Proving continuity and monotonicity of $t\mapsto t^x, t>0$ with minimal assumptions. I'm trying to prove that 

The function $t\mapsto t^x,\, x\in \Bbb R,\, t>0$ is continuous and monotonic.


Suppose $+, \cdot\,:\Bbb R^2\to \Bbb R$ (addition and multiplication) have already been defined (via the standard dedekind cuts construction).
If $a\in \Bbb Z$, we define $t^a=t^{a-1}\cdot t$, $t^{1}=t$.
If $r=\frac a b\in \Bbb Q$, we define $t^r=\sup \{x: x^b<t^a\}$.
If $x\in\Bbb R$, we define, for $t>1$, $t^x=\sup\{t^r: r<x, r\in\Bbb Q\}$.  And if $t<1$ we swap $\sup$ by $\inf$ in the previous definition.

I want to avoid usage of the exponential function $e^x=\sum \frac {x^n}{n!}$, series and further concepts, as I want to build the foundation for those first.
 A: 1. The construction of $t \mapsto t^x$ for $t>1$ and $x \in \Bbb R$
If $n = 0$, define $t^n := 1$; if $n \in \Bbb N \setminus \{ 0 \}$, define $t^n := t^{n-1} t$. If $n \in \Bbb Z \setminus \Bbb N$, define $t^n := \left( \frac 1 t \right) ^{-n}$.
Asume now that $t>1$. If $n \in \Bbb N \setminus \{ 0 \}$, define $t^{\frac 1 n} := \sup \{ x \mid x^n \le t \}$. Notice that $\{ x \mid x^n \le t \} \ne \emptyset$ because it surely contains $1$. Notice, too, that $t>1$ implies $t^2 > t$ and $t^2$ is an upper bound for this set, which implies that the $\sup$ is finite.
If $t<1$ replace $\sup$ by $\inf$ in the above. If $t=1$ then things are clear.

Assume now $t > 1$. Since $t ^{\frac 1 n} = \sup \{ x \mid x^n \le t \}$, let $(t_n) \subset \{ x \mid x^n \le t \}$ be any sequence such that $t_n \to t$ (there always exist sequences in a set tending to the $\sup$ or $\inf$ of that set). Then $t_n ^n \le t$, so passing to the limit produces $\left( t ^{\frac 1 n} \right) ^n \le t$.
It is easy to prove that, $\sup \{ x \mid x^n \le t \} = \inf \{ x \mid x^n \ge t \}$. Taking, then, $(t_n) \subset \{ x \mid x^n \ge t \}$ with $t_n \to t$, one gets that $t_n ^n \ge t$ so passing to the limit implies $\left( t ^{\frac 1 n} \right) ^n \ge t$.
Since we have obtained inequalities in both directions, it follows that $\left( t ^{\frac 1 n} \right) ^n = t$.
If $0<t<1$, just exchange $\inf$ and $\sup$ above, or use the proof above for $\frac 1 t$ and reverse the fractions.

Let us show that $\left( st \right) ^{\frac 1 n} = s^{\frac 1 n} t^{\frac 1 n}$. Assuming $st > 1$, then either $s>1$ or $t>1$. To make a choice, let's asume $t>1$, the proof being identical if $s>1$. We have
$$\left( st \right) ^{\frac 1 n} = \sup \{ x \mid x^n \le st \} = \sup \left\{ x \left| x^n \le \left( s ^{\frac 1 n} \right) ^n t \right. \right\} = \sup \left\{ x \left| \left( \frac x {s ^{\frac 1 n}} \right) ^n \le t \right. \right\} = \dots$$
and, if you make the change of variable $x = y \ s ^{\frac 1 n}$, the above continues as
$$\dots = \sup \left\{ y \ s ^{\frac 1 n} \Big| y^n \le t \right\} = s ^{\frac 1 n} \sup \{ y \mid y^n \le t \} = s ^{\frac 1 n} t ^{\frac 1 n} .$$
If $st = 1$ then $t = \frac 1 s$ and things are easy.
If $st < 1$ then at least one of them is $<1$. To make a choice, assume $t<1$ and use the same idea as above, but replace $\sup$ by $\inf$.

If $a, b \in \Bbb N \setminus \{0\}$, let us prove by induction on $a$ that $(t^a) ^{\frac 1 b} = \left( t ^{\frac 1 b} \right) ^a$. If $a=1$ then you just have the definition. If $a>1$ then $\left( t ^{\frac 1 b} \right) ^a = \left( t ^{\frac 1 b} \right) ^{a-1} \ \left( t ^{\frac 1 b} \right) = (t^{a-1}) ^{\frac 1 b} \ t^{\frac 1 b} = (t^{a-1} t) ^{\frac 1 b} = (t^a) ^{\frac 1 b}$.
If $a,b>0$ with $\gcd (a,b) = 1$, define $t^{\frac a b} := (t^a) ^{\frac 1 b} = \left( t^ {\frac 1 b} \right) ^a$. If $\gcd (a,b) = d$, let $a' = \frac a d, \ b' = \frac b d$ and define $t^{\frac a b} := t^{\frac {a'} {b'}}$. Then, if $q \in \Bbb Q _+$, $t^q$ does not depend on how you represent $q$ as a fraction.
If $q \in \Bbb Q, q<0$, then define $t^q = \left( \frac 1 t \right)^{-q}$.

If $t>1$ and $x>0$ or $t<1$ and $x<0$ then define $t^x := \sup \{ t^q \mid q \in \Bbb Q, \ q \le x\} $. If $t>1$ and $x<0$ or $t<1$ and $x>0$ then use $\inf$ instead of $\sup$. If $t=1$ or $x=0$ then things are trivial.

2. The monotonicity of $t \mapsto t^x$
We shall assume in the first part of the proof that $x \ge 0$.
Let $n \in \Bbb N$ and $1 < s < t$. Let us show by induction on $n$ that $s^n < t^n$. If $n=1$ then this is obvious. If $n>1$ then $s^n = s^{n-1} s < t^{n-1} s < t^{n-1} t = t^n$. So far, we have proved that the power function is strictly increasing for non-zero natural powers.
Suppose that $s ^{\frac 1 n} > t ^{\frac 1 n}$. Using the above paragraph, we obtain that $\left( s ^{\frac 1 n} \right) ^n > \left( t ^{\frac 1 n} \right) ^n$. Using the fact that $\left( t ^{\frac 1 n} \right) ^n = t$, this becomes $s>t$, which contradicts the choice $s<t$. Therefore, $s ^{\frac 1 n} < t ^{\frac 1 n}$, so the power function is strictly increasing when the power is the inverse of some natural number.
Let now $q = \frac a b >0$. Then, chaining the result obtained so far,
$$s^q = s^{\frac a b} = (s^a) ^{\frac 1 b} < (t^a) ^{\frac 1 b} = t^{\frac a b} = t^q .$$
This shows that the power function is strictly increasing for strictly positive rational powers.
Finally,
$$s^x = \sup \{ s^q \mid q\in \Bbb Q, \ q \le x \} \le \sup \{ t^q \mid q\in \Bbb Q, \ q \le x \} = t^x ,$$
so $t \mapsto t^x$ is increasing for $x \ge 0$ on $(1, \infty)$.
If $0<s<t<1$, then $1 < \frac 1 t < \frac 1 s$ and applying all of the above for $\frac 1 t$ and $\frac 1 s$, we deduce that $\left( \frac 1 t \right)^x \le \left( \frac 1 s \right)^x$, which amounts to $s^x < t^x$.
Finally, if at least one of $s,t$ is $1$, things are simpler and I shall not do that (as authors say: I'll leave it as an exercise to the reader :) ).
So far, we have proved that if $x \ge 0$, then $t \mapsto t^x$ is increasing. (In fact, it can be shown that it is strictly so, but I'll keep things simple.)
If $x < 0$, then according to the above $t \mapsto t^{-x}$ is increasing, so $t \mapsto \frac 1 {t^{-x}} = t^x$ will be decreasing.
3. The continuity of $t \mapsto t^x$ with respect to $t$
Let us first show that for $n \in \Bbb N \setminus \{0\}$ the function $t \mapsto t^n$ has the inverse $t \mapsto t^{\frac 1 n}$. We already know that $\left( t ^{\frac 1 n} \right) ^n = t$. It is easy to show that $(st)^n = s^n t^n$, so on the other hand,
$$(t^n) ^{\frac 1 n} = \sup \{ x \mid x^n \le t^n \} = \sup \left\{ x \left| \left( \frac x t \right) ^n \le 1 \right. \right\} = \sup \{ t y \mid y^n \le 1 \} = t \sup \{ y \mid y^n \le 1 \} = \\
t \ 1 ^{\frac 1 n} = t ,$$
where we have used again an argument already presented above with the "change of variable" $x = y t$.
Having an inverse and being monotonic, $t \mapsto t^n$ must be strictly monotonic. Since it is also continuous (a consequence of $\lim \limits _{t_n \to t} t_n ^p = (\lim \limits _{t_n \to t} t_n) ^p = t^p$), its inverse $\left( t ^{\frac 1 n} \right) ^n = t$ must also be continuous.
It is immediate, now, that $t \mapsto t^{\frac a b} = (t^a) ^{\frac 1 b}$ is also continuous, as the composition of the continuous mappings $t \mapsto t^a$ and $s \mapsto s ^{\frac 1 b}$.
It remains to prove continuity for the power being a real number. You will find it here, short and clearly explained, done with $\varepsilon$ and $\delta$.
