$$ \left\{ x \in\mathbb{R}\; \middle\vert\; \tfrac{x}{|x| + 1} < \tfrac{1}{3} \right\}$$
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.
Best Answer
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 $0$.