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$$ \left\{ x \in\mathbb{R}\; \middle\vert\; \tfrac{x}{|x| + 1} < \tfrac{1}{3} \right\}$$

What is the supremum and infimum of this set? I thought the supremum is $\frac{1}{3}$. But can we say that for any set $ x < n$ that $n$ is the supremum of the set? And for the infimum I have no idea at all. Also, let us consider this example:

$$ \left\{\tfrac{-1}{n} \;\middle\vert\; n \in \mathbb{N}_0\right\}$$

How can I find the infimum and supremum of this set? It confuses me a lot. I know that as $n$ gets bigger $\frac{-1}{n}$ asymptotically approaches $0$ and if $n$ gets smaller $\frac{-1}{n}$ approaches infinity, but that's about it.

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up vote 1 down vote accepted

Let $x$ be negative. then $\dfrac{x}{|x|+1}$ is negative, and in particular $\lt \dfrac{1}{3}$.

So there is no infimum. (But some people allow the symbol $-\infty$ as an infimum.)

For the supremum, note that there are positive $x$ such that $\dfrac{x}{x+1}\lt \dfrac{1}{3}$.

Since $\dfrac{x}{x+1}=1-\dfrac{1}{x+1}$, our function is increasing for positive $x$. Solve $1-\dfrac{1}{x+1}=\dfrac{1}{3}$. We get $x=\dfrac{1}{2}$.

Thus $\dfrac{1}{2}$ is an upper bound for our set.

However close we are to $\dfrac{1}{2}$, but below $\dfrac{1}{2}$, we will have $\dfrac{x}{x+1}\lt\dfrac{1}{3}$. So there is no cheaper upper bound than $\dfrac{1}{2}$.

Remark: For your other question, I think you intend to look at $\{-\frac{1}{n}\}$, where $n$ ranges over the positive integers. The smallest element of this set is $-1$. It is therefore the infimum. For the supremum, note that our numbers are all $\lt 0$, but can be made arbitrarily close to $0$ by choosing $n$ large enough. So the supremum is $0$.

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Would this be valid way of doing it? $\dfrac{x}{|x| + 1} = \dfrac{1}{3}$ So: $3x = |x| + 1 $ and we so that $x = \dfrac{1}{2}$. If this is correct, 2 follow-up questions: Can we always find a supremum like this? If so, why, what is the logic behind it? And also, can we just subtract the |x| from the 3x? – JohnPhteven Dec 26 '12 at 21:44
Nevermind, I see the logic, I believe. $0.5$ is the point where the 2 intersect, so any point further along the line is not a supremum but is an upper bound and any point previous on the line is too small to be the supremum, correct? – JohnPhteven Dec 26 '12 at 21:49
To remove the absolute value sign, you need to break into cases, (i) $x\ge 0$ and (ii) x\lt 0. For $x\ge 0$, $|x|=x$, and you can subtract. For $x\lt 0$, $|x|=-x$, you get $3x=-x+1$, giving $x=1/4$, not negative, so discarded. There is no universal way to find a sup. Here the function was increasing, that's what made the procedure work. – André Nicolas Dec 26 '12 at 21:49
Your last statement is correct, you now know what's going on in this situation. – André Nicolas Dec 26 '12 at 21:51
Just thought about the function, how it behaves for large negative $x$ (negative, no inf), for large positive $x$ (near $1$), for smallish positive $x$ (can be less than $1/3$), so breakpoint at the only place it is exactly $1/3$. So the preliminary analysis was a picture in the head, then filling in the words is simple. – André Nicolas Dec 26 '12 at 22:11

For $x\geq 0$, the condition $\frac{x}{|x|+1}<1/3$ is equivalent to $x<x/3+1/3$, which in turn is equivalent to $x<1/2$. So the supremum of your set is $1/2$.

For $x\leq 0$, we have that $\frac{x}{|x|+1}<1/3$ is equivalent to $x<-x/3+1/3$, and then to $4x<1$, which is true for every nonpositive number. So the infimum of your set is $-\infty$.

Another way to put it is to observe that the set under consideration is $]-\infty,1/2[$.

For the second set $\{-1/n|n\geq 1\}=\{-1,-1/2,-1/3,\ldots\}$, the infimum is a mimimum and is equal to $-1$, while the supremum is the limit of this increasing sequence, namely $0$.

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Considering separately cases $x\geqslant{0}$ and $x<0$ you can find range of $x$ for which $f(x)=\dfrac{x}{|x| + 1} - \dfrac{1}{3} <0.$

For $x\geqslant {0}$ $$f(x)=\dfrac{x}{x+1}-\dfrac{1}{3}=\dfrac{x+1-1}{x+1}-\dfrac{1}{3}=\dfrac{2}{3}- \dfrac{1}{x+1},$$ For $x<0$ $$f(x)=\dfrac{x}{1-x}-\dfrac{1}{3}=\dfrac{x-1+1}{1-x}-\dfrac{1}{3}= \dfrac{1}{1-x} -\dfrac{4}{3} $$ therefore, the function $f(x)$ increases on $\mathbb{R}.$ Then for $x \in [0, \,+\infty) $ the inequality $f(x)<0$ holds for $0 \leqslant x < \dfrac{1}{2}.$ Solutions of the inequality $\dfrac{1}{1-x} -\dfrac{4}{3}<0$ in the second case lies in $(- \infty, \, 0) \cap \left(- \infty, \, \dfrac{1}{4}\right)= (- \infty, \, 0) .$

Thus $f(x)<0$ for $x \in (- \infty, \, 0)\cup \left[0, \, \dfrac{1}{2}\right)= \left(- \infty, \, \dfrac{1}{2} \right) \Rightarrow \sup \left \lbrace x\vert \; f(x)<0 \right \rbrace = \dfrac{1}{2} ; \;\; \inf \left \lbrace x\vert \; f(x)<0 \right \rbrace = -\infty .$

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But the question is not about the infimum and supremum of $f(\mathbb{R})$. It is about the infimum and supremum of $f^{-1}(]-\infty,0[$. – 1015 Dec 26 '12 at 22:01
@julien Thanks for remark. My anwer was inaccurate and i edit it. – M. Strochyk Dec 26 '12 at 23:23

Hint: Find the range of $$f(x)=\frac{x}{\left|x\right|+1}-\frac13$$ This function is continuous and increasing.

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