I have ben facing this hurdle for a long time: First it was how ordinary properties for $x^r$ ($r$ rational) extend to $x^p$ ($p$ real); but thanks to Tao's Analysis I, I have overcome most of that difficulty.
But this one remains: (And, from yesterday's my engagement in chatroom, I got to know that very many have same problem, and have to take it granted for lack of availability of proof)
How to show that $f:[0,+\infty)\to\mathbb R$ defined as $f(x)=x^p$, is continuous for a fixed $p\in\mathbb R$?
Is this reasoning of mine correct:
Let us consider $\ln (x^p)$ instead. Continuity of $\ln x$ implies $p\ln x=\ln(x^p)$ continuous. Now, baby Rudin's exer 26, chapter 4 says that: "Suppose $X,Y,Z$ are metric spaces, and $Y$ compact. Let $f$ map $X$ into $Y$, let $g$ be a continuous one-to-one mapping $Y$ into $Z$, and put $h(x)=g(f(x))$ for $x\in X$. Then $f$ continuous if $h$ continuous." So, if we restrict domain to $[a,b]$ for $a,b\in[0,\infty)$ for $a<b$, then we see $x^p$ continuous on $[a,b]$. And hence, $x^p$ continuous on $(0,+\infty)$.
Is this my answer correct? How to extend this to $[0,+\infty)$?