Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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.

share|improve this question

4 Answers 4

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 $0$.

share|improve this answer
    
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? –  ZafarS 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? –  ZafarS 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
1  
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 $0$.

share|improve this answer

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

share|improve this answer
    
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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.