Well-Ordering Theorem & Countably Finite Sets: Analysis

ehrenfest
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Homework Statement


Isn't the solution at the site http://www.kalva.demon.co.uk/putnam/psoln/psol849.html incomplete because the author assumes he can map the set to Z and we were not given that the set was countably finite? The well-ordering theorem (that states any set can be well-ordered) does not allow you to add indices like that to the set, right?


Homework Equations





The Attempt at a Solution

 
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ehrenfest said:

Homework Statement


Isn't the solution at the site http://www.kalva.demon.co.uk/putnam/psoln/psol849.html incomplete because the author assumes he can map the set to Z and we were not given that the set was countably finite? The well-ordering theorem (that states any set can be well-ordered) does not allow you to add indices like that to the set, right?

Where does he map the set to Z? He's given a set with n elements, hence a finite set, so what he does is perfectly well justified.
 
d_leet said:
Where does he map the set to Z? He's given a set with n elements, hence a finite set, so what he does is perfectly well justified.

You're right. However, if you were not given that the set were finite or even countable infinite, would you still be allowed to use indices like that? Using the indices i is basically an injection from your set to Z, right?
 
ehrenfest said:
You're right. However, if you were not given that the set were finite or even countable infinite, would you still be allowed to use indices like that? Using the indices i is basically an injection from your set to Z, right?
If you well-order the set, then I believe you can pull this off using transfinite recursion.
 

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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