$x^{1/n}$ is a positive real number?

Question

Terence Tao, Analysis I, 3e:

Lemma 5.6.6. Let $x,y \ge 0$ be non-negative reals, and let $n,m \ge 1$ be positive integers.

(a) ...

(b) ...

(c) $x^{1/n}$ is a positive real number.

Statement (c) does not seem correct to me:

Choose $x = 0$. Then, by definition, $x^{1/n} = \text{sup}\{ y \in \mathbb{R} : y \ge 0 \; \wedge \; y^n \le 0 \} = \text{sup}\{0\}$. Assume there is a least upper bound $\delta > 0$. Then there is an upper bound $\delta/2$ smaller than $\delta$. This contradicts the assumption that $\delta$ is a least upper bound.

Hence, there is a least upper bound $0$, and I would expect $x^{1/n}$ to be a non-negative real number.

Answer 1

To give an answer, as greelioushas pointed out in a comment, the Errata to the third edition (hardback) says

Page 123: Lemma 5.6.6(c) should read “$x^{1/n}$ is a non-negative real number, and is positive if and only if $x$ is positive”.