Why Isn’t d a Metric for Continuous Functions in R?

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Homework Statement



If d= sup{|f(x)-g(x)|} where f, g are contuous function from R to R, why is d not a metric

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The Attempt at a Solution

 
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Take f(x)=x, g(x)=0. then d(f,g)=infinity. So d is not a map into R.
 
why infinity is not a map into R?
 
I said that d is not a map into R. And this is because if you take f(x)=x and g(x)=0, then d(f,g)=sup{|x|:x in R} is not an element of R. Indeed, the real number sup{|x|:x in R}, if it exists, is an upper bound for the set {|x|:x in R}. But this set is unbounded from above and hence possesses no upper bound so sup{|f(x)|:x in R} does not exists.
 
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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