Relation between the norms on X and Y and the induced product norm on X x Y 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. 
 A: 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.
A: 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.  
