A function without a maximum and a minimum

James LeBron
Messages
23
Reaction score
0

Homework Statement



Give an example of a non-continuous function on [0,1] that has no maximum and no minimum.

Homework Equations



Well, a continuous function on a non-empty compact set will have a maximum and a minimum, so I guess this is why we need a non-continuous function.

The Attempt at a Solution



Does this work?

f(x) =

1/2 if x is a rational number
x if x is an irrational number.

Does this work? Breaking this up into 1/2s (if rational) and x's (if irrational) hopefully makes it non-continuous, and I suppose there is always some larger number that could be formed as we approach 1 (so no maximum). What do you think?
 
Physics news on Phys.org
I would agree.
 
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