Verify if $\sup(A) = +\infty$, where $A=\{\frac{2}{x-1} : x > 1\}$.

Question

Verify if $\sup(A)=+\infty$, where $$A=\{\frac{2}{x-1} : x > 1\},$$ using the definition and no limit.

My steps:

$$\frac{2}{x-1}=1$$ $$\frac{2}{x-1}-1=0$$ $$\frac{2-x+1}{x-1}=0$$ $$\frac{3-x}{x-1}=0$$ D: $x\neq1$ so $3-x=0$ and $x=3$

Are my steps right of there is a different way to solve this type of exercise?

Answer 1

I'm not sure what your steps meant, but give my opinions.

To show $\sup \{\frac{2}{x-1}:x>1\}=\infty,$ what you need to verify is $\forall r>0,~\exists x>1$ such that $\frac{2}{x-1}>r.$

Answer 2

You are trying to show that given a real number $r \in \mathbb R$ there is some $a\in A$ such that $a \ge r$

Since every $a\in A$ is of the form $\frac{2}{x-1}$ for $x > 1$ you are looking for an $x>1$ such that $\frac{2}{x-1} \ge r$

Answer 3

Let $x_n=1+1/n$. Then $ \frac{2}{x_n-1}=2n$.

Consequence: $2n \in A$ for all $n \in \mathbb N$.

Your turn !