I went to a functional analysis course this year and my lecturer wrote down this Theorem. Lots of students pointed out it is incorrect, but she insisted it was. I am stating it now and hope someone can give me a good explanation about the confusion.

Here is the statement of the Theorem:

For any normed vector space $(V,\|\cdot\|_V)$, there exists another normed space $(Z,\|\cdot\|_Z)$ such that a map $i:V\rightarrow Z$ such that

(1) $i$ is an isometrical isomorphism

(2) $i(V)$ is dense in $Z$

where $Z$ is unique up to isomorphism.

In this Theorem, nothing said about $Z$ being complete. The students asked if $V$ is $\mathbb{Q}$, what $Z$ should be? $\mathbb{Q}$ fits the definition for $Z$, because $\mathbb{Q}$ is dense in $\mathbb{Q}$, but the lecturer said $\mathbb{Q}$ is not dense. She said something on the line dense means something different in a normed space. Can someone explain to me what this is about? In the Theorem, do we have to have say $Z$ is complete?

  • $\begingroup$ She forgot to say that Z is complete, else Z could be V. $\endgroup$ – DanielWainfleet Feb 21 '16 at 20:07

She was wrong. It’s clear that if $V=\Bbb Q$, then both $Z=\Bbb Q$ and $Z=\Bbb R$ satisfy the stated conclusion. You get uniqueness by requiring $Z$ to be complete. Indeed, that’s the whole point of the theorem: if $V$ is any normed linear space, it has a unique completion (which is a Banach space).

There are actually two important notions of dense subset of $\Bbb R$, but they coincide. One may say that a subset $D$ of $\Bbb R$ is dense with respect to the order $\le$ on $\Bbb R$ if for all $x,y\in\Bbb R$ with $x<y$ there is a $d\in D$ such that $x<d<y$; that’s the order-theoretic notion of dense subset. One may also say that $D$ is dense in $\Bbb R$ if $\operatorname{cl}D=\Bbb R$, where the closure is taken in the usual Euclidean topology of $\Bbb R$; that’s the topological notion of dense subset. But as I said, they coincide, and by either notion $\Bbb Q$ is dense both in itself and in $\Bbb R$.

Added: One normally deals with linear spaces over the complete fields $\Bbb R$ and $\Bbb C$, and I strongly suspect that the theorem was intended to be limited to such spaces. The conclusion should still have specified that $Z$ is complete, however, as uniqueness fails trivially without it: just take any infinite-dimensional normed linear space $V$ that is not complete, and note that her version of the theorem is satisfied both by $V$ and by its completion, which are distinct.

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    $\begingroup$ $\mathbb{Q}$ is not vector space neither over $\mathbb{R}$ nor $\mathbb{C}$. $\endgroup$ – Norbert Oct 20 '12 at 11:37
  • $\begingroup$ @Norbert: True, but I was addressing the teacher’s response. The students’ example was poorly chosen, and I intend to add to my answer to address that issue as well. $\endgroup$ – Brian M. Scott Oct 20 '12 at 11:40
  • $\begingroup$ $\mathbb{Q}$ is a vector space over the field $\mathbb{Q}$? $\endgroup$ – Lost1 Oct 20 '12 at 11:46
  • $\begingroup$ @Yufan: Yes, it is. And so is $\Bbb R$. $\endgroup$ – Brian M. Scott Oct 20 '12 at 11:47
  • $\begingroup$ @Yufan This strange to develop functional analysis over $\mathbb{Q}$, I think you are using fields $\mathbb{R}$ and $\mathbb{C}$ in your course. $\endgroup$ – Norbert Oct 20 '12 at 11:49

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