Proof of Equality for B=Int(B)∪bd(B) in Metric Space A

  • Thread starter Thread starter gotjrgkr
  • Start date Start date
  • Tags Tags
    Proof
gotjrgkr
Messages
84
Reaction score
0
Let A be a metric space and B be a compact subset of A.
Is it true that an equality B = Int(B) \cup bd(B) holds??
 
Physics news on Phys.org
In a metric space every compact set is closed (actually in every topological space). What does that tell you about its boundary points?
 
Boundary point is also limit point, right?
Then it belongs to B since B is closed?
 
HallsofIvy said:
In a metric space every compact set is closed (actually in every topological space).

Hausdorff topological space, no? Which does include metric spaces, I admit.
 
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