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When we consider some $x=${$x_n$} in the $l^\infty$ space, we know that we must have, for each x, sup $|x_n|$ < $\infty$. This implies that x is bounded. However, does this mean that there must exist some constant $C$ such that $|x_n|$ $\leq$ $C$ $\forall$ $x\in l^\infty$, $n\in \mathbb{N}$?

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    This would mean that, as a normed vector space, $\ell^\infty$ would be a bounded metric space, which is impossible. – 2017-02-23
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    @Aweygan I think I understand now, thank you! – 2017-02-23

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The answer is no. If such a constant $C$ would exist, then the constant sequence $x_n=C+1$ would not belong to $l^\infty$.

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    Does it mean, then, that there exists some $s_n$ such that $|x_n|$ $\leq$ $s_n$ for all $x\in X$, $n\in \mathbb{N}$? – 2017-02-23
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    No, and for the same reason. Take $x_n=s_n+1$. – 2017-02-23
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    @NickPeterson what exactly does it mean for $x\in l^\infty$ to be bounded then? – 2017-02-23
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    Exactly what it means for a sequence to be bounded. But don't confuse the fact that EACH element of $\ell^{\infty}$ is bounded with the claim that they are _uniformly_ bounded. – 2017-02-23
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    Each sequence in $l^\infty$ is bounded. This doesn't mean that all of the sequences are bounded by the same bound(and actually this only happens in $l^\infty(\{0\}$) Think about it like this... If I hand you a sequence from l^\infty, you should be able to find a bound for that one. That doesn't mean though that you can find a bound for all of them simultaneously, just each one had its own bound. – 2017-02-23
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    @NickPeterson But we can consider a sequence to simply be a function $f:\mathbb{N} -> \mathbb{R}$, no? Then, $f$ bounded simply means $f$ is pointwise bounded (for each n, we can find some function which is always greater than $f$). – 2017-02-23
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    Right. But the claims that you are making are not about a SINGLE sequence being bounded; they are about being able to find one thing that bounds EVERY sequence. Which you can't do in this space. – 2017-02-23
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    @SeanEnglish I'm sorry if my notation has been confusing - I think in my first reply I meant something other than what I typed. I'm interpreting any given $x\in l^\infty$ to be bounded by some function. Then, some other $y\in l^\infty$ is bounded by a function, albeit not necessarily the same function as the one bounding x. Is this correct? – 2017-02-23
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    My original interpretation stemmed from the definition my professor gave in class. That is, $l^\infty$ = {$x$={$x_n$}$_{n\in \mathbb{N}}$ : $x_n\in \mathbb{R} \forall n\in \mathbb{N}$. Thus, when we choose $n=1$, we're singling out an element $x_1$, where x={$x_n$}. – 2017-02-23
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    What you have typed is not the definition of $l^\infty$, but instead is just the set of real sequences. You need to add the condition that for each $x$, there exists some constant $C\in\mathbb{R}$ such that $|x_n|\leq C$ for all $n$. – 2017-02-23
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    @mizichael Be careful with the order of quantifiers. It is true that for each sequence there exists a $C$ such that the required condition happens. It is not true that there exists a constant such that for each sequence the condition happens. – 2017-02-23
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    @mizichael The difference between the two statements is that we can find a constant which depends on the sequence, but we cannot find a constant which works for all sequences at once. – 2017-02-23
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    @SeanEnglish Yes, that was part of the definition. Sorry - I didn't type it out because I just wanted to emphasize that the definition of $x$ and the role of $n$ was the cause of my confusion originally. – 2017-02-23