Understanding the Usual Metric on R - {0}: A Question from a Homework Statement

  • Thread starter Thread starter tylerc1991
  • Start date Start date
  • Tags Tags
    Metric
tylerc1991
Messages
158
Reaction score
0

Homework Statement



I was working on a problem and think I might have run across an issue. Is the usual metric defined on R - {0}? (Where R is the real numbers) Reworded, can I say that I have a space R - {0} with the usual metric on it? Thank you.
 
Physics news on Phys.org
I think you do almost whatever you want really ;)

however as the metric still satisfies all the metric consditions, it is still a metric space
http://en.wikipedia.org/wiki/Metric_space

however it will no lonegr be complete as it doesn't contain all its limit points
 
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