Solution Set Inequality Continuous Function, Probability Theory I am trying to understand the Tchebychev's inequality, which turns into understanding the cardinality of solutions sets.
Here is what i've got so far. Tchebychev's inequality states that for $c > 0$:
$$
\mathbb{P}(|X - E[X]| > c) \leq \frac{Var(X)}{c^2}
$$
This comes from application of Markov inequality on $(X - E[X])^2 > c^2$:
$$
\mathbb{P}((X - E[X])^2 > c^2) \leq \frac{Var(X)}{c^2}
$$
and taking the square root:
$$
\sqrt{(X - E[X])^2} > \sqrt{c^2} \Rightarrow |X - E[X]| > c
$$
because $\sqrt(x^2) = |x|$ and $a < b \Rightarrow \sqrt{a} < \sqrt{b}$ for $a,b \in \mathbb{R}$. Now my understanding is that we require:
$$
\{|X - E[X]| > c\} \subseteq \{(X - E[X])^2 > c^2\}
$$ so that
$$
\mathbb{P}(\{|X - E[X]| > c\}) \leq \mathbb{P}(\{(X - E[X])^2 > c^2\}) \leq  \frac{Var(X)}{c^2}
$$
otherwise it could not be guaranteed that
$$
\mathbb{P}(\{|X - E[X]| > c\}) \leq  \frac{Var(X)}{c^2} 
$$
So my question really is why does:
$$
\{|X - E[X]| > c\} \subseteq \{(X - E[X])^2 > c^2\}
$$
hold?

If $f:\mathcal{Y}\subseteq \mathbb{R}\to\mathbb{R}$ is a strictly increasing function,
$$ 
y>c \Leftrightarrow f(y)>f(c).
$$
Thus, for a random variable $Y$,
$$
\{\omega: Y(\omega)>c\} =\{\omega: f(Y(\omega))>f(c)\}.
$$
In your case, $Y=|X-\mathsf{E}X|$ and $f(y)=y^2$ which is strictly increasing on $\mathcal{Y}=\mathbb{R}_{>0}$.
