If I want to show that $\ell^\infty$ (the set of all bounded sequences of real numbers) is a vector space does it suffice to show that since $\ell^\infty \subset \mathbb{R}_\infty$ (the set of all sequences of real numbers) and $\mathbb{R}_\infty$ is a vector space that the following are true:

(a.) $0\in \ell^\infty$ (this is trivial because the 0 sequence is bounded and so is in $\ell^\infty$)

(b.) For $u,v \in \ell^\infty$ we have that $u+v \in \ell^\infty$.

We can show this from the definition of bounded sequences, $u = \left\{x_1,x_2,\dots\right\}$ is bounded if $\exists M>0$ such that $|x_n|<M$ for all $N\in\mathbb{N}$. Similarly for $v = \left\{y_1,y_2,\dots\right\}$ bounded by $T>0$. Therefore:

$$|(u+v)_n|=|x_n +y_n| \leq |x_n| + |y_n| < M+T$$

Therefore for any two bounded sequences $u,v$ the sum $u+v$ is also bounded and so $\ell^\infty$ is closed under addition.

(c.) A similar proof shows that $c\cdot u \in \ell^\infty$ for all $c \in \mathbb{R}$

This shows that $\ell^\infty$ satisfies all the necessary conditions to be a subspace of $\mathbb{R}_\infty$. Therefore I think I have shown $\ell^\infty$ is a vector space, but then the problem has me show that $\mathbb{R}^\infty$ (the set of sequences that are eventually zero) is a subspace of $\ell^\infty$ and so I think I may have messed up.

  • $\begingroup$ Your proof looks essentially correct to me. I think you may mean $|x_n + y_n| \le |x_n| + |y_n|$ in the body of your proof. $\endgroup$
    – eepperly16
    Dec 6, 2016 at 3:23
  • $\begingroup$ Thank you, I copied the Latex from my homework so thanks a ton for pointing it out to me. $\endgroup$
    – Ben Ray
    Dec 6, 2016 at 3:24

1 Answer 1


Your method is perfectly correct. The same criteria can be used to show that $\mathbb R^\infty$ is a subspace of $\ell^\infty$.

  • $\begingroup$ Awesome. I took a walk and considered the proof for $\mathbb{R}^\infty$, since it's a finite sequence there exists a natural number $N \in \mathbb{N}$ such that for $n > N$ we have that $x_n = 0$ therefore we can take the upper bound to be $\max x_n$ and go about the same procedure to prove it is a subspace of $\ell^\infty$. Does this seem right (with more attention to detail added when I write the proof, of course) $\endgroup$
    – Ben Ray
    Dec 6, 2016 at 3:28

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