(path) connected, open, closed

**TXGirl** · May 11th 2008, 04:22 PM

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?

**ThePerfectHacker** · May 11th 2008, 06:34 PM

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?

**TXGirl** · May 12th 2008, 03:08 AM

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...

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