# Does $f \sim g$ imply $f \asymp g$ in certain conditions?

I got a good answer to this question over on MathOverflow a while ago. Harald Hanche-Olsen claimed that, if $f, g: D\to \mathbb{R}^+$, then $$f(x) \sim g(x) \implies f(x) \asymp g(x) \qquad \qquad (*)$$ holds whenever $D\subseteq\mathbb{C}$ is closed in $\mathbb{C}$, and is false whenever it is not closed.

However, something bugs me about this. Most instances I've seen this, it does hold when $D$ is the strictly positive reals, which I believe is not closed in $\mathbb{C}$.

So this is my question. Does $(*)$ hold when $f, g: \mathbb{R}^+ \to \mathbb{R}^+$? That is, when $f$ and $g$ are positive real-valued and defined on the positive reals?

(Edit: For a detailed explanation of which definitions I am using, see the link. I believe they are standard.)

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Claimed in the sense of proved. – Did Jun 10 '11 at 17:48

No, for basically the same reason it fails for other non-closed sets. Just because $\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}=1$, it doesn't mean that $\lim_{x\rightarrow 0} \frac{f(x)}{g(x)}$ even exists.
So you could, for example, take $f(x) = \frac{1}{x}$ and $g(x) = \frac{1}{x+1}$. Then certainly $\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}=1$, but there is no $A$ such that $f(x) < Ag(x)$ for all $x$, since $f$ has a vertical asymptote at 0.