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For example, is $f(z) = 1/z$, on the set $0<z<1$ "bounded away from zero"?

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    $\begingroup$ It probably means that there is an open interval around $0$ which the function doesn't hit. $\endgroup$ – Arthur Jun 26 '15 at 22:08
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    $\begingroup$ They’re apparently talking about the values of the function, not the domain. Indeed, $1/z\in\{w:|w|>1\}$ on that domain, certainly a set bounded away from zero. The wording is ambiguous, though. Poor writing on the part of your author! $\endgroup$ – Lubin Jun 26 '15 at 22:18
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    $\begingroup$ @Lubin: I would not say that the author is to blame here: "bounded away from zero" is a common enough expression, with a well-defined meaning. At least it was in my day. $\endgroup$ – TonyK Jun 26 '15 at 22:43
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    $\begingroup$ @TonyK, “spoken words are light as air; written words are everywhere.” In lectures, we get away with all sorts of sloppinesses and infelicities. If it is from a lecture, I grit my teeth and let it pass. In a book, though, it should be edited to be perfectly precise. $\endgroup$ – Lubin Jun 26 '15 at 22:50
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    $\begingroup$ @Lubin: What are you talking about? Is that Emily Dickinson? Anyway, the expression belongs to the written language of mathematics $-$ we didn't go around saying things like "the number of beers I've drunk is bounded away from zero". Honestly we didn't. So you can ungrit your teeth. And google "bounded away from zero" if you like. $\endgroup$ – TonyK Jun 26 '15 at 22:53
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If a set $S \subset \mathbb R$ is bounded away from zero, it means that there exists $m > 0$ such that $|x| > m$ for all $x \in S$.

If a function $f$ is bounded away from zero, it means that its range is bounded away from zero: there exists $m > 0$ such that $|f(x)| > m$ for all $x$.

Edited to clarify: When we say a set is bounded away from zero, we are not saying that away from zero, it is bounded. What would that even mean? We are saying that its distance from zero is bounded below by a strictly positive number. I see now that this is not self-evident, but that's what it means.

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  • $\begingroup$ Hi @TonyK - you mean |f(x)| < m for all x, right? (not > ). Then, would you say that 1/z is bounded away from zero, even though we can pick z as close to zero as we want, on the domain 0< z < 1. So really 1/z is not bounded ... yet we say it is bounded away from zero, I think. $\endgroup$ – User001 Jun 26 '15 at 22:24
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    $\begingroup$ @LebronJames No, the point is that $f$ stays away from $0$, so $\lvert f(x)\rvert > m$ (or $\geqslant m$) is exactly what one wants. $\endgroup$ – Daniel Fischer Jun 26 '15 at 22:27
  • $\begingroup$ Ok, got it. Thanks so much, @DanielFischer. $\endgroup$ – User001 Jun 26 '15 at 23:01
  • $\begingroup$ Thanks for the clarification, @tonyk. $\endgroup$ – User001 Jun 26 '15 at 23:03
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Not to start an argument, but to give the way that I've heard this phrase used:

$f$ is bounded away from zero if there is $\varepsilon > 0$ such that the range of $f$ does not meet $(-\varepsilon, \varepsilon)$, or equivalently, there is some open set containing zero which the range of $f$ does not meet.

In this sense, the function $f(z) = 1/z$ is bounded on $0 < z < 1$, as the range of $f$ does not meet, say, $(-\frac{1}{2},\frac{1}{2})$.

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  • $\begingroup$ got it - thanks, @qaphla. $\endgroup$ – User001 Jun 26 '15 at 23:04
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i emphasize, that bounded away from zero means correctly, that:

For all $\epsilon>0$: $f([\epsilon,1])$ is bounded

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