Is this proof of the existence of the general $n$-th root in $\mathbb{R}$ legitimate?

Question

I am trying to prove the existence of the $n$-th root of a positive real number in $\mathbb{R}$ and Pugh suggests doing this, in his book Real Mathematical Analysis, by first showing that the power function $f(x) = x^n, n \in \mathbb{N}$ is continous.

Therefore I begin by proving the "lemma", i.e. the continuity using the complete induction as follows: For the base case of $n = 1$ we have $$ \forall \epsilon > 0:\exists \delta=\epsilon \implies |x-a|<\delta \implies |x^1 - a^1|<\epsilon \text{,} $$ then for the inductive case let $\epsilon',\epsilon > 0 \text{; } \epsilon < 1$ and let $\forall k\in\{1, 2, \ldots, n\}: \exists \delta_k > 0 \implies |x-a|<\delta_k \implies |x^k-a^k|<\epsilon$, so we look at $$ |x^{n+1}-a^{n+1}| = |x-a|\cdot\left|\sum_{k=0}^{n}x^{n-k}a^k\right| \\ < \epsilon \cdot \sum_{k=0}^{n}|x^{n-k}a^k|\text{,} $$ where we notice that $|x^{n-k}a^k| = |x^{n-k}a^k - a^n + a^n| = |(x^{n-k} - a^{n-k})a^k + a^n|$. Therefore $|x^{n-k}a^k| \leq |x^{n-k} - a^{n-k}||a|^k + |a|^n$, where by the inductive assumption $|x^{n-k}a^k| < \epsilon|a|^k + |a|^n$.

Hence we get $$ |x^{n+1}-a^{n+1}| < \epsilon^2\cdot s_n + \epsilon(n+1)|a|^n\text{,} $$ where $s_n = \sum_{k=0}^n|a|^k$ and, since $\epsilon < 1$, $$ |x^{n+1}-a^{n+1}| < \epsilon^2\cdot s_n + \epsilon(n+1)|a|^n < \epsilon(s_n + (n+1)|a|^n) < \epsilon' \text{.} $$ Therefore $\exists m \in \mathbb{N}$ so that we can choose $$ \epsilon = \frac{\epsilon'}{m \cdot (s_n + (n+1)|a|^n)} \leq \frac{\epsilon'}{s_n + (n+1)|a|^n} \qquad and \qquad 0 < \epsilon < 1 \text{.} $$

Therefore we have shown there exists such $\delta = \min(\delta_1, \delta_2, \ldots, \delta_n)$ that $|x-a|<\delta \implies |x^{n+1}-a^{n+1}|<\epsilon'$, which completes the proof of the continuity.

Now, moving on to the proof of the root existence, let $x>0$ and $n\in\mathbb{N}$ and let $$ A = \{r\in\mathbb{R}: r \leq 0 \quad or \quad r^n < x\}\text{.} $$ Then $0 \in A$, so $A \neq \emptyset$. Similarly $(x+1)^n > x$, hence is $x+1$ an upper bound of $A$, so $A$ is bounded above. Therefore there exists $y=\sup(A)\in \mathbb{R}$.

By trichotomy of $\mathbb{R}$ only one of the following can be true:

1. $y^n < x$: By continuity for $\epsilon = x - y^n > 0$ there exists $\delta > 0$ so that $\exists \alpha \in(y, y + \delta)$ for which $|\alpha^n - y^n|<\epsilon$ or $\alpha^n < y^n + \epsilon = x$, therefore $\alpha \in A$, but $y < \alpha$, so y is not the supremum of $A$, a contradiction.

2. $y^n > x$: Similarly, for $\epsilon = y^n - x$ there exists $\delta > 0$ so that $\exists \alpha \in (y-\delta,y)$ and for which $|\alpha^n - y^n|<\epsilon$ or $\alpha > y^n - \epsilon = x$, therefore $\alpha$ is an upper bound of $A$, but $\alpha < y$, so $y$ is not the supremum of $A$, which is again a contradiction.

3. y^n = x, which is how the n-th root of x is defined as and this is the only option left.$\square$

PS: I am merely a freshman undergraduate math student, so while I do have some experience with proofed based math, I am still not proficient in writing formal proofs. I am also aware of the fact that my proof of the continuity is pretty much "all around the place" without properly formaly defining some of the variables and so on, but I think you get the idea of the proof. I am pretty sure the proof is correct, but wanted some feedback, as well as practice my LaTeX skills, so call it hitting two birds with one stone.

Thank you for your feedback in advance and I apologise for any violation of of the forum rules, as I am new this place and my "question" isn't really a question other than asking for feedback and suggestions.

Answer 1

Feedback it is!

1. You're right about your induction proof being all over the place, and its readability suffers for it. Consequently, I haven't checked every step, but you look like you have the right ingredients to form a proof, even if they were mixed somewhat haphazardly.

2. If you want to clean up your induction proof, focus first on what you want to prove: given $\varepsilon > 0$, you want to find a $\delta > 0$ such that $$|x - a| < \delta \implies |x^{n+1} - a^{n+1}| < \varepsilon.$$ I would structure the induction step as follows: "Suppose $\varepsilon > 0$. By the induction hypothesis, there must exist a $\delta_k$ such that $$|x - a| < \delta_k \implies |x^k - a^k| < [\text{some function of $\varepsilon$ and $k$}]$$ for all $k = 1, \ldots, n$. Let $\delta = \min\{\delta_1, \ldots, \delta_n\}$, and suppose $|x - a| < \delta$. Then ... [logic, logic, logic] ... thus $|x^{k+1} - a^{k+1}| < \varepsilon$." This structure makes more sense to me.

3. If you really want to clean up your induction proof, I suggest using algebra of limits. You know $x \mapsto x^n$ is continuous and $x \mapsto x$ is continuous, so $x \mapsto x^n \cdot x = x^{n+1}$ is also continuous. Even if you don't want to use algebra of limits, certainly using the proof that the product of two continuous functions is continuous could really help make the proof more straight forward.
4. The second part of the proof is essentially a special case of the proof of intermediate value theorem. If you can guarantee some $a$ and $b$ such that $a^n < x < b^n$, then the intermediate value theorem guarantees the existence of $c$ between them such that $c^n = x$.

Answer 2

Good work. I have a few remarks.

1. After proving by the definition that $x\mapsto x$ is continuous, the natural way (at least for) of proving that $x\mapsto x^n$ is continuous consists in using induction and the fact that the product of two continuous functions is continuous.
2. After proving that $x\mapsto x^n$ is continuous, you could use the intermediate value theorem to prove what you want to prove: if $0\leqslant r\leqslant 1$, then, since $0^n=0\leqslant r$ and $1^n=1\geqslant r$, there is some $y\in[0,1]$ such that $y^n=r$. And if $r\geqslant1$, then, since $1^n=1\leqslant r$ and $r^n\geqslant r$, there is some $y\in[1,r]$ such that $y^n=r$.