# Definition of a bounded sequence

My professor gave the following definition: A sequence $\{x_n \}$ is said to be bounded if $\exists M > 0$ such that $|x_n| \le M$ for all $n \in \mathbb N^+.$

But then what about the sequence $(0, 0, ...)$? In that case, can't $M$ be $0$?

Wikipedia provides the following definition, which seems more reasonable to me:
A sequence $\{x_n \}$ is said to be bounded if $\exists M \in \mathbb R$ such that $|x_n| \le M$ for all $n \in \mathbb N^+.$

Is my professor's definition inaccurate or imprecise in any way?

• The important is the existence of $M$, not the "smallest" $M$. Feb 21, 2015 at 11:11
• The condition $M>0$ in the definition is just as entirely superfluous as it is innocuous. You might as well ask $M>-3$ or $M>e^{100!}$. The point is that $M$ can be as large as you like; how small it can be is not the point. For $\epsilon$ it is usually the opposite, but it really depends on how the variable is used. Feb 21, 2015 at 13:37
• To add on to Marc's comment: the critical constraint is that $M < \infty$, which is implied by $M \in \mathbb{R}$. Feb 22, 2015 at 0:01
• Why wiki says $M \in \mathbb{R}$? Is it not evident that $M \geq 0$, for if $M<0$, the inequality $|x_n|\leq M$ makes no sense right. Jun 9, 2021 at 1:01

The definition of your teacher is right. And the one from the Wikipedia is right, too. They are equivalent.

It is true that for the sequence $(0,0,\ldots)$ we have $|x_n|\le 0$ for every $n\in \Bbb N$, but this does not contradict your teacher's definition, since it says that a sequence is bounded if there exists some $M>0$ such that $|x_n|<M$.

In other words, your teacher's definition does not say that a sequence is bounded if every bound is positive, but if it has a positive bound. The sequence $(0,0,\ldots)$ has indeed a positive bound: $1$, for example (in fact, every positive real number is a bound for this sequence!)

No,it's fine. If you have the zero sequence $\{a_n\}$ then for every $M>0$ you have $a_n\leq M$. We define $M>0$ so we can use it sometimes to a fraction like ,let $\epsilon =\frac {1}{M}$. etc...

• To have a more concrete bound: The zero sequence is bounded by $M=1$. Feb 21, 2015 at 11:12

Both your Prof. and wikipedia is right infact your Prof. is more precisely telling you where that $M$ lies.

The only difference between both the definitions is that your Prof. take $M\gt 0$ and wikipedia is taking $M\in \mathbb R$ but don't you think wikipedia definition also indirectly says that $M\ge 0$ because $|x_n| \ge0 ,$thus it can't be made smaller than a negative number. And $M=0 \Rightarrow |x_n| \le 0 \Rightarrow x_n=0$ and if sequence is constant zero then any $M\gt 0$ will also work.