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Let {$f_{n}$} be given by $f_{n}(x)=x^n$. Prove the sequence is not cauchy

My 'proof': Since F is complete, we can prove it doesn't converge. The sequence converges pointwise to f given by f(1)=1 and f(x)=0 otherwise. Since uniform convergence implies pointwise convergence, if this sequence is goin to converge, it will do so to f.

two cases to consider.

x=1; Then $sup{|f_{n}(x)-f(x)|}=sup({0})=0$ so certainly whatever epsion is chosen, we can choose N=1 and we are fine.

x in [0,1); then $sup{|{f_n}(x)-f(x)|}=sup({x^n})=1$ so choosing epsilon =0.5 shows that $f_n{x}$ doesn't converge to f(x) for x not 1.

Hence sequence is not cauchy.

Not sure if this is right. If it is, is there a better way.