misconception about pointwise/uniform convergence

Question

Can anyone clear up the misconception in my argumentation. If a sequence of functions $f_k(x)$ is pointwise convergent towards some limit $f(x)$ for every $x$, then at every point $x$ we can choose an arbitrary distance $\varepsilon$ such that from a follow index $n_0$ the distance of $f_k(x)$ with $k\ge n_0$ and $f(x)$ is smaller than $\varepsilon$. So why can't we take the maximum $K$ of all the $n_0$ regarding to all the $x$ and say from this $K$ we have $|f_k(x) - f(x)| < \varepsilon$ for every $x$ which means that $\sup_x |f_k(x) - f(x)|<\varepsilon$ for $k>K$? Is the reason that since the set of all these indices $n_0$ regarding to all the $x$ is overcountable and so a maximum does not exist?

Answer 1

Consider the sequence of functions $f_n(x)=x^n$ defined on $(0,1)$, with limit function $f(x)=0$. Say we try to satisfy the definition of uniform convergence with $\epsilon=0.1$. We have $|f_n(x)-f(x)|=x^n$ and so we want $x^n<0.1$.

For $x=0.9$ we can take $n=22$ since then $(0.9)^{22}<0.1$.

For $x=0.99$ we can take $n=230$ since $(0.99)^{230}<0.1$.

For $x=0.999$ we can take $n=2302$ since $(0.999)^{2302}<0.1$.

Clearly there is no $N$ we can take which would work for all $x\in(0,1)$. That is, given $x$ we need $n(x)>\log_{x}0.1$ and as $x$ approaches $1$ the values of $n(x)$ approach infinity.

Another way of putting this is that, as you say, we cannot take the maximum of the set of all $n$'s, simply because that set is infinite. A maximum is guaranteed to exist for finite sets, but not for infinite ones, and indeed the maximum does not exist in this example, as the above calculation shows.