# normed space is complete if and only if the closed unit ball is complete

Show that a normed space is complete if and only if the closed unit ball is complete. I know theorem about this: If a normed space X has the property that the unit closed ball is compact, then X is finite dimensional space. Is there any relation between the theorems? Thank you for your helping .

• A closed subset of a complete metric space is complete. For the other direction, take a Cauchy sequence and shift the values by some vector so that they all lie eventually in the closed unit ball. Mar 24 '15 at 16:24

A closed subset of a complete space is complete. So if $X$ is complete, its unit ball is too.

On the other hand, if the unit ball is complete and $\{x_k\}$ is a Cauchy sequence in $X$, there must exist a constant $M$ such that $\|x_k\| \le M$ for all $k$: Cauchy sequences are bounded. Let $y_k = \frac{x_k}{M}$ so that $\{y_k\}$ is Cauchy and $\|y_k\| \le 1$ for all $k$. Thus $y_k \to y$ in the unit ball, so that $x_k \to My$.

Hints:

"$\Rightarrow$:"

1. Let $(x_k)_{k \in \mathbb{N}}$ be a Cauchy sequence in the closed unit ball $B[0,1]$. Show that there exists $x \in X$ such that $x_k \to x$.
2. Conclude from $\|x_k\| \leq 1$ that $\|x\| \leq 1$.

"$\Leftarrow$:"

1. Let $(x_k)_{k \in \mathbb{N}}$ be a Cauchy sequence in $X$. Show that there exists a constant $M>0$ such that $$\|x_k\| \leq M \qquad \text{for all} \, k \in \mathbb{N}.$$
2. Show that the sequence $\tilde{x}_k := x_k/M$ is a Cauchy sequence in $B[0,1]$.
3. Conclude.