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 $\mathcal{Y}=\mathbb{R}_{>0}$.