Let $\sigma$ denote the sigmoid function $\sigma(x) = \frac{1}{1+e^{-x}}$, let $x,y \in \mathbb{R}$. Given the following two conditions: $|\sigma(-x) - \sigma(y)| < \epsilon$ and $x - y > c > 0,$ where $\epsilon$ can be regarded as a small positive number and $c$ as a large positive number.

Revised question: can we show that $x > \alpha c + f(\epsilon)$ and $y < - \alpha c + g(\epsilon)$, where $\alpha$ is some positive constant, $f$ and $g$ are some functions.

Original question (which has been answered by losif Pinelis): Can we draw the conclusion that $x > \frac{c}{2} - f(\epsilon), y <-\frac{c}{2} + f(\epsilon)$, where $f(\epsilon)$ is some function of $\epsilon$.

Let $\sigma$ denote the sigmoid function $\sigma(x) = \frac{1}{1+e^{-x}}$, let $x,y \in \mathbb{R}$. Given the following two conditions: $|\sigma(-x) - \sigma(y)| < \epsilon$ and $x - y > c > 0,$ where $\epsilon$ can be regarded as a small positive number and $c$ as a large positive number.

Revised question: can we show that $x > \alpha c + f(\epsilon)$ and $y < - \alpha c + g(\epsilon)$, where $\alpha$ is some positive constant, $f$ and $g$ are some functions. (Intuitively I want to show $x$ is bounded above from 0 and $y$ is bounded below from 0)

Original question (which has been answered by losif Pinelis): Can we draw the conclusion that $x > \frac{c}{2} - f(\epsilon), y <-\frac{c}{2} + f(\epsilon)$, where $f(\epsilon)$ is some function of $\epsilon$.

Let $\sigma$ denote the sigmoid function $\sigma(x) = \frac{1}{1+e^{-x}}$, let $x,y \in \mathbb{R}$. Given the following two conditions: $|\sigma(-x) - \sigma(y)| < \epsilon$ and $x - y > c > 0,$ where $\epsilon$ can be regarded as a small positive number and $c$ as a large positive number. Can we draw the conclusion that $x > \frac{c}{2} - f(\epsilon), y <-\frac{c}{2} + f(\epsilon)$, where $f(\epsilon)$ is some function of $\epsilon$.

Let $\sigma$ denote the sigmoid function $\sigma(x) = \frac{1}{1+e^{-x}}$, let $x,y \in \mathbb{R}$. Given the following two conditions: $|\sigma(-x) - \sigma(y)| < \epsilon$ and $x - y > c > 0,$ where $\epsilon$ can be regarded as a small positive number and $c$ as a large positive number.

Revised question: can we show that $x > \alpha c + f(\epsilon)$ and $y < - \alpha c + g(\epsilon)$, where $\alpha$ is some positive constant, $f$ and $g$ are some functions.

Original question (which has been answered by losif Pinelis): Can we draw the conclusion that $x > \frac{c}{2} - f(\epsilon), y <-\frac{c}{2} + f(\epsilon)$, where $f(\epsilon)$ is some function of $\epsilon$.