Showing that the given sequence of functions converges to a given function on the given set as follows

Question

Let $(f_n): \Bbb R \to \Bbb R$ be a sequence of functions defined by $f_n(x) := x^n$ for $x \in \Bbb R$ and $n \in \Bbb N$. Let $f:(-1,1] \to \Bbb R$ be a function defined by \begin{align*} f(x):= \begin{cases} 0, &\text{if} \quad -1<x<1 \\ 1, &\text{if} \quad x=1 \end{cases}. \end{align*} Then, the sequence $(f_n)$ converges to $f$ on the set $(-1,1]$.

Here's what I did to prove the above using the definition given below. Any advices and corrections are very welcome.

Definition. A sequence $(f_n)$ of functions on $A \subseteq \Bbb R$ to $\Bbb R$ converges to a function $f:A_0 \to \Bbb R$ on $A_0 \subseteq A$ if and only if for each $\varepsilon>0$ and each $x \in A_0$, there exists $n_0 \in \Bbb N$ (depending on $\varepsilon$ and $x$) such that for all $n \ge n_0$, $|f_n(x) - f(x)| < \varepsilon$.

1. For $x=1$: Let $\varepsilon>0$ be arbitrary and $x=1$. Choose $n_0=1 \in \Bbb N$ such that for all $n \ge n_0$, we have \begin{equation*} |f_n(x)-f(x)| = |f_n(1)-f(1)| = |1^n-1| = |1-1|=0<\varepsilon. \end{equation*} Hence, proved by definition. $\quad \Box$
2. For $x=0$: Let $\varepsilon>0$ be arbitrary and $x=0$. Choose $n_0=1 \in \Bbb N$ such that for all $n \ge n_0$, we have \begin{equation*} |f_n(x)-f(x)| = |f_n(0)-f(0)| = |0^n-0| =0<\varepsilon. \end{equation*} Hence, proved by definition. $\quad \Box$
3. For $0<x<1$: Let $\varepsilon>0$ and $x \in (0,1)$ be arbitrary. Choose $n_0 \in \Bbb N$ with $n_0 > \frac{\ln \varepsilon}{\ln x}$ such that for all $n \ge n_0$, we have \begin{equation*} |f_n(x)-f(x)| = |x^n| = |x|^n = x^n \le x^{n_0} <\varepsilon. \end{equation*} Hence, proved by definition. $\quad \Box$
4. For $-1<x<0$: Let $\varepsilon>0$ and $x \in (-1,0)$ be arbitrary. Choose $n_0 \in \Bbb N$ with $n_0 > \frac{\ln \varepsilon}{\ln (-x)}$ such that for all $n \ge n_0$, we have \begin{equation*} |f_n(x)-f(x)| = |x^n| = |x|^n = (-x)^n \le (-x)^{n_0} <\varepsilon. \end{equation*} Hence, proved by definition. $\quad \Box$

From the cases above, we can conclude that the sequence $(f_n)$ converges to $f$ on the set $(-1,1]. \quad \Box$

1. Does the above correct?
2. Should I give a proof that indeed for the cases $x=-1$ and $|x|>1$, $(f_n)$ is diverge on the set $(-1,1]$?

Thanks in advanced.

Answer 1

Here's what I tried with another proof (by using Bernoulli's Inequality).

Notice that for $0<x<1$, we can write $x=\frac{1}{1+y}$ where $y:=\frac{1}{x}-1$ so that $y>0$. By Bernoulli's Inequality, we have $(1+y)^n \ge 1+ny$. Now, given any $\varepsilon>0$ and any $0<x<1$. By the Archimedean Principle, we know that there exists a natural number $n_0$ such that $\frac{1}{n_0} < \varepsilon y$. Hence, for any $n \ge n_0$, we have \begin{equation*} \left|f_n(x) - f(x)\right| = \left|x^n - 0\right| = |x^n| = x^n = \frac{1}{(1+y)^n} \le \frac{1}{1+ny} < \frac{1}{ny} \le \frac{1}{n_0y} < \varepsilon. \end{equation*} Thus, proved by definition that $(f_n) \to f$ on the set $(0,1)$.

Now, the proof of the case for $x \in (-1,0)$ is similar and goes as follows: Notice that for $-1<x<0$, we can write $x=-\frac{1}{1+y}$ where $y:=\frac{1}{-x}-1$ so that $y>0$. By Bernoulli's Inequality, we have $(1+y)^n \ge 1+ny$. Now, given any $\varepsilon>0$ and any $-1<x<0$. By the Archimedean Principle, we know that there exists a natural number $n_0$ such that $\frac{1}{n_0} < \varepsilon y$. Hence, for any $n \ge n_0$, we have \begin{equation*} \left|f_n(x) - f(x)\right| = \left|x^n - 0\right| = |x^n| = (-x)^n = \frac{1}{(1+y)^n} \le \frac{1}{1+ny} < \frac{1}{ny} \le \frac{1}{n_0y} < \varepsilon. \end{equation*} Thus, proved by definition that $(f_n) \to f$ on the set $(-1,0)$.

Since $(f_n(x)) \to f(x)$ for the cases $x=0$ and $x=1$ immediately, we can conclude that $(f_n)$ converges pointwise to $f$ on the set $(-1,1]. \quad \Box$