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Hello. This is an improvement on a previous post, "Continuity of y^2". My original plan was to first prove that y and y^2 were continuous and then prove by induction that y^n was continuous; however, in the process of doing so I think I found a better way. This proof is for rudimentary practice of proof construction and should be taken in context of an undergraduate Real Analysis I course. How does this look?

---Start Proof---

Define ##(f_n)_{n \in \mathbb{N}}## as a sequence of functions where ##f_n:\mathbb{R} \to \mathbb{R}## maps ##y \mapsto y^n##. I intend prove pointwise continuity for any ##f_n## by showing that:

Choose any ##n \in N, e > 0,## and ##u##.

Case 1: ##u = 0##

Case 2: ##u \neq 0##

Thanks in advance!

---Start Proof---

Define ##(f_n)_{n \in \mathbb{N}}## as a sequence of functions where ##f_n:\mathbb{R} \to \mathbb{R}## maps ##y \mapsto y^n##. I intend prove pointwise continuity for any ##f_n## by showing that:

##\forall n \in \mathbb{N} \;\; \forall e > 0 \;\; \forall u \;\; \exists \delta: \forall y (|y-u| < \delta \implies |f_n(y) - f_n(u)| = |y^n - u^n| < \epsilon)##

Choose any ##n \in N, e > 0,## and ##u##.

Case 1: ##u = 0##

Pick ##\delta = \epsilon^{\frac{1}{n}}## and then any ##y##. It is clear that ##|y-0

| < \epsilon^{\frac{1}{n}} \implies |y^n - 0| < \epsilon##.

| < \epsilon^{\frac{1}{n}} \implies |y^n - 0| < \epsilon##.

Case 2: ##u \neq 0##

Choose a ##\delta < |y|## and ##\delta < \frac{\epsilon}{(2^{n}-1)|u|^{n-1}}.## A ##\delta## meeting this criteria can be found since both ##|y|## and ##\frac{\epsilon}{(2^{n}-1)|u|^{n-1}}## are positive, and there exists no smallest positive number.

Note the following:

It follows that:

. Note the following:

- ##|y^n-u^n| = |y-u| \left |\sum_{i=0}^{n-1} u^{n-1-i}y^i \right |##. This is obvious from evaluating ##\frac{y^n-u^n}{y-u}##.
- ##|y-u| < \delta < |y| \implies |u| < 2|y|##

Proof of 2.

##|y-u| < |y|## so ##u - y < |y|## and ##u < |y| + y \leq 2|y|##. At the same time ##y-u < |y|##,which implies ##-u < |y| - y \leq 2|y|##. Both ##u## and ##-u## are ##<2|y|##; therefore ##|u|<2|y|##. - ##\left |\sum_{i=0}^{n-1} u^{n-1-i}y^i \right | < \sum_{i=0}^{n-1} 2^i|u|^{n-1} = (2^n-1)|u|^{n-1} ##. This follows from 2. and the evaluation of the geometric series.

It follows that:

##\begin{array}

&|y^n-u^n| &=& |y-u| | \sum_{i=0}^{n-1} u^{n-1-i}y^i | & \text{1.}\\

&<& \delta (2^{n}-1)|u|^{n-1} & \text{By assumption that } |y-u| < \delta \text{ and 3.}\\

&<& \frac{\epsilon}{(2^{n}-1)|u|^{n-1}}(2^{n}-1)|u|^{n-1} & \delta\text{ was chosen for this}\\

&=& \epsilon

\end{array}##

&|y^n-u^n| &=& |y-u| | \sum_{i=0}^{n-1} u^{n-1-i}y^i | & \text{1.}\\

&<& \delta (2^{n}-1)|u|^{n-1} & \text{By assumption that } |y-u| < \delta \text{ and 3.}\\

&<& \frac{\epsilon}{(2^{n}-1)|u|^{n-1}}(2^{n}-1)|u|^{n-1} & \delta\text{ was chosen for this}\\

&=& \epsilon

\end{array}##

QED

---End Proof---Thanks in advance!

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