Let $ c_0 = \{ x = (x_n)_{n \in \mathbb N} \in l^\infty : \lim_{n \to \infty} x_n = 0\}$. Show that $c_0$ is a Banach space with the norm $\rVert \cdot \lVert_\infty$

I am capable of showing the space where the limit of $x_n$ exists is normed linear space but am having trouble with showing that the limit of Cauchy sequences must converge to 0.

Let $(x^{(n)})_{n \in \mathbb N}$ be a Cauchy sequence in $c_0$ such that $x^{n} = (x^n_1, x^n_2,...)$. Fix $k \in \mathbb N$ consider the sequence $(x^n_k)_{n \in \mathbb N}$ in $\mathbb F$. For any $n,m \in \mathbb N$

$\lvert x^n_k - x^m_k \rvert \le \sup_{k \in \mathbb N} \lvert x^n_k - x^m_k \rvert = \lVert x^n - x^m \rVert_\infty \lt \epsilon $ (1)

Thus $x^n_k$ is Cauchy in $\mathbb F$ and so has limit $y_k$ such that $y = (y_1,y_2,...)$ and y is the limit of $x^n$

To show that such a y exists we look at the value of $\lvert y_n - y_m \rvert \le \lvert y_n - x^N_n \rvert + \lvert x^N_n - x^N_m \rvert + \lvert x^N_m - y_m \rvert \lt \epsilon$ for all $n,m \ge N$ (2)

The middle expression on RHS of (2) is $\lt \epsilon/3$ by (1)

The other two are also $\lt \epsilon/3$ follow from $x^N_k$ being Cauchy and converging to $y_k$

This shows that $\lim_{n \to \infty} y_n$ exists but we still have not shown that $y \in c_0$.

I know that to show y tends to 0 i should show that $\lvert y_k \rvert \lt \epsilon$ for $k \ge N$

This is where I am stuck. Perhaps $\lvert y_k \rvert = \lvert \lim_{n \to \infty} x^n_k \rvert$ and then we can take the limit function outside the absolute value sign by continuity? Then we might say due to it being a Cauchy sequence $x^n_k \lt \epsilon$. I know this last bit isn't at all convincing so I could do with some help.

  • 1
    $\begingroup$ First show $(x^{(n)})$ converges to the pointwise limit $y$ in $\ell_\infty$. Given $\epsilon>0$, choose $N$ so that $\Vert x^{(N)}-y\Vert<\epsilon/2$. Then choose $M$ so that $|x^{(N)}(i)|<\epsilon/2$ for $i>M$. Then $|y(i)|<\epsilon$ for $i>M$ by the triangle inequality. $\endgroup$ May 10, 2015 at 16:48

2 Answers 2


Suppose $x^k \in c_0$ and $x^k \to x$. Let $\epsilon>0$ and pick $N$ such that $\|x^k-x\|_\infty < {1 \over 2 } \epsilon$ for $k \ge N$. Since $x^N \in c_0$, there is some $N'$ such that $|x_i^N| < {1 \over 2 } \epsilon$ for $i \ge N'$. Then $|x_i| \le |x_i^N| +|x_i-x_i^N| \le |x_i^N| +\|x-x^N\|_\infty <\epsilon$. Hence $x_i \to 0$ and so $x \in c_0$.

Hence $c_0$ is a closed subspace of $l_\infty$.

It follows that $c_0$ is Banach since $l_\infty$ is Banach (any Cauchy sequence in $c_0$ is Cauchy in $l_\infty$ hence converges to some point an closedness shows that this point lies in $c_0$).

  • $\begingroup$ Theres one thing I don't understand about this answer, namely, to show the space is complete, don't we need to show every Cauchy sequence in the space converges to a limit point within the space? Why is it enough to start with taking a convergent sequence in the first place. $\endgroup$
    – kam
    Apr 3, 2020 at 15:07
  • $\begingroup$ Clearly convergence implies Cauchy, but the implication the other way round is not always true. $\endgroup$
    – kam
    Apr 3, 2020 at 15:07
  • $\begingroup$ A closed subspace of a complete space is complete. The hard work was done in the other space. Read the last sentence of the answer. $\endgroup$
    – copper.hat
    Apr 3, 2020 at 15:58
  • $\begingroup$ So you began with a limit point x which is in $l^\infty$ , and showed that it is in $c_0$ right? copper.hat $\endgroup$
    – user652838
    Dec 10, 2020 at 13:51
  • 1
    $\begingroup$ @user726608 Yes. $\endgroup$
    – copper.hat
    Dec 10, 2020 at 16:17

We note that $$ |y_k| \leq |y_k - x^n_k| + |x^n_k| = \lim_{m→∞}|x^m_k - x^n_k| + |x^n_k| \leq\lim_{m→∞} ‖x^m - x^n‖ + |x^n_k|$$ which holds for arbitrary $n$. (the superscript is which sequence, the subscript is the index of the sequence) As $x^n∈ c_0$, choose $K_n$ large such that $|x^n_k |<ε$ for $k>K_n$.

Pick $N(ε)$ large such that $\|x^m - x^n||<ε $ for $n,m>N$. Taking limits($\star$) gives $$ \lim_{m→∞}‖x^m - x^n‖ \leq ε \quad ∀ n \geq N$$

So putting the two bounds together: With $ε>0$ given, there is $K = K_{N(ε)}$ such that $$ |y_k| \leq \lim_{m→∞} ‖x^m - x^{N(ε)}‖ + |x^{N(ε)}_k| \leq 2ε $$ for all $k > K$.

The step marked $(\star)$ is easy to see if you notationally suppress the $n$: $$ a_m < ε ∀ m > N \implies \lim a_m \leq ε $$ as (you have shown) this limit will exist.


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