Confusion with uniform convergence of $f_n(x)=\frac{x}{n}$ on $\mathbb{R}$ vs $D \subset \mathbb{R}$

Question

I think I understand the proof why $\lim_{n \to \infty}f(x_n)=\frac{x}{n}$ doesn't converge to $f(x)=0$ for $x \in \mathbb{R}$. This is because $\lim_{x \to \infty}\frac{x}{n} = \infty$ and $\lim_{x \to -\infty} \frac{x}{n}=-\infty$ and therefore I can't define an $\epsilon$-ball around $f(x)$ such that $d(f_n(x),f(x))$ is within this ball.

So what if instead I define some finite subset $D \subset \mathbb{R}$ such that $x \in D$? Will then $f_n(x)$ converge to $f(x)$ uniformly? I think it will because I can vary $x \in D$ as I like and $\forall x$ this $\epsilon-$ball will be found for $n> N$.

So here's my confusion: why doesn't this work with the case $x \in \mathbb{R}$? If infinity is not a real number, then I can take any real number and apply the same idea.

Yet it is wrong. Why?

The problem is you cannot chose the same $N$ for every $x$. The bigger $x$ is the bigger you have to chose $N$. — user127.0.0.1, Jan 08 '14 at 20:13

score 1 · Accepted Answer · answered Jan 08 '14 at 20:17

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When talking about uniform convergence, you are concerned with the sequence of $\sup\limits_{x \in D} |f_n(x)-f(x)|$ (which in our case is $\sup\limits_{x \in D} \dfrac{|x|}{n}$) and seeing if it converges to zero as $n \to \infty$.

If $D$ is bounded, then intuitively, you can see that this supremum goes to zero as $n \to \infty$, so it is indeed uniformly convergent for all $x \in D$. However, if $D$ is unbounded, then the supremum is infinite for any $n$ as you pointed out, and so it cannot possibly go to zero.

answered Jan 08 '14 at 20:17

angryavian

89,882

so the main idea is that $\mathbb{R}$ doesn't have a supremum, and therefore $\frac{x}{n}$ can't be bounded from the above? – Alex Jan 08 '14 at 20:42
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@Alex Yes, that's about it. – angryavian Jan 08 '14 at 20:51

Confusion with uniform convergence of $f_n(x)=\frac{x}{n}$ on $\mathbb{R}$ vs $D \subset \mathbb{R}$

1 Answers1