Probability of infimum less than supremum of probability

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I'm reading Martin Hairer's Advanced Stochastic Analysis notes and have a question about the proof of Lemma 6.2 (statement on page 22; proof on page 23). Specifically, why does \begin{align*} \mathbb{P}\big(\inf_{k\leq N}\langle \eta_k,M\eta_k\rangle\leq 4\epsilon\big) \leq C\epsilon^{2-2n}\sup_{\lvert\eta\rvert=1}\mathbb{P}\big(\langle \eta,M\eta\rangle\leq 4\epsilon\big) \end{align*} hold? The relevant notation (which I think should suffice to answer my question):

  • $M$ is a symmetric postive semidefinite $n\times n$ matrix-valued random variable
  • $\mathbb{P}$ is the probability measure of the probability space on which $M$ is defined
  • $\epsilon\in (0,1]$ is fixed
  • $\{\eta_k\}_{1\leq k\leq N}$ is a subset of $S^{n-1}$ such that for every $\eta\in S^{n-1}$ there exists $k$ with $\lvert \eta_k-\eta\rvert\leq \epsilon^2$
  • In the above, one can (and does) choose $N$ such that $N\leq C\epsilon^{2-2n}$ for some constant $C>0$
  • $\langle \cdot,\cdot\rangle$ the usual inner product on $\mathbb{R}^n$
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This is a union bound: The event $\{\inf_{k\leq N}\langle \eta_k,M\eta_k\rangle\leq 4\epsilon\}$ can be written as $$ \bigcup_{k \leq N} \{\langle \eta_k,M\eta_k\rangle\leq 4\epsilon\}. $$ Therefore, \begin{align*} \mathbb{P}\big(\inf_{k\leq N}\langle \eta_k,M\eta_k\rangle\leq 4\epsilon\big) &= \mathbb{P}\big(\bigcup_{k \leq N} \{\langle \eta_k,M\eta_k\rangle\leq 4\epsilon\}\big) \\ &\leq \sum_{k \leq N} \mathbb{P}\big(\langle \eta_k,M\eta_k\rangle\leq 4\epsilon\big) \\ &\leq \sum_{k \leq N} \sup_{|\eta|=1} \mathbb{P}\big(\langle \eta,M\eta\rangle\leq 4\epsilon\big) \\ &\leq C\epsilon^{2-2n}\sup_{\lvert\eta\rvert=1}\mathbb{P}\big(\langle \eta,M\eta\rangle\leq 4\epsilon\big). \end{align*}