Proof of Properties of Real Powers of Nonnegative Real Numbers

Each result cited is universally quantified over the data appearing in its own statement, and is applied here to the data named in the statement above. Throughout, $a$ and $b$ are positive real numbers and $s,t\in\mathbb{R}_{+}$; by Real Power of a Nonnegative Real Number §power each of $s,t$ is either positive, in which case its powers are those of Real Power of a Positive Real Number, or equal to $0$, in which case every power with positive exponent is $0$.

Claim 1. If $0<t$ then $t^{a}=\exp(a\log t)$, which is positive by claim 2 of Basic Properties of the Exponential Function; in particular $0\le t^{a}$ and $t^{a}\ne0$. If $t=0$ then $t^{a}=0$, so again $0\le t^{a}$. Since a nonnegative real number is positive or $0$ and not both, this proves $0\le t^{a}$ for all $t$, and also that $t^{a}=0$ holds exactly when $t=0$.

Claim 2. Let $0<t$. Then $t^{1}=t$, the real power $t^{n}$ agrees with the natural power for every $n\in\mathbb{N}$, and $t^{1/2}=\sqrt{t}$, all by claim 1 of Real Powers Through the Exponential, and Elementary Asymptotic Tools: Monotonicity, Null Sequences of Negative Powers, Exponential Domination, Integer Rounding, and Square-Root and Exponential Inequalities. Let $t=0$. Then $0^{1}=0$ by the definition, and the natural power $0^{n}$ equals $0$ for every $n\in\mathbb{N}$ by claim 4 of Properties of Natural Number Powers in a Field, so the two readings agree. Finally $0$ is nonnegative and $0^{2}=0$, so by the uniqueness in Existence and Uniqueness of the Nonnegative Square Root we get $\sqrt{0}=0=0^{1/2}$.

Claim 3. If $0<s$ and $0<t$ then $0<st$ by claim 5 of Elementary Order Arithmetic in an Ordered Field, and $(st)^{a}=s^{a}t^{a}$ by claim 1 of Real Powers Through the Exponential, and Elementary Asymptotic Tools: Monotonicity, Null Sequences of Negative Powers, Exponential Domination, Integer Rounding, and Square-Root and Exponential Inequalities. Otherwise $s=0$ or $t=0$, so $st=0$ and $(st)^{a}=0$; and by claim 1 the corresponding factor $s^{a}$ or $t^{a}$ is $0$, so $s^{a}t^{a}=0$ as well.

Claim 4. Note that $a+b$ and $ab$ are positive, by claim 1 of Elementary Arithmetic in an Ordered Field and claim 5 of Elementary Order Arithmetic in an Ordered Field respectively, so all powers written below are defined. If $0<t$, both identities are contained in claim 1 of Real Powers Through the Exponential, and Elementary Asymptotic Tools: Monotonicity, Null Sequences of Negative Powers, Exponential Domination, Integer Rounding, and Square-Root and Exponential Inequalities. If $t=0$, then $0^{a+b}=0$ and $0^{a}0^{b}=0\cdot0=0$; and $(0^{a})^{b}=0^{b}=0=0^{ab}$, using claim 1 above to see that $0^{a}=0$ is again a nonnegative base.

Claim 5. Suppose $s\le t$. If $s=0$ then $s^{a}=0\le t^{a}$ by claim 1. If $0<s$ then $0<s\le t$, and claim 2 of Real Powers Through the Exponential, and Elementary Asymptotic Tools: Monotonicity, Null Sequences of Negative Powers, Exponential Domination, Integer Rounding, and Square-Root and Exponential Inequalities gives $s^{a}\le t^{a}$. Now suppose $s<t$, so that $0<t$. If $s=0$ then $s^{a}=0<t^{a}$ by claim 1. If $0<s$ then $0<s<t$, and the same claim 2 gives $s^{a}<t^{a}$.

Claim 6. Since $0<a$, the inverse $1/a$ is positive by claim 7 of Elementary Order Arithmetic in an Ordered Field, so $P_{1/a}$ is defined, and both $P_{a}$ and $P_{1/a}$ map $\mathbb{R}_{+}$ into $\mathbb{R}_{+}$ by claim 1. By claim 4 and claim 2, for every $t\in\mathbb{R}_{+}$,

Hence $P_{a}$ is injective, since $P_{a}(s)=P_{a}(t)$ gives $s=P_{1/a}(P_{a}(s))=P_{1/a}(P_{a}(t))=t$, and surjective onto $\mathbb{R}_{+}$, since every $t\in\mathbb{R}_{+}$ equals $P_{a}(P_{1/a}(t))$ with $P_{1/a}(t)\in\mathbb{R}_{+}$. So $P_{a}$ is a bijection of $\mathbb{R}_{+}$ onto $\mathbb{R}_{+}$ whose inverse is $P_{1/a}$.

For the comparisons, let $s,t\in\mathbb{R}_{+}$. If $s^{a}\le t$, then applying claim 5 with the exponent $1/a$ gives $s=(s^{a})^{1/a}\le t^{1/a}$; conversely if $s\le t^{1/a}$, then claim 5 with the exponent $a$ gives $s^{a}\le(t^{1/a})^{a}=t$. If $t<s^{a}$, then the strict part of claim 5 with the exponent $1/a$ gives $t^{1/a}<(s^{a})^{1/a}=s$; conversely if $t^{1/a}<s$, the strict part with the exponent $a$ gives $t=(t^{1/a})^{a}<s^{a}$.

Claim 7. By Continuity Between Metric Spaces is Equivalent to Sequential Continuity it is enough to show that $P_{a}$ is sequentially continuous. Let $(t_{m})_{m\in\mathbb{N}}$ be a sequence in $\mathbb{R}_{+}$ converging to $t\in\mathbb{R}_{+}$ in $(\mathbb{R}_{+},d_{+})$; since $d_{+}$ is the restriction of $d_{\mathbb{R}}$, the same sequence converges to $t$ as a sequence of real numbers.

Suppose first $0<t$. Applying the definition of a limit with the positive number $t/2$, there is $M\in\mathbb{N}$ such that $|t_{m}-t|<t/2$ for every $m\ge M$, and hence $t/2<t_{m}$ for such $m$ by claim 6 of Properties of the Absolute Value in an Ordered Field; in particular $0<t_{m}$ for $m\ge M$. Define a sequence $(u_{m})_{m\in\mathbb{N}}$ by $u_{m}=t_{m}$ for $m\ge M$ and $u_{m}=t$ for $m<M$. Then every $u_{m}$ is positive and $(u_{m})$ converges to $t$, because it agrees with $(t_{m})$ from the index $M$ on and convergence depends only on the terms beyond any fixed index. By claim 3(g) of Real Powers Through the Exponential, and Elementary Asymptotic Tools: Monotonicity, Null Sequences of Negative Powers, Exponential Domination, Integer Rounding, and Square-Root and Exponential Inequalities the sequence $(u_{m}^{a})$ converges to $t^{a}$; since $t_{m}^{a}=u_{m}^{a}$ for every $m\ge M$, the sequence $(t_{m}^{a})$ converges to $t^{a}$ as well.

Suppose now $t=0$, so that $t^{a}=0$. Let $\varepsilon$ be a positive real number. Then $\varepsilon^{1/a}$ is positive by claim 1, so there is $M\in\mathbb{N}$ with $|t_{m}-0|<\varepsilon^{1/a}$ for every $m\ge M$; since $t_{m}$ is nonnegative this says $t_{m}<\varepsilon^{1/a}$. By the strict part of claim 5, $t_{m}^{a}<(\varepsilon^{1/a})^{a}$, and $(\varepsilon^{1/a})^{a}=\varepsilon$ by claim 6, so $t_{m}^{a}<\varepsilon$; and $t_{m}^{a}$ is nonnegative by claim 1, so $|t_{m}^{a}-0|<\varepsilon$ for every $m\ge M$. As $\varepsilon$ was arbitrary, $(t_{m}^{a})$ converges to $0=t^{a}$.

In both cases $(P_{a}(t_{m}))$ converges to $P_{a}(t)$, so $P_{a}$ is sequentially continuous and therefore continuous.