$\sup$s of $|f_n-f|$ with $f_n=x^n$ on $[0,1]$ and on $[0,1)$

Question

Consider the following sequence of functions $f_n : [0,1] \to \mathbb{R}$ $$f_n(x)=x^n\,\,\,\,\,\,\,\,\, n \geq 1$$ The sequence converges in $[0,1]$ to the function $f: [0,1] \to \mathbb{R}$ $$f(x)=\begin{cases} 0 & x\in [0,1) \\ 1 & x=1 \end{cases}$$

My question is: how can I prove the following?

$$\sup_{x \in [0,1]} |f_n(x)-f(x)|=\sup_{x \in [0,1)} |f_n(x)-f(x)|$$

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In particular, the problem here is that the function $f_n(x)-f(x)$ is not continous, since

$$f_n(x)-f(x)=\begin{cases} x^n & x\in [0,1) \\ 0 & x=1 \end{cases}$$

Therefore in general it is not true that the $\sup$ of $|f_n(x)-f(x)|$ does not change if I go from a subset (in this case the interval $[0,1)$) to its closure (in this case $[0,1]$) and viceversa. Nevertheless in this case it should be true.

Answer 1

You use the fact that $$\sup_{x\in A\cup B} g(x)=\max\left\{\sup_{x\in A}g(x),\,\sup_{x\in B}g(x)\right\}$$ with $A=[0,1)$ and $B=\{1\}$ and observe that, for all $n$, $\sup\limits_{x\in\{1\}} \lvert f_n(x)-f(x)\rvert=0$.