Writing a formal proof for pointwise and uniform convergence Let $f_k:[0,1]\to\mathbb{R}$,
$$
  f_k(x) =
  \begin{cases}
                                   k-k^2x & \text{if $x\in(0,1/k)$} \\
                                   0 & \text{if $x\in[1/k,1]\cup\{0\}$}. \\
  \end{cases}
$$
By using definition how can we show that $(f_k)$ converges pointwise to $f=0$ but not converges uniformly. I think we should use Archimedean Principle but i can not write a formal proof.  
 A: Let $x \in [0,1]$ such that $x>0$.
Then exists $N \in \Bbb{N}$ such that $\frac{1}{N} < x ,\forall n \geq N$
Thus $f_n=0,\forall n \geq N \Longrightarrow f_n(x) \to 0$
Take the sequence $x_k=\frac{1}{k^2}$
Then $\sup_{x \in [0,1]} |f_k(x)| \geq f_k(x_k)=k,\forall k \in \Bbb{N}$
A: For $x=0$ we get $f_n(x)=0 \to 0$ trivially. For $x >0$ note that $\frac 1 k <x$ whenever $k >\frac  1 x$. So $(f_n(x))$ consists only on zeros after some stage. This proves pointwise convergence.
Suppose $f_n \to 0$ uniformly. Then there exists $n_0$ such that $|f_n(x)| <1$ for all $x$ for all $n \>n_0$. Now put $x=\frac 1 {2n}$ in this to get $\frac n 2 <1$ for all $n>n_0$ which is a contradiction. Hence the convergence is not uniform. 
A: First, observe that
$$
f_k\Big(\frac{1}{2k}\Big)=k-k^2\cdot\frac{1}{2k}=\frac{k}{2}\ge \frac{1}{2}.
$$
If $f_k:[0,1]\to\mathbb R$ converged to 0, uniformly, then for every $\varepsilon>0$, there would exist a $ k_0\in\mathbb N$, such that
$$
k\ge k_0\Longrightarrow |f_k(x)|<\varepsilon, \quad \text{for all $x\in [0,1]$}.
$$
The above is clearly violated for $\varepsilon=1/2$.
