Epsilon-Delta Proof regarding Induction/Continuity My question is the same as the one asked here: An $\epsilon$-$\delta$ proof for difference between powers of real numbers
I see that for the $n = 1$ case, we can set delta equal to epsilon and observe that the statement holds true. How would I go about doing that for $n = 2$ and for the inductive case?
I've been stuck on this problem for $2$ days and just haven't been able to really understand what steps it requires.
 A: To recap what is happening in the post you linked to, we are trying to establish continuity of the map $f(x) = x^n$ on $\mathbb R$, where $n$ is a number equal to $1,2,\dots$. The "hint" is to write down a proof in the cases $n = 1, 2$, and to  establish the identity $u^n-y^n = (u-y)(u^{n-1} + u^{n-2}y + \dots + uy^{n-2} + y^{n-1})$ to solve the general $n$ case.
Here is what happens for $n = 2$. We want to prove that $f(x) = x^2$ is continuous, so we fix an arbitrary point $u$, and a challenge number $\varepsilon > 0$. We want to find a $\delta > 0$ so that whenever $y$ is a number satisfying $|u-y| < \delta$, we have $|u^2-y^2|<\varepsilon$. As is typical with $\varepsilon,\delta$-proofs, we assume $|u-y| < \delta$ (at this point $\delta$ is still undetermined) and we want to use the freedom to pick $\delta$ to try to make the expression $|u^2-y^2|$ small (how small? We are aiming for $< \varepsilon$) with only two things (actually three things):

*

*the assumption that $|u-y| < \delta$, and

*the freedom to choose $\delta$, and

*(knowledge about the function $f(x) = x^2$)

So, one piece of information (one of the only data points we have from the list above) is to use that $|u-y| < \delta$. To use this piece of data in order to establish $|u^2 - y^2| < \varepsilon$, we would like to relate the expression $|u^2-y^2|$ to the expression $|u-y|$. Here is where we use the special third data point, which gives this particular $\varepsilon,\delta$ problem its unique flavor. We know about the identity $u^2-y^2 = (u-y)(u+y)$, (which is the $n=2$ case of the identity that relates $u^n-y^n$ to $u-y$). If we use this identity, then we can say
$$
|u^2-y^2| = |u-y|\cdot|u+y| < \delta\cdot|u+y|.
$$
Now, we have hopefully gotten closer to the expression we are after ($|u^2-y^2| < \varepsilon$) by achieving an expression involving $\delta$ (one of the things we are allowed to choose, which should help us in the end), and the expression $|u+y|$, which no longer involves the function $f(x) = x^2$ anymore. At this point, we should try to use the first data point again (that $|u-y|<\delta$) in order to transform the expression $|u+y|$ into something involving $\delta$ and the particular point $u$ we chose. In the answer that you linked to, they write
$$
|u + y| \le |u| + |y| \quad(\text{triangle inequality}),
$$
and then
$$
|y| = |u + y-u|\le |u| + |y-u|\quad (\text{an example of the useful idea of adding $0$ in a helpful way}).
$$
Putting these two steps together with our first step of writing $|u^2-y^2| < \delta\cdot|u+y|$, we get
$$
|u^2-y^2| \le \delta\cdot(|u| + |u| + |y-u|).
$$
Notice that this work was made with the intention of transforming the expression $|u+y|$ into something related to $|u-y|$, and the outcome of the work was the "transformation"
$$
|u+y|\leadsto |u| + |u| + |y-u|.
$$
Let us see if this work was fruitful or not ($\varepsilon,\delta$ proofs are kind of like solving integrals—there is not any guarantee that what you will try is going to work, but there are decisions that are more natural than others, and experience is the only way to get better at solving such problems).
If we use the assumption $|u-y| < \delta$ (and the equality $|y-u| = |u-y|$), then we see that we have achieved
$$
|u^2-y^2| < \delta\cdot (2|u| + \delta).
$$
At this point, we are in the ideal position. We have replaced the expression $|u^2-y^2|$ with an expression only involving $u$, the point we started with, and $\delta$ (as an aside, in a proof of uniform continuity, we would like to do even better, and end up with an expression only involving $\delta$, and not the particular point $u$, hence the uniformity of the continuity—the $\delta$ we pick will be able to be chosen independently of the eccentric personality of the particular $u$ we began with).
So our work will be worth it if we can now choose $\delta$ to be so small that $\delta(2|u| + \delta) < \varepsilon$. Well, let us do one more rewrite:
$$
\delta\cdot (2|u| + \delta) = 2|u|\delta + \delta^2,
$$
so we replace our expression with a sum of two terms. A common goal now is to make each individual term less than $\varepsilon/2$, so that the sum is less than $\varepsilon$. In order to make the first term $2|u|\delta< \varepsilon/2$, simply "solve for $\delta$" and choose
$$
\delta < \frac{\varepsilon}{2\cdot 2|u|} = \frac{\varepsilon}{4|u|},
$$
but only if $u \ne 0$ so we don't divide by $0$. If $u = 0$, then the first term is actually already equal to $0$, which is less than $\varepsilon/2$. Now we want to make $\delta^2 < \varepsilon/2$, so solve for $\delta$ again, to see that we can take $\delta < (\varepsilon/2)^{1/2}$. How can we guarantee that we can choose $\delta$ so small as to ensure both terms are less than $\varepsilon/2$? We can simply pick $\delta$ to be smaller than the minimum of both expressions, so we finally say, if $u\ne 0$,
$$
\text{"let $\displaystyle\delta < \min\left(\frac{\varepsilon}{4|u|},\left(\frac{\varepsilon}{2}\right)^{1/2}\right)$,"}
$$
and if $u = 0$, we say
$$
\text{"let $\displaystyle\delta < \left(\frac{\varepsilon}{2}\right)^{1/2}$."}
$$
This choice of $\delta$ guarantees that $|u^2-y^2| < \varepsilon/2 + \varepsilon/2 = \varepsilon$ (or $|0^2-y^2| < \varepsilon/2 < \varepsilon$), so we are done.
The work we did to handle the term $|u+y|$ is an example of an essential problem in analysis, that of "finding estimates" (which in my opinion gives analysis a lot of its excitement, in an entirely analogous way that people familiar with this site know that solving integrals brings excitement). The general $n$ case is similar, but involves more intricate "estimates" because we have to manage the more complex term $u^{n-1} + u^{n-2}y + \dots + uy^{n-2} + y^{n-1}$, but the work boils down to using the data points and doing our best to make decisions that are most natural at every turn. For the case of $n$, at some point we would like to use the additional data point, which is the inductive hypothesis that $f(x) = x^{k}$ for $1\le k \le n-1$ is continuous. To do so, we would like to transform the expression $u^{n-1} + u^{n-2}y + \dots + uy^{n-2} + y^{n-1}$ into something involving the function $f(x) = x^k$ for possibly some or all $1\le k \le n-1$, and you can see an example of how that is done in the post that you linked to. It is perhaps important to note that this is an example of "strong induction" because we assume $f(x) = x^k$ is continuous for all $1 \le k \le n-1$, instead of merely for $k = n-1$. Best of luck and happy estimates!
