Is the Subset D={(x,y)| x≠0 and y≠0} an Open Set in R^2?

  • Thread starter Thread starter Deadward1994
  • Start date Start date
Deadward1994
Messages
2
Reaction score
0

Homework Statement


Show that the subset D={(x,y)| x≠0 and y≠0} is an open set in R^2
.

Homework Equations


Open set: U is a subset of R^n. U is an open set when for every point X1, contained within U, there exists some open disk centered around X1 with radius r>0, that is completely contained within U. Or for simplicity's sake, a set U is open if it does not contain any of its boundary points.

The Attempt at a Solution


I have an understanding of what makes up an open set and know why this set is open, but i have no idea as to how I am meant to prove this, graphically or analytically( ideal method).
 
Last edited:
Physics news on Phys.org
Deadward1994 said:

Homework Statement


Show that the subset D={f(x; y)| x≠0 and y≠0} is an open set in R^2
.

Homework Equations





The Attempt at a Solution


I have an understanding of what makes up an open set and know why this set is open, but i have no idea as to how I am meant to prove this, graphically or analytically( ideal method).
What does the notation f(x;y) mean?
 
Sorry about that, I miss quoted the question
 
First, tell us the definition of an open set.
 
So let (x, y) be a point in D. Can you find a radius r such that all points in the open ball of radius r around this point are in D?

First you may want to try drawing this, then proving it analytically. Note that for most points it's pretty trivial - the interesting case is for when (x, y) is close to (0, 0).
 
CompuChip said:
So let (x, y) be a point in D. Can you find a radius r such that all points in the open ball of radius r around this point are in D?

First you may want to try drawing this, then proving it analytically. Note that for most points it's pretty trivial - the interesting case is for when (x, y) is close to (0, 0).

Also, presuming you can always find ##r## for arbitrary ##(x,y) ≠ (0,0)##, what happens at ##(0,0)## exactly?

Get reeeeeally close to ##0##. Can you always find an ##r## such that ##(0,0)## is not contained any neighborhood of your point?
 
Last edited:
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top