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Is there any standard name for this concept that is weaker than local one-to-one-ness?

In some open neighborhood of $x_0$ there is no point $x\ne x_0$ such that $f(x)=f(x_0)$.

Or, if you like: In some neighborhood of $x_0$, for every point $x$ in that neighborhood, $f(x)=f(x_0)$ only if $x=x_0$.

Might one simply say that "$f$ is weakly locally one-to-one at $x_0$"?

Trivial question, it might seem, if it's only about mathematics. Maybe it's about psychology of learning mathematics. I recently came across this error: If $f\;'(a)>0$, the $f$ is strictly increasing in some neighborhood of $a$. (Much less recently, I made that mistake myself, when, as an undergraduate, I was frustrated by my inability to write what I thought must be a simple $\varepsilon$-$\delta$ proof of that proposition.) Would people be less likely to make that mistake if somewhere in the recesses of their brain they subconsciously remembered hearing that "If $f\;'\ne 0$ at a point, then $f$ is weakly locally one-to-one at that point."?

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    I've completely revised it. I realized it didn't say what I meant and came back to fix it, expecting a hundred downvotes and 50 indignant comments. Apparently I got here before they did.2011-11-21
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    I was busy trying to figure out why weakly locally 1-1 wasn't equivalent to 1-1. :)2011-11-21
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    Your last comment about $f'\ne0$ at a point is interesting, from a pedagogical standpoint (and in and of itself). I think your statement can be proven without MVT. Then you could ask if $f'\ne0$ at a point implies monotonicity near the point. Show that this is false, and then ask "what extra condition would be needed to get monotonicity"?2011-11-21
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    To quote my discrete math professor: if you try to prove something and fail, and then find out this something is false then you did a good job not proving a false claim.2011-11-21
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    @DavidMitra : Certainly the following can be proved without MVT. If $f\;'(a)>0$, then in some neighborhood of $a$, for all $x$ in that neighborhood, if $x>a$ then $f(x)>f(a)$ and if $x then $f(x). All you need to do is set $\varepsilon=|f(a)|$, then let $\delta>0$ be so small that for $x$ in the $\delta$-neighborhood of $a$, the difference quotient $(f(x)-f(a))/(x-a)$ is within $\varepsilon$ of $f\;'(a)$. That implies the difference quotient is positive, and the result easily follows. No MVT needed.2011-11-22

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