I'm currently reviewing some concepts of continuity from Hubbard & Hubbard's Vector Calculus, Linear Algebra and Differential Forms.
There is a proof that I'm having a hard time accepting as valid: https://photos.app.goo.gl/zsLWfDuHEVEV6hu3A
My problem is that, when proving the converse, the authors say that for the choice of ${\delta _n}=1/n$, you can choose ${{\mathbf{x}}_{n}}$ such that $\left| {{\mathbf{f}}({{\mathbf{x}}_n}) - {\mathbf{f}}({{\mathbf{x}}_0})} \right| \geqslant {\varepsilon _0}$. However, the premise to arrive at the contradiction is that, for the given delta, there is some $\mathbf{x}$ such that $\left| {{\mathbf{f}}({{\mathbf{x}}}) - {\mathbf{f}}({{\mathbf{x}}_0})} \right| \geqslant {\varepsilon _0}$. Case in point, consider the sequence $\left\{ {{x_n}} \right\} = \frac{1}{{\sqrt n }}$ and ${x_0} = 0$. Clearly there is no point satisfying the first inequality of 1.5.55 given this convergent sequence, thus making the proof invalid in general. Am I missing something?
Note that $x_n$ is chosen such that $|x_0-x_n|<\delta_n = \frac{1}{n}$.
The author assumes, for the sake of contradiction, that $f$ is not continuous, i.e. there exists $\epsilon_0>0$, such that for all $\delta>0$, there exists $x$ such that $|x-x_0|<\delta$ and $|f(x)-f(x_0)|>\epsilon_0$.
This means that for $\delta=\delta_n=\frac{1}{n}$, you can choose an $x_n$ such that $|x_n-x_0|<\delta = \delta_n =\frac{1}{n}$ and $|f(x_n)-f(x_0)|>\epsilon_0$.