2
$\begingroup$

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 .

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ Mar 24 '15 at 16:24
4
$\begingroup$

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 $x_k \to My$.

$\endgroup$
2
$\begingroup$

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.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.