11
$\begingroup$

Let $V,W$ be two normed vector spaces (over a field $K$). Then their product $V \times W$ with the norm $\|(x,y)\| = \|x\|_V + \|y\|_W$ is a normed space.

Using this norm it's easy to show that if $V,W$ are complete then so is $V \times W$. To see this, let the limit of the sequence $(x_n , y_n)$ be $(x,y) = (\lim x_n, \lim y_n)$. Then for $n$ large enough, both $\|x - x_n\|_V$ and $\|y - y_n\|_W$ are less than $\varepsilon / 2$ and hence $\|(x,y) - (x_n, y_n)\|< \varepsilon$.

The other direction does (probably) not hold. Can someone show me an example of a space $V \times W$ that is complete but either $V$ or $W$ (or both) are not?

$\endgroup$
7
  • 4
    $\begingroup$ The vector space structure is completely irrelevant here. If $(X,d_X)$ and $(Y,d_Y)$ are non-empty metric spaces, you can equip their product with the metric $d_{X \times Y}((x,y),(x',y')) = d_X(x,x') + d_Y(y,y')$ and exactly the same argument as Davide's shows that $X \times Y$ is complete if and only if $X$ and $Y$ are complete. $\endgroup$
    – t.b.
    Jul 10, 2012 at 16:09
  • $\begingroup$ @t.b. Thanks for pointing that out! $\endgroup$
    – Rudy the Reindeer
    Jul 10, 2012 at 16:33
  • $\begingroup$ @t.b. I had this very fundamental and very stupid confusion about keeping topological and linear properties apart. I'm not saying I'm cured but I finally get that if I have two normed linear spaces $X,Y$ I can define any $\ell^p$ norm on them and it will induce the same topology as the product topology (I showed that just now for $p=1$). I was confusing linear with topological properties because on finite dimensional spaces I know that all norms are equivalent. But the other way doesn't work in general. But in the case of a finite product and an $\ell^p$ norm it does. $\endgroup$
    – Rudy the Reindeer
    Aug 16, 2012 at 9:11
  • 1
    $\begingroup$ Sorry: I don't understand what you're trying to say in your last two sentences. Is there a question? $\endgroup$
    – t.b.
    Aug 16, 2012 at 10:41
  • 1
    $\begingroup$ Just use that if you have $n$ factors then $\max_i\{\lVert x_i \rVert_{X_i}\} \leq \sum_{i} \lVert x_i \rVert_{X_i} \leq n \cdot \max_i\{\lVert x_i \rVert_{X_i}\}$ and the maximum gives the product topology. If you have infinitely many factors matters are different. $\endgroup$
    – t.b.
    Aug 16, 2012 at 10:58

2 Answers 2

Reset to default
13
$\begingroup$

If $V$ is not complete and $V\times W$ complete, take $\{v_n\}$ a Cauchy sequence which doesn't converge in $V$. Then $(v_n,0)$ is a Cauchy sequence in $V\times W$ and converges to $(v,w)\in V\times W$. We have $$\lVert (v_n,0)-(v,w)\rVert_{V\times W}=\lVert v_n-v\rVert+\lVert w\rVert\to 0$$ hence $v_n\to v$ in $V$, a contradiction.

Hence $V\times W$ is complete if and only if so are $V$ and $W$.

$\endgroup$
4
  • 2
    $\begingroup$ I hope you don't mind but I will accept the other answer because both are equally deserving but you have two more votes and I'd like both of you to get more or less the same amount of points. $\endgroup$
    – Rudy the Reindeer
    Jul 10, 2012 at 16:31
  • 1
    $\begingroup$ No problem!{}{} $\endgroup$
    – Davide Giraudo
    Jul 10, 2012 at 18:13
  • $\begingroup$ Actually, does $w$ not have to be $0$? Or else, the argument stays the same if we replace $0$ with $w_n$, I think. $\endgroup$
    – Rudy the Reindeer
    Jul 10, 2012 at 19:23
  • 1
    $\begingroup$ It's necessarily $0$ because $\lVert w\rVert\leq \lVert v_n-v\rVert+\lVert w\rVert$ for all integer $n$ and the RHS converges to $0$. $\endgroup$
    – Davide Giraudo
    Jul 10, 2012 at 20:28
11
$\begingroup$

I believe that the spaces $V$ and $W$ must be complete whenever $V\times W$ is complete.

Closed subspace of a complete normed space is complete.

The space $V$ is isometrically isomorphic to the closed subspace $V\times\{0\}$ of $V\times W$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .