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It is easy to come up with a metric space whose completion has strictly larger cardinality. Something like $\mathbb Q$ with completion $\mathbb R$ will do. Or more generally any subset of $\mathbb R^n$ has a countable dense subset we can use.

But for normed spaces over $\mathbb R$ this kind of funnybusiness is not allowed. $\mathbb Q$ is still dense in $\mathbb R$ but it is not a normed space over $\mathbb R$ like we want. The matter is complicated further by how all incomplete normed spaces bust have infinite dimension.

The standard example is the space $c_{00}$ of eventually-zero real sequences with completion equal to the space of sequences converging to zero. Here we use the norm $\|a\| = \sum_{i=1}^\infty |a_i|$. Unfortunately both these spaces have cardinality $\frak c$.

Can it happen that the completion is strictly larger than the original normed space?

CJ Dowd
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Daron
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  • Unfortunately both these spaces have cardinality $\frak c$. --- Probably an even more natural such example is the completion of the continuous functions (on a suitable space) being the $L^1$ functions on that space. Note that while there are $2^{\frak c}$ many Lebesgue integrable functions (e.g. on $[0,1]$), there are only $\frak c$ many equivalence classes of integrable functions (equivalence is equality up to measure zero), this being due to the fact that the metric completion of a space with $\frak c$ many elements has cardinality ${\frak c}^{\aleph_0} = \frak c.$ – Dave L. Renfro Jul 27 '23 at 09:24
  • There might be a way around this by using a normed space over $\mathbb R$ whose cardinality $b$ is such such that $b < b^{\aleph_0}$ (see my comments to this MSE answer). [In case it's not clear why ${\frak c}^{\aleph_0} = \frak c$ is relevant, and why we're considering $b^{\aleph_0},$ note that the cardinality of the metric space completion of a metric space is at most the cardinality of the set of (countable length) sequences of elements in that metric space.] – Dave L. Renfro Jul 27 '23 at 09:36
  • After writing the above comments, I saw your question Is the cardinality of $L^1[0,1]$ greater than $\frak c$? (one way to answer is in my above comments; FYI, this question also occurred to me around 1984-1985 -- see this 11 January 2007 sci.math post) in which geetha290krm mentions (in a comment) the paper Cardinalities of metric completions by Ronald C. Freiwald (1973). Freiwald's paper might be useful here. – Dave L. Renfro Jul 27 '23 at 09:54
  • @DaveL.Renfro Would you believe I wrote my answer before reading your first comment? I think the idea is a good one. Do you see anything wrong with the answer? – Daron Jul 27 '23 at 12:19

1 Answers1

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I think so.

Let $|\Omega|\ge \frak c$ and let $c_{00}(\Omega)$ be the vector space of functions $\Omega \to \mathbb R$ that are zero on all but finitely many elements, equipped with the $\sup$ norm. To see the density is at least $|\Omega|$ consider the family of open balls of radius $1/2$ centred at the functions $\{c_x: x \in \Omega\}$ with entries $c_x(x) = 1$ and $c_x(y) = 0$ for $y \ne x$. These balls are pairwise disjoint so every dense subset needs at least one element per ball.

In the other direction, we claim the density is at most $|\Omega|$ because this is the cardinality of the space. To prove that, we construct a function $\mathcal P_{\text{fin}}(\Omega) \times \mathbb R^{\aleph_0} \to c_{00}(\Omega)$. The left-hand-side has cardinality $|\Omega| \times \frak c = |\Omega| $.

To construct the function, fix a total ordering on $\Omega$. Given $(P,a_1,a_2,\ldots) \in P_{\text{fin}}(\Omega) \times \mathbb R^{\aleph_0}$ write $P = \{p_1,p_2,\ldots p_n\}$ with $p_1 < p_2 < \ldots < p_n$ and assign it to the function $f$ with $ f(p_i) = a_i$ and $f(x) = 0$ for $x \notin P$.

This shows $c_{00}(\Omega)$ has density exactly $|\Omega|$. Since $c_{00}(\Omega)$ is dense in its completion, and the completion is a metric space, $c_{00}(\Omega)$ has the same density as the completion.

The completion is a Banach space, and therefore has cardinality exactly $|\Omega|^{\aleph_0}$ . One way to see this is to use how every Banach space is homomorphic to some Hilbert space $\ell^2(K)$ and then show directly $\ell^2(K)$ has density $|K|$ and cardinality $|K|^{\aleph_0}$.

In particular let $|\Omega|$ be a set have countable cofinality. Then Konig's theorem says $|\Omega|^{\aleph_0} > |\Omega|$. But the LHS is the cardinality of $c_{00}(\Omega)$ and the RHS is the cardinality of the completion.

Daron
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