-1

This is general related to typical $\epsilon > 0$ arguments. For an instance, if I have: For every $\epsilon > 0$ there exists an $N(\epsilon) > 0$ such that $|f_n(x)-f(x)| < \epsilon$, whenever $n > N(\epsilon)$ for all $x \in \mathbb{R}$, what is an explanation this implying $$\lim_{n \rightarrow \infty} f_{n}(x)=f(x)~.$$

Can I just say, this follows since $\epsilon > 0$ is arbitrary, by letting $\epsilon \rightarrow 0^{+}$.

  • 3
    Well, that is the definition of $\lim_{n \to \infty} f_n(x) = f(x)$. https://en.wikipedia.org/wiki/Uniform_convergence Right, as Tyler mentions, you are missing $n \geq N$ in the statement. – Nicholas Roberts Dec 07 '19 at 02:28
  • Check your statement again, because you neglect to mention what $N(\epsilon)$ is doing in the statement. Also, epsilon definitions are definitions, so there's no proof necessary. –  Dec 07 '19 at 02:28

1 Answers1

1

In the example on your post, this is just the definition of the limit.

I believe you are referring to cases like this, where the author says:

Taking supremum over $x,y\in\operatorname{cl}(E)$, one has that $$\operatorname{diam}(\operatorname{cl}(E))=\sup_{x,y\in\operatorname{cl} E}d(x,y)\leq\operatorname{diam}(E)+\varepsilon.$$ Since $\varepsilon>0$ is arbitrary, it follows that $$\operatorname{diam}(\operatorname{cl}(E))\leq\operatorname{diam}(E).$$

In the general case, suppose we want to prove that two quantities, say $a$ and $b$, are equal. In other words, $a-b=$.

But I can only proof that, for every $\varepsilon >0$, $$|a-b|\le \varepsilon$$.

But $|a-b|$ is a non-negative real number that is smaller than any positive real number, because we can pick $\varepsilon$ as small as we want. The only number that satisfies these conditions is $0$. Therefore, $|a-b|=0$.

Célio Augusto
  • 1,559