Are all elements $x$ in $c_0$ such that $\|x\|_\infty < \infty$

Question

It seems to me this is true, but it's a bold statement, so I would like to verify. First some definitions:

Let $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. Let $c_0$ be the sequence space of sequences whose limit is $0$: $$ c_0 = \left\{x=(x_1 , x_2 , ...) : \lim_{j \to \infty}|x_j|=0, \,\, x_j \in \mathbb{K} \,\, \forall j \right\} $$ And consider the norm $$\|x\|_\infty = \sup_j |x_j|$$

My question is:

Given $x \in c_0$ must it be the case that $\|x\|_\infty < \infty$?

I want to say yes, it is true, since convergent sequences are bounded. But I cannot convince myself with absolute rigor, only with this "educated guess"

Answer 1

An exercise:

Assume $\sup \{x_i|i\in \mathbb{Z^+}\} \not =L$, real.

Then $(x_i)$ is not bounded above.

One can construct a subsequence $(x_{i_k})$ s.t. $(x_{i_k}) \rightarrow \infty$.

Contradiction, since every subsequence of a convergent sequence $(x_i)$ converges to the same limit $(0)$.

Answer 2

For $x=(x_1 , x_2 , ...) \in c_0$ we have $x_n \to 0$, hence $x$ is a bounded sequence.

(Each convergent sequence is bounded !)

Answer 3

Yes. $\lim_n |x_n|=0$ guarantees that $\exists N \in \mathbb{N}$ so that $|x_n|\leqslant 1 \forall n > N$. This guarentees that $$\lVert x \rVert \leqslant \max\{|x_1|,\ldots,|x_N|,1\}$$