Why is this sequence of functions not uniformly convergent?

Question

For the following sequence of functions and its limit function, we can see that $f_n(x)$ is clearly pointwise convergent

$$f_n(x) = x^n\text{ }\forall x\in[0,1]\text{ and }\forall n\in\mathbb N^*\\ f(x) = \begin{cases}0&\text{if } x\in[0,1)\\1&\text{if } x=1\end{cases}$$

However, I was wondering why this is not uniformly convergent. The condition for uniform convergence is:

$$|f_n(x) - f(x)| < \epsilon,\ \ \ \forall x \text{ when } n > N$$

Now most sources present an argument along the lines of: assume that $f_n(x)$ is uniformly convergent and that $0 < x < 1$, this means that $x^n<\epsilon$ whenever $n>N$. Specifically, this would mean $x^{N+1}<\epsilon$ for some fixed $N$. But if we now pick $x$ such that $1 > x > ε^{\frac{1}{N+1}}$, then this would lead to a contradiction, therefore $f_n(x)$ is not uniformly convergent.

However, I was wondering why couldn't we take $n$ to infinity. If $0 < x < 1$, then $\lim_{n\rightarrow \infty} |f_n(x) - f(x)| = 0$ (which is less than $\epsilon$). Now since $|f_n(x) - f(x)|$ will always be $0$ if $n > \infty$, then wouldn't this be uniformly convergent?

Answer 1

You can't just take $n=\infty$. In effect what you have done is to verify pointwise convergence, not uniform convergence.

Uniform convergence fails because, for every $n$, $\sup_{x\in[0,1]} |f_n(x)-f(x)| = 1$, which you can see by continuity.

Alternatively, uniform convergence must fail because each $f_n$ is continuous, $f$ is not continuous, and a uniform limit of continuous functions is continuous.

Answer 2

It's a matter of negation of logical quantifiers. Let $(f_n)$ be a sequence of functions defined on a set $A \subset \mathbf{R}$, $f$ be a function defined on the same domain.

Definition of Uniform convergence: $\forall \epsilon > 0$, $\exists N \in \mathbf{N}$, $\forall n \geq N$ and $x \in A$ , $f_n(x)-f(x) < \epsilon$ holds.

Definition of non Uniform convergence: $\exists \epsilon > 0$, $\forall N \in \mathbf{N}$, $\exists n \geq N$ and $x \in A$, $f_n(x)-f(x) >= \epsilon$ holds

So $\forall N$, you can choose $x=\frac{1}{\sqrt[n]{2}}$ and $\exists n \geq N$, $f_n(x)-f(x) \geq \frac{1}{2}$ holds. See also here.