Are bounded sequences always strictly less than some fixed number $M$?

Definition: A sequence $(x_n)$ is bounded if there exists a number $M \gt 0$ such that $\left| x_{n} \right| \leq M$ for all $n \in \mathbb{N}$.

Question: Is there any reason why one can't replace $\left| x_{n} \right| \leq M$ with the strict inequality $\left| x_{n} \right| < M$? Since $M$ is a fixed number it seems that $1$ could always be added to make the inequality strict, but maybe I am missing something... If I am right, and the inequality can always be strict, then why didn't they write it that way in the first place?

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You want to allow yourself the benefit of a "maximal element" as the bound, which will result a weak inequality. – Asaf Karagila May 14 '11 at 15:44
The answer to the question in the title is: yes.The answer to the first question in the post is: no. The answer to the second question in the post is: they could have. – Did May 14 '11 at 15:47

You are absolutely right, one could also write "there exists $M$ such that $|x_n| < M$ for all $n$"; the two statements are equivalent. But if they had written it that way, you'd be asking why they didn't write "$x_n \le M$"...
There's a slight advantage to writing $\le$: if the sequence is bounded, then there is always a smallest $M$ such that $|x_n| \le M$ for all $n$. (Namely, $M = \sup_n |x_n|$.) But there may not be a smallest $M$ such that $|x_n| < M$ for all $n$. (Consider the constant sequence $x_n = 0$...)