Independent of path; conservative vector field

• Nov 7th 2007, 06:24 AM
kittycat
Independent of path; conservative vector field
Hi,

I need your help for the below concepts.

1) A region D is OPEN if it doesn't contain any of its boundary points.
Can you give me an example of an OPEN region?

Can you also give me an example of open, simply connected region?

Thank you very much.
• Nov 7th 2007, 06:46 AM
topsquark
Quote:

Originally Posted by kittycat
Hi,

I need your help for the below concepts.

1) A region D is OPEN if it doesn't contain any of its boundary points.
Can you give me an example of an OPEN region?

Can you also give me an example of open, simply connected region?

Thank you very much.

I believe that the interval of the real line (0, 1) satisfies both requirements.

If I am right, then the circular region of the plane (the "open unit disc") $A = \{ (r, \theta) | 0 < r < 1, \theta ~\text{anything } \}$ will be another example of both.

For open and not simply connected: $(-1, 0) \cup (0, 1)$. (I don't think I can do open and connected, though I'm sure they exist.)

(These examples are taken using the usual topology on the real line and $\mathbb{R}^2$, of course.)

-Dan
• Nov 7th 2007, 07:14 AM
kalagota
i read your topic title.. i think you are to consider paths here..

the region enclosed by $C_1: y=x^2$ and $C_2: y=2x$ from (0,0) to (2,4) to (0,0) again, taking C_1 to C_2 but does not include (0,0) is simply connected which is open.

$D = {(x,y) | x^2 + y^2 < 1}$ is open..
• Nov 7th 2007, 07:30 AM
Plato
Here some unbounded examples.
The first quadrant, x>0 & y>0, is connected open set.
The upper half-plane, y>0, is connected open set.
The vertical plank, -2<x<2, is connected open set.