Basis for Vector Space: Understanding the Exceptional Case

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Homework Statement



My notes has the following statement, but I seem to have forgotten to write down the conclusion of the statement before my professor erased it from the board.

"Any vector space V there will be a basis except for 1 type of space: "

Any ideas as to what that 1 type of space is?

Thanks
 
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perhaps "infinite dimensional"? They don't have a basis unless you assume the axiom of choice.
 
Probably the space consisting of just the 0 vector.
 

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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