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Let $(X,||\cdot||_X)$, $(Y,||\cdot||_Y)$ be a pair of normed linear spaces, and $(X \times Y, ||\cdot||_{X \times Y})$ the induced product space and norm.

If $(x,y)$ is an element in $X \times Y$, is it true that $||(x,y)||_{X \times Y} \lt \delta$ implies that $||x||_{X \times Y} \lt \delta$, where $(||x||_{X\times Y}=||(x,0)||_{X\times Y}$)?

This has been true with every product norm that I have encountered, but I am not sure if it is true in general and/or if there is an easy way to show it.

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    @User1736: Or perhaps that the *projections* $X\times Y\to X$ and $X\times Y\to Y$ are continuous?2010-10-10

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No, this need not be true, unless I am missing some assumption implicit in your use of the phrase "the induced norm" (emphasis added). Here is an example where $X=Y=\mathbb{R}$. All norms on $X\times Y=\mathbb{R}^2$ are equivalent. The norm $\|(x,y)\|^2=2x^2-2xy+y^2=x^2+(x-y)^2$ does not satisfy your property because, for example $\|(x,x)\|=|x|$ while $\|(x,0)\|=\sqrt{2}|x|$. (This norm actually comes from an inner product on $\mathbb{R}^2$.)


The following was a result of misreading the question. I'll leave it here, for now at least, if for no other reason than leaving Arturo's correction comprehensible:

What is true is that the projection map $(x,y)\mapsto x$ is continuous. This implies as a special case the following (which turns out to actually be equivalent to continuity):

For all \epsilon>0, there is a \delta>0 such that for all $x\in X$ and $y\in Y$, $\|(x,y)\|<\delta$ implies $\|x\|<\epsilon$.

In fact, the projection is Lipschitz continuous, and $\delta$ can be taken to be $\epsilon$ divided by the Lipschitz constant. For the reason Qiaochu Yuan gave in a comment on Ross Millikan's answer, this is the best you could hope for.

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    I've now corrected it, I think.2010-10-10
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Your question doesn't show up well, but the wikipedia entry Normed vector space claims that if you have the product of a finite number of normed vector spaces, the sum of the norms in the individual spaces is a norm in the product.

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    In fact, you can 'combine' the two norms and $X$ and $Y$ using any norm on $\mathbb R^2$, for example lp-norms. For $p=1$, you obtain you original product norm.2010-12-03