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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?

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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. –  t.b. Jul 10 '12 at 16:09
    
@t.b. Thanks for pointing that out! –  Matt N. Jul 10 '12 at 16:33
    
@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. –  Matt N. Aug 16 '12 at 9:11
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Sorry: I don't understand what you're trying to say in your last two sentences. Is there a question? –  t.b. Aug 16 '12 at 10:41
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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. –  t.b. Aug 16 '12 at 10:58
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2 Answers 2

up vote 8 down vote accepted

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 $V\times W$.

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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 $W$.

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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. –  Matt N. Jul 10 '12 at 16:31
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No problem!{}{} –  Davide Giraudo Jul 10 '12 at 18:13
    
Actually, does $w$ not have to be $0$? Or else, the argument stays the same if we replace $0$ with $w_n$, I think. –  Matt N. Jul 10 '12 at 19:23
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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 $0$. –  Davide Giraudo Jul 10 '12 at 20:28
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