require that $U\cup V = \mathbb{R}^2\setminus \mathbb{R}\times \{ 0 \}$ and $U\cap V = \emptyset$ with $U,V$ open.
The sets are indeed disjoint and do unionize to give the whole set. It thus only remains to show that they are open.
Can you see why they are open sets?
Yes, because we are dealing with strict inequality, the epsilon-neighborhood of every point in say, U is also in U (analogously for V).

Regarding open/close of $\mathbb{R}^2 \setminus \mathbb({R}\times \{ 0 \})$ in $\mathbb{R}^2$, am I correct in the following:

$\mathbb{R}^2 \setminus \mathbb({R}\times \{ 0 \})$, being the union of open sets, is open in $\mathbb{R}^2$.

and because $\mathbb({R}\times \{ 0 \})$ is in the boundary of $\mathbb{R}^2 \setminus \mathbb({R}\times \{ 0 \})$ but not in $\mathbb{R}^2 \setminus \mathbb({R}\times \{ 0 \})$, it follows that $\mathbb{R}^2 \setminus \mathbb({R}\times \{ 0 \})$ is not closed?

I'm really not certain about the "NOT closed" part...