Set of Integers: Open or Closed?

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Homework Statement




is the set of integers open or closed

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The Attempt at a Solution



I thought not closed
open because R/Z=Union of open intervals
like ...U(-1,0)U(0,1)U...
 
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You got it backwards!
 
With what topology? The topology inherited from the reals? (That is, the metric topology with d(m,n)= |m- n|.)

Morphism's point is that since R\Z is a union of a union of open intervals, The complement of Z is open and so Z itself is ?

However, don't think that "open" or "closed" are all the options. It is possible for a set to be neither open nor closed. It is even possible for a set to be both open and closed.
 
ow sorry I switched open and closed
I meant that it was a closed set and not open
 
Yes, then that's right. The set is closed and not open.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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...
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