This is the exercise problem 6.2.2(a) in Abbott's Understanding Analysis text. I'd like someone to verify if my proof is correct. Also, is my exposition clear, or is it over-engineered?
Define a sequence of functions on $\mathbf{R}$ by
$$ f_n(x) = \begin{cases}
1 & \text{ if } x = 1,\frac{1}{2},\frac{1}{3},\ldots,\frac{1}{n}\\
0 & \text{ otherwise }
\end{cases}$$
Is each $f_n$ continuous at zero? Does $f_n \to f$ uniformly on $\mathbf{R}$? Is $f$ continuous at zero?
Proof.
(a) Is each $f_n$ continuous at zero?
Pick an arbitrary $\epsilon > 0$. Let $N \in \mathbf{N}$ be arbitrary.
Let $(x_k) \to 0$ be an arbitrary convergent sequence. As $(x_k)$ is convergent, there exists K such that $|x_k| < \frac{1}{N}$, for all $k \geq K$.
Immediately, it follows that $f_N(x_k)=0$ for all $k \geq K$. Consequently, the image sequence $f_N(x_k) \to 0$.
As $(x_k)$ was arbitrary, this is true for all sequences approaching zero. By definition, $f_N(x)$ is continuous at zero.
As $N$ was arbitrary, this is true for all $f_n(x)$, where $n \in \mathbf{N}$.
QED.
Although not asked for, I investigate the pointwise convergence of $(f_n)$.
Is $(f_n)$ pointwise convergent to some $f$?
$(f_n)$ is the zero sequence for all $x \neq \frac{1}{n}$, $n \in \mathbf{N}$. Let us explore pointwise convergence at points of the form $x = \frac{1}{n}$ in the interval $(0,1)$.
Pick an arbitrary $N\in \mathbf{N}$ and fix $x = \frac{1}{N}$.
The tail $f_n(x)=1$ for all $n \geq N$. Consequently,
$$f(x) = \begin{cases}
1 & \text{ if } x = \frac{1}{n}, n \in \mathbf{N} \\
0 & \text{ otherwise }
\end{cases}$$
QED.
Does $f_n \to f$ uniformly on $\mathbf{R}$?
Let sample the functions $f_n(x)$ at different points.
\begin{align*}
(f_n(1)) &= (1,1,1,1,\ldots)\\
(f_n(\frac{1}{2})) &= (0,1,1,1,\ldots)\\
(f_n(\frac{1}{3})) &= (0,0,1,1,\ldots)\\
(f_n(\frac{1}{4})) &= (0,0,0,1,\ldots)\\
\vdots
\end{align*}
No. Pick $\epsilon = \frac{1}{2}$. Let $N \in \mathbf{N}$ be arbitrary. And let $x = \frac{1}{N+1}$.
For $n = N$, we have,
$|f_n(x) – f(x)| = |f_N(x) – f(x)| =|0 – 1| > \epsilon$.
Thus, there exists an $\epsilon > 0$ and some points $x \in \mathbf{R}$ such that for all $N \in \mathbf{N}$, $|f_n(x) – f(x)| > \epsilon$ for some $n \geq N$. So, $|f_n(x) – f(x)|>\epsilon$ for infinitely many $n$'s.
That is, $(f_n)$ does not uniformly converge to $f$.
QED.
Is $f$ continuous at zero?
No $f$ is not continuous at zero. Pick $(x_n) = 1/n$. Then, as $(x_n) \to 0$, $f(x_n) \to 1$, but $f(0) = 0$.
Best Answer
It is correct. You wrote that you were going to investigate the pointwise convergence of $(f_n)_{n\in\Bbb N}$ “Although not asked for”, but you had to study the pointwise convergence of the sequence in order to get the only possible function to which $(f_n)_{n\in\Bbb N}$ could possibly converge uniformly.
And a minor detail: asserting that $(f_n)_{n\in\Bbb N}$ does not converge uniformly to $f$ meanse that there is some $\varepsilon>0$ such that, for every $N\in\Bbb N$, you have the inequality $|f(x)-f_n(x)|\geqslant\varepsilon$ for some $n\geqslant N$ and for some $x\in\Bbb R$. So, you could have taken $\varepsilon=1$ instead of $\varepsilon=\frac12$.