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Let $Y = \{1/(1 + |x|): x \in \mathbb{R}\}$. This means that $Y$ is the collection of all the numbers of form $1/(1 + |x|)$, where $x$ is real. Does $Y$ have a maximum element (i.e. is there a $y_0$ in $Y$ such that $y \leq y_0$ for every $y$ in $Y$)? Determine the greatest real number $T$ such that $T \leq y$ for every $y$ in $Y$ (i.e. the greatest lower bound), and justify your answer

The solution starts simple enough:

$1/(1 + |x|) \leq 1$ as $0 \leq |x|$. The upper bound $1$ is attained when $x = 0$

Therefore $1$ is the maximum element of $Y$.

All elements of $Y$ are positive. If $t > 0$, then make $1/(1 + |x|) < t$

This is where I got confused, the solution then says "by choosing $x > 1/t$"

I don't understand this, is this correct? Why isn't it when $|x| > (1 - t)/t$?

How did they get $x > 1/t$? Is it wrong?

N. Bar
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Sam Connell
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    You could have done it your way as well, but it is easier to word the argument their way. By letting $x>\frac{1}{t}$ we have that $\frac{1}{1+x}<\frac{1}{x}<\frac{1}{1/t}=t$, remembering that $0<a<b\iff 0<\frac{1}{b}<\frac{1}{a}$ and that here we had by our choice that $0<\frac{1}{t}<x<x+1$ – JMoravitz Aug 10 '20 at 00:02

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