Exercise involving pointwise and uniform convergence: why does it converge and why it does not converge? (a) Let $(f_{n})_{n=1}^{\infty}$ be a sequence of functions from one metric space $(X,d_{X})$ to another $(Y,d_{Y})$, and let $f:X\to Y$ be another function from $X$ to $Y$. Show that if $f_{n}$  converges uniformly to $f$, then $f_{n}$ also converges pointwise to $f$.
(b) For each integer $n\geq 1$, let $f_{n}:(-1,1)\to\textbf{R}$ be the function $f_{n}(x) = x^{n}$. Prove that $f_{n}$ converges point wise to the zero function, but does not converge uniformly to any function $f:(-1,1)\to\textbf{R}$.
(c) Let $g:(-1,1)\to\textbf{R}$ be the function $g(x) = x/(1-x)$. With the notation as in $(b)$, show that the partial sums $\sum_{n=1}^{N}f_{n}$ converges point-wise as $N\to+\infty$ to $g$, but does not converge uniformly to $g$ on the open interval $(-1,1)$.
MY ATTEMPT
(a) According to the definition of uniform convergence, for $\varepsilon > 0$ there is a natural number $N\geq 1$ such that for every $x\in X$ we have that
\begin{align*}
n\geq N \Rightarrow d_{Y}(f_{n}(x),f(x)) < \varepsilon
\end{align*}
Since it holds for every $x\in X$, it converges point wise for each $x_{0}\in X$ separately.
(b) Let us tackle the problem in three parts: $x\in(-1,0)$, $x = 0$ and $x\in(0,1)$.
When $x\in(0,1)$, we have that
\begin{align*}
0 < x < 1 \Rightarrow 0 < x^{2} < x < 1 \Rightarrow 0 < x^{3} < x^{2} < x < 1 \Rightarrow \ldots
\end{align*}
That is to say, $x^{n}$ is decreasing and bounded below by $0$. Consequently, it converges to some real number $L$. More precisely, we have that
\begin{align*}
L = \lim_{n\rightarrow\infty}x^{n+1} = \lim_{n\rightarrow\infty} x\times x^{n} = x\times\lim_{n\rightarrow }x^{n} = xL \Longleftrightarrow L(1 - x) = 0
\end{align*}
Given that $x\in(0,1)$, we conclude that $L = 0$, and we are done.
On the other hand, if $x = 0$, then $f_{n}(x) = 0$. Hence $x^{n}$ converges to $0$.
Finally, we must consider $x\in(-1,0)$, that is to say, $-x\in(0,1)$. Since the series
\begin{align*}
\sum_{n=1}^{\infty}x^{n} = \sum_{n=1}^{\infty}(-1)^{n}(-x)^{n}
\end{align*}
converges according to the Leibniz test, we conclude that $x^{n}$ converges to 0.
Gathering all the previous results, we conclude that $f_{n}$ converges point-wise to the zero function on $(-1,1)$.
However I am not able to prove that it does not converge uniformly to any function $f$ defined on $(-1,1)$.
Can someone help me with this?
(c) Once again, we have that
\begin{align*}
\sum_{n=1}^{N}f_{n}(x) = x + x^{2} + \ldots + x^{N} = \frac{x(1 - x^{N})}{1-x}
\end{align*}
which is well defined for every $x\in(-1,1)$. Since $x^{N}\to 0$ when $x\in(-1,1)$, $\sum f_{n}\to g$, and the desired result follows.
Once again, I am not able to prove that it does not converges uniformly to $g$ on $(-1,1)$.
Can someone help me with this?
Any comments our alternative solutions are welcome as well.
 A: For the remainder of part b), choose some $\varepsilon \in (0, 1)$ (you can choose any such number). If $x^n$ converges uniformly to $0$, then we expect there to be some $n$ such that $x^n \in (0 - \varepsilon, 0 + \varepsilon)$ for all $x \in (-1, 1)$. (Indeed, we actually expect this to be true for all sufficiently large $n$, but we only need one value of $n$ for this argument.)
The problem is, as you head closer to $1$, it takes longer and longer for $x^n$ to lie in this interval. In particular, if we consider:
$$x_0 = \varepsilon^{\frac{1}{n+1}} \in (0, 1),$$
then $x_0^n = \varepsilon^{\frac{n}{n+1}} > \varepsilon$, which contradicts our choice of $n$. So, there can never be even a single $n$ so that $x^n$ is uniformly closer than $\varepsilon$ to $0$, which strongly contradicts uniform convergence.
For the remainder of part c), suppose we have uniform convergence. We will use this supposition to show that $x^n \to 0$ uniformly, contradicting part b).
Fix $\varepsilon > 0$. Then we have an $N$ such that, for all $x \in (-1, 1)$,
$$n \ge N \implies \left|\sum_{i=1}^n x^i - \frac{x}{1 - x}\right| < \frac{\varepsilon}{2}.$$
Choose any $n \ge N + 1$. Then $n - 1 \ge N$, hence for all $x \in (-1, 1)$,
\begin{align*}
\varepsilon &= \frac{\varepsilon}{2} + \frac{\varepsilon}{2} \\
&> \left|\sum_{i=1}^n x^i - \frac{x}{1 - x}\right| + \left|\sum_{i=1}^{n - 1} x^i - \frac{x}{1 - x}\right| \\
&\ge \left|\left(\sum_{i=1}^n x^i - \frac{x}{1 - x}\right) - \left(\sum_{i=1}^{n-1} x^i - \frac{x}{1 - x}\right)\right| \\
&= |x^n| = |x^n - 0|.
\end{align*}
So, $n \ge N + 1 \implies |x^n - 0| < \varepsilon$, i.e. $x_n \to 0$ uniformly, contradicting b).
A: To show $f_n(x), x \in E$ do not uniformly converge to $f(x)$, the standard trick is to show $\sup_{x \in E} |f_n(x) - f(x)|$ fails to converge to $0$ as $n \to \infty$. 
With this in mind, to show $x^n$ do not uniformly converge to $0$, we can proceed as follows:
\begin{align}
\sup_{|x| < 1} |x^n| \geq \left(1 - \frac{1}{n}\right)^n \to e^{-1} \neq 0 \text{ as } n \to \infty. \quad (\text{since } |1 - n^{-1}| < 1)
\end{align}
Using the same trick, I believe you can complete part (c) easily.
