Let $V$ be a vector space over $\mathbb{R}$.

Let $M \subset V$ (as a set), and let $d:M \times M \rightarrow \mathbb{R}$ be such that $(M,d)$ is a complete metric space.

Let $W$ be a vector subspace of $V$, and assume that $W \subset M$. Moreover, let $|| \cdot ||: W \rightarrow \mathbb{R}$ be a norm on $W$, and assume that it coincides with $d$; i.e. for any $u,v \in W$, we have $d(u,v) = ||u-v||$.

So, we have a normed vector space $W$ residing in a complete metric space $M$ contained in a vector space $V$.

Let $\overline{W}$ denote the closure of $W$ in $M$. My question is the following: is $\overline{W}$ closed under addition (where addition takes place in $V$)?

My thought process is that any element $x$ of $\overline{W}$ can be described as the limit of a Cauchy sequence $(x_n)_n$ of elements in $W$. Then if we have $x,y \in \overline{W}$ and $(x_n)_n, (y_n)_n \subset W$ with $x_n \rightarrow x$ and $y_n \rightarrow y$, then the sequence $(x_n + y_n)_n$ is Cauchy. By completeness of $M$, it converges to some element $s$. However, I don't know how to show $s = x+y$, or if this is even true.

I suspect, in fact, that the result is not true. If so, I would like to see a counter-example, and possibly extra restrictions that could make it true.

EDIT: As an example, consider $V$ to be the vector space of functions $\{f:\mathbb{R} \rightarrow \mathbb{R}\}$, $M = L^1(\mathbb{R})$ with the metric induced by the norm, and $W$ be the set of step functions with finite support, with the $L^1$ norm.

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    $\begingroup$ @J.R. I am asking that on $W$, the metric be equal to the one induced by the norm. I think that is a way of making the metric compatible with the vector space structure. $\endgroup$
    – Sambo
    May 25 '18 at 21:31
  • $\begingroup$ Could it be proved such a $\;W\;$ is closed or, at least, complete? $\endgroup$
    – DonAntonio
    May 25 '18 at 21:35
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    $\begingroup$ My thought is that you would show $\text{lim}_{n\rightarrow \infty} x_n+y_n=x+y$ by looking at the limit of $||x+y-(x_n+y_n)||\leq ||x-x_n|| + ||y-y_n||$, where the the right hand side goes to zero by convergence of $x_n$ and $y_n$. Maybe this misses some subtlety which you have thought of already? $\endgroup$
    – mcwiggler
    May 25 '18 at 22:23
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    $\begingroup$ @mcwiggler Sadly, it does. There was actually a (now-deleted) answer with the same mistake: you can't have $x+y$ inside the norm, since you don't know that it belongs to $M$. $\endgroup$
    – Sambo
    May 25 '18 at 22:25
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    $\begingroup$ @Sambo Of course, very obvious once it's mentioned! (: I will keep puzzling... $\endgroup$
    – mcwiggler
    May 25 '18 at 22:27

The assumptions don't seem to require any relation between the metric on $M\setminus W$ with the vector space structure on $V$, so it would seem we can fabricate a counterexample by brute force:

Let $W$ be your favorite non-complete normed vector space -- such as, for concreteness, the space of real sequences with finite support under the sup norm.

Let $M$ be the metric completion of $W$.

Let $V$ be the direct sum of $W$ and the space of finite formal linear combinations of elements of $M\setminus W$. ($M$ embeds into this in the obvious way...)

Then, obviously, $\overline W=M$, but $M$ is not closed under addition.

  • $\begingroup$ Hmm, this is actually identical to Rhys Steele's earlier answer, just terser ... $\endgroup$ May 25 '18 at 23:28
  • $\begingroup$ Stupid question: how do we know that $M \neq V$? $\endgroup$
    – user211599
    May 25 '18 at 23:30
  • $\begingroup$ What is the metric completion of $c_{00}$? Is it not the same as its norm completion $c_0$, which is indeed a vector space? $\endgroup$ May 25 '18 at 23:39
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    $\begingroup$ @mechanodroid $c_0$ is a completion of $c_{00}$ but it doesnt remain a vector space when you give it the additive structure described here. The formal linear combination of $x$ and $y$ denoted by $x+y$ is not the same thing as their sum in $c_{0}$. $\endgroup$ May 25 '18 at 23:47
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    $\begingroup$ @JefferyOpoku-Mensah the space of finite formal linear combinations of elements of $M \setminus W$ (the free vector space over $M \setminus W$) is strictly larger than $M \setminus W$. $\endgroup$ May 25 '18 at 23:49

Let $W$ be a subspace of a Banach space $M$ such that $W$ is not closed in $M$ and let the metric on $M$ be the one induced by its norm. Concretely, take e.g. $W$ to be the space of polynomials on $\mathbb{R}$ and $M$ to be the space of continuous functions with the $\sup$-norm.

In particular $M$ is a complete metric space and the restriction of the metric to $W$ coincides with the norm on $W$.

My idea is to enlarge $M$ and change the additive structure on $M \setminus W$ in a way that is suitably incompatible with the original addition there.

Let $F$ be the free vector space over the set $M \setminus W$. Now let $V = F \oplus W$ be the vector space direct sum of $F$ and $W$. I make the obvious identification of $W$ with $\tilde{W} = \{0\} \oplus W$ as a vector space and of $M$ with $\tilde{M} = [(M \setminus W) \oplus \{0\}] \cup [\{0\} \oplus W]$ as a metric space.

Notice that the vector space structure induced on $\tilde{W}$ by $V$ is the same as its original structure. From here on, all topological operations take place in the topology induced by the metric on $\tilde{M}$ and vector space operations take place in $V$.

Finally, take $x_n, y_n \in W$ such that $x_n \to x \in M \setminus W$, $y_n \to y \in M \setminus W$. Translating this all in to $V$ we have $(0, x_n) \to (x,0) \in (M \setminus W) \oplus \{0\}$ and $(0,y_n) \to (y,0) \in (M \setminus W) \oplus \{0\}$.

However $(x,0) + (y,0) = (x+y,0) \not \in \tilde{M}$ since the addition takes place in the free vector space over $M \setminus W$ so that $x+y \not \in M \setminus W$. Obviously, $\overline{\{0\} \oplus W} \subset \tilde{M}$ and $(x,0), (y,0) \in \overline{\{0\} \oplus W}$ so we see that the closure isn't closed under addition in $V$.


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