I am studying more deeply the first chapters of Rudin's Principles of Mathematical Analysis. What interests me are theorems' proofs strategies.
Theorem 1.21 states:
For every real $x > 0$ and every integer $n > 0$ there is one and only one real y such that $y^n=x$.
The one and only one part is soon proved; in order to prove the actual existance of $y$, the set
$E$ consisting of all positive real numbers $t$ such that $t^n<x$
is defined and it's then proved that $y=sup(E)$ exists in $R$. It remains to prove that $y^n=x$ and it's done in the last (and quite lenghty) part of the proof.
What I don't understand is why this last part is needed. I would say: once we defined $E$ as above we can consider the set $S$ containing all the upper bounds $f$ of $E$ such that $f^n \geq x$ right? Among those upper bounds we can consider the smallest one which is indeed $y=sup(E)$ (we know it exists) and being the smallest it is the one such that $y^n = x$.
Am I being incorrect?
Best Answer
Yes, you are being incorrect. If $f$ is an upper bound of $E$, then, yes, $f^n\geqslant x$. But it is not obvious at all that if $f$ is the least upper bound, then $f^n=x$. That equality requires a proof.