Find the least upper bound and greatest lower bound for {$x : x^3 \ge 8 $}

sequences-and-seriesupper-lower-bounds

The notation {$x : x^3 \ge 8 $} confuses me a bit. Am I understanding this right? It's asking which x is the smallest that are great than or equal to 8?

If I'm understanding the question then there is no lub right because there are an infinite set of numbers greater than 8. And would the greatest lower bound (glb) be 2 since $8^\frac13$ would be the smallest number?

Best Answer

Your set $S=\{x: x^3 \geq 8\}$ is unbounded above, so mathematicans would tend to write $sup(S)=\infty$.

We also have $inf(S)=2$ because $min(S)=2$. (Every element of $S$ is $\geq2$, and $2 \in S$, hence $min(S)$ exists and $=2$.)