Prove a sequence in $\ell^\infty$ is bounded but not Cauchy

Question

$e_n$ in $\ell^\infty$ is the sequence whose $n$th entry is 1, and all others are 0. Show that ${e_n}$ from $n=1$ to infinity is bounded but not Cauchy.

I'm really not certain of the whole concept of $\ell^\infty$. Anyone know what I do here??

Answer 1

$l^{\infty}$ is the vector space of bounded sequences over some scalar field, usually $\mathbb R$ or $\mathbb C$. Vectors are added as componentwise addition of sequences, the scalar multiplication is componentwise scalar multiplication.

The norm of a vector is the supremum of the absolute value of each entry in the sequence.

The distance between two vectors is just the norm of the difference. So, if you look at $e_n-e_m$ for $n\ne m$, you'll have a 1 in the n'th entry of the sequence, a -1 in the m'th entry, and a 0 everywhere else...so the norm of $e_n-e_m$ is $1$ for all $n\ne m$. Thus it can't be cauchy, because you can never get the terms closer than 1 to each other. It's clearly bounded, as each term has norm 1

Answer 2

The sequence $(e_n)_{n=1}^\infty$ is a sequence of elements in $\ell^\infty$, where each element is itself a sequence of real (or possibly complex) numbers. The sequence $e_n$ of real numbers is defined by $$ e_n(k) = \begin{cases} 1 & k=n\\ 0 & \text{otherwise} \end{cases} \quad \text{ for any } k \in \Bbb N $$
This can be confusing at first, but you get used to it.

Given a sequence $(x_n) \subset \ell^\infty$, we say that the sequence is bounded if there is a constant $M > 0$ such that every $x_n$ satisfies $\|e_n\|_\infty \leq M$. Try to find such an $M$ for our sequence $(e_n)$.

Similarly, I'm sure you can find the definition of "Cauchy". In order to show that this sequence is not Cauchy, verify that any two distinct elements are of distance $1$ from each other, which is to say that $\|e_m - e_n\|_\infty = 1$ for $m \neq n$.