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Suppose $(Y,\Vert\cdot\Vert)$ is a complete normed linear space. If the vector space $X\supset Y$ with the same norm $\Vert\cdot\Vert$ is a normed linear space, then is $(X,\Vert\cdot\Vert)$ necessarily complete?

My guess is no. However, I am not aware of any examples.

Side interest: If X and Y are Banach (with possibly different norms), I want to make $X \times Y$ Banach. But I realize that in order to do this, we cannot use the same norm as we did for $X$ and $Y$ because it's not like $X \subseteq X \times Y$ or $Y \subseteq X \times Y$. What norm (if there is one) on $X \times Y$ will garuntee us a Banach space?

I'm sure these questions are standard ones in functional analysis. I just haven't come across them in my module. Thanks in advance.

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    Yes you are right as well.2012-05-28

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You have to look at infinite dimensional Banach spaces. For example, $X=\ell^2$, vector space of square-summable sequence of real numbers. Let $Y:=\{(x_n)_n, \exists k\in\Bbb N, x_n=0\mbox{ if }n\geq k\}$. It's a vector subspace of $X$, but not complete since it's not closed (it's in fact a strict and dense subset).

However, for two Banach spaces $X$ and $Y$, you can put norms on $X\times Y$ such that this space is a Banach space. For example, if $N$ is a norm on $\Bbb R^2$, define $\lVert(x,y)\rVert:=N(\lVert x\rVert_X,\lVert y\rVert_Y)$.

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    I think X and Y have to be over the same field for this to work.2012-05-28