Analysis - compactness and sequentially compactness

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quasar987 said:
1. Homework Statement
According to the dfn, a subset A of a metric space is sequentially compact is every sequence in A has a subsequence that converges to a point in A.

An example of a sequentially compact set that comes to mind is R itself.

Then, Bolzano-Weierstrass's thm says that sequentially compactness and compactness are equivalent.

Finally, Heine-Borel's thm says that in R^n, compactness and closed+bounded are equivalent.

Thus, in R^n, closed+bounded and sequentially compactness are equivalent. But R is not bounded. What's going on?
Why would R "come to mind" as an example of a sequentially compact set? In particular what do you say the sequence {n}= {1, 2, 3, 4,...} converges to?
 

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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