State for the set $S = \left\{ \frac{1}{1 + e^{-x}} \mid x \in \ \mathbb{R} \right\}$ if it has a max or min.

Question

State for the set $$S = \left\{ \frac{1}{1 + e^{-x}} \mid x \in \ \mathbb{R} \right\}$$ its Sup and Inf and if it has a max or min, prove your answer.

Issue: I know the properties of this set, but I am explicitly trying to work on my rigour and preciseness when writing proofs. This being the case I have the niggling feeling that the following proof isn't sufficient or even correct. If I could get feedback on the attempt I would appreciate the help.

Attempted Solution:

$Sup(S) = 1$

$Inf(S) = 0$

Claim: The set does not contain a max or min.

Proof(going to do the minimal point): Suppose $S$ had a minimal point this would mean that there exists a $x_1 \in \mathbb{R}$ and $x_1 < 0$ such that $$\frac{1}{1 + e^{-x_{1}}} \leq s $$ for all $s \in S$.

but since $x_1 \in \mathbb{R}$ and $x_1 < 0$ this means $x_{1} - 1 < x_{1}$

This means $$\frac{1}{1 + e^{-x_{1} + 1}} < \frac{1}{1 + e^{-x_{1}}}$$

But we assumed $$\frac{1}{1 + e^{-x_{1}}}$$ was the minimal element. This is a contradiction, hence $S$ has no minimal element.

Comment: If this idea didn't work I had a similar approach where I would instead use the fact that "we know the infimum of the set is 0 and I would compare my element to the infimum in the the form as: $$\frac{1}{1 + e^{-x_{1}}} \leq \inf(S) = 0$$

And draw a conclusion from that. But still not sure if that is the right approach.

Answer 1

Option:

$y:=e^{-x}$, $x$ real.

Then $y \in (0,\infty)$.

$S=$ { $\dfrac{1}{1+y}| y >0$ };

$0 < \dfrac{1}{1+y} <1.$

Lower bound: $0$; In fact $\inf (S) =0;$

Upper bound $1$; In fact $\sup (S)=1;$

$f(y):=\dfrac{1}{1+y}$, $y>0$, is a strictly decreasing function.

Hence no local min or max.

P.S. Not so difficult to prove that $\inf (S)=0$, and $\sup (S)=1$

Answer 2

It's easy to show that $0< \frac{1}{1+e^{-x}}<1$.

Now, $\lim_{x\to\infty}\frac{1}{1+e^{-x}}=1$ and $\lim_{x\to-\infty}\frac{1}{1+e^{-x}}=0$, this shows that in fact $1=\sup\{ \frac{1}{1+e^{-x}}\}$ and $0=\inf\{ \frac{1}{1+e^{-x}}\}$.

And these infimum and supremum are not attained by the previous mentioned, $0< \frac{1}{1+e^{-x}}<1$. So, there is not min or max.

Answer 3

Obviously, $0< \frac{1}{1+e^{-x}}<1$. Now for any $0<a<1$ the value $a$ is attained by $\frac{1}{1+e^{-x}}$ iff $x=-\ln\bigl(\frac1a-1\bigr)$.