I am having difficulties understanding an argument related to the consistency of an estimator. The argument is:
$$ \{A \leq \frac{\epsilon}{2}\} \cap \{B \leq \frac{\epsilon}{2}\} \subseteq \{A + B \leq \epsilon\} $$
where no further information about $A, B$ is provided, however the context is that if $\hat{\theta}_{n,i} = (\hat{\theta}_{n,1},...,\hat{\theta}_{n,p})$ is a consistent estimator with $p$ components for $\theta_{n,i} = (\theta_{n,1},...,\theta_{n,p})$ : $$ \mathbb{P}(||\hat{\theta}_{n,i} - \theta_{n,i}||_2 > \epsilon) \leq \sum_{i=1}^p \mathbb{P}(|\hat{\theta}_{n,i} - \theta_{n,i}| > \epsilon/\sqrt{p}) $$
Now:
$$ \mathbb{P}(||\hat{\theta}_{n,i} - \theta_{n,i}||_2 > \epsilon) = \mathbb{P}( \sqrt{\sum_{i=1}^p(\hat{\theta}_{n,i} - \theta_{n,i})^2} > \epsilon) $$
and: $$ \mathbb{P}( \sum_{i=1}^p(\hat{\theta}_{n,i} - \theta_{n,i})^2 > \epsilon^2) \leq \sum_{i=1}^p\mathbb{P}( (\hat{\theta}_{n,i} - \theta_{n,i})^2 > \frac{\epsilon^2}{p}) = \sum_{i=1}^p\mathbb{P}( |(\hat{\theta}_{n,i} - \theta_{n,i})| > \frac{\epsilon}{\sqrt{p}}) $$
where the first argument for p values was uesed: $$ \cap^{i=1,...,p}\{A_i \leq \frac{\epsilon}{p}\} \subseteq \{\sum_{i=1}^p A_i \leq \epsilon\} $$
Let $\hat{\theta}_{n,i} - \theta_{n,i} =: Z_i$, then why is:
$$ \mathbb{P}( \sum_{i=1}^p Z_i^2 > \epsilon^2) \leq \sum_{i=1}^p\mathbb{P}( Z_i^2 > \frac{\epsilon^2}{p}) = \sum_{i=1}^p\mathbb{P}( |Z_i| > \frac{\epsilon}{\sqrt{p}}) $$
Edit:
It is direct that:$$\omega\in\left\{A\leq\frac12\epsilon\right\}\cap\left\{B\leq\frac12\epsilon\right\}\iff A(\omega)\leq\frac{\epsilon}2\text{ and }B(\omega)\leq\frac{\epsilon}2$$$$\implies A(\omega)+B(\omega)\leq\epsilon\iff (A+B)(\omega)\leq\epsilon\iff\omega\in\{A+B\leq\epsilon\}$$
Taking complements we also have:$$\{A+B>\epsilon\}\subseteq\left\{A>\frac12\epsilon\right\}\cup\left\{B>\frac12\epsilon\right\}$$
or more generally: $$\left\{\sum_{i=1}^p A_i>\epsilon\right\}\subseteq\bigcup_{i=1}^p\left\{A_i>\frac1p\epsilon\right\}$$implying:$$\Pr\left(\left\{\sum_{i=1}^p A_i>\epsilon\right\}\right)\leq\sum_{i=1}^p\Pr\left(\left\{A_i>\frac1p\epsilon\right\}\right)$$ does that help?