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An Eberlein compact space $X$ is a topological space homeomorphic to a weakly compact subset of a Banach space.

An equivalent characterization is that $X$ is homeomorphic to a compact subset of some $c_0(\Gamma)$ for some set $\Gamma$. Here, $c_0(\Gamma)$ is the space of functions $f:\Gamma\to\mathbb R$ "vanishing at infinity", that is, for each $\epsilon>0$, $\{x\in\Gamma:|f(x)|>\epsilon\}$ is finite. The topology on $c_0(\Gamma)$ is the topology of pointwise convergence, induced from the product topology on $\mathbb R^\Gamma$.

I have not checked the equivalence myself, but one good reference seems to be [MR].

If $X$ is the Alexandroff one-point compactification of a infinite discrete space, it is supposed to be Eberlein compact. That's easy to see if $X$ is countably infinite, as $X$ is homeomorphic to a converging sequence in $\mathbb R$, and that's clearly a compact subset of $\mathbb R=c_0(\Gamma)$ for $\Gamma$ a singleton.

Now suppose $Z$ is a discrete space of uncountable cardinality and $X=Z\cup\{\infty\}$ is the one-point compactification of $Z$ (for example, the Fort space on the real numbers). How would you show directly that $X$ is Eberlein compact, with each of the characterization?

My idea for the second characterization:

Take $\Gamma=Z$ and let $W=c_0(Z)\subseteq\mathbb R^Z$. For each $z\in Z$, let $e_z:Z\to\mathbb R$ be the characteristic function of the singleton $\{z\}$ (i.e., $e_z(z)=1$ and $e_z(t)=0$ for $t\ne z$). Each $e_z$ has finite support, hence belongs to $c_0(Z)$. It is easy to check that the set $A=\{e_z:z\in Z\}\subseteq c_0(Z)$ is discrete, and that its closure is $\overline A=A\cup\{\mathbf 0\}$ (where $\mathbf 0$ is the zero function), with $\overline A$ homeomorphic to the one-point compactification of $A$ (every nbhd of $\mathbf 0$ contains all but finitely many of the $e_z$). So $A$ is compact and homeomorphic to $X$. This shows that $X$ is Eberlein compact.

Is this all there is to it, or am I missing something?

Can someone show the same thing with the first definition? I.e., exhibit a Banach space and a weakly compact subset that is homeomorphic to $X$.

And as an extra, if you could give an idea of why the two characterizations of Eberlein compact are equivalent in general, that would be greatly appreciated.

[MR] Michael, Ernest; Rudin, Mary Ellen, A note on Eberlein compacts, Pac. J. Math. 72, 487-495 (1977).

PatrickR
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  • I mean, this *is* a weakly compact subset of a Banach space? The weak topology on the unit ball of $c_0(Z)$ is the same as the topology of pointwise convergence. – David Gao May 29 '25 at 04:57
  • @DavidGao The quoted fact about the unit ball of $c_0(Z)$, is that part of the Banach-Alaoglu theorem? – PatrickR May 29 '25 at 05:15
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    Not really. It’s an approximation argument like this: take any net $x_\lambda$ in the unit ball. If $x_\lambda\to x$ weakly, then $x_\lambda\to x$ pointwise because evaluating at each point is a bounded linear functional. Conversely, assume $x_\lambda\to x$ pointwise. The dual space of $c_0$ is $\ell^1$ through the pairing $((a_i),(b_i))=\sum_ia_ib_i$. Then for any $y\in\ell^1$ that’s finitely supported, you get $(x_\lambda,y)\to(x,y)$ as the pairing just gives a finite linear combination of pointwise coefficients. Now use a $3\epsilon$ argument to get the same holds for all $y\in\ell^1$. – David Gao May 29 '25 at 05:24
  • Thank you for the explanation. – PatrickR May 29 '25 at 05:45
  • @DavidGao Perhaps you can post this as an answer? – Dean Miller May 29 '25 at 13:38
  • @DeanMiller Just did. – David Gao May 30 '25 at 23:07

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As suggested, here are my comments, turned into an answer:

The set $\overline{A}$ you wrote down is already a weakly compact subset of a Banach space, namely, $c_0(Z)$. This is due to the fact that the weak topology and the topology of pointwise convergence coincide on the unit ball of $c_0(Z)$. Indeed, this follows from an approximation argument that's quite standard in functional analysis: Take any net $x_\lambda$ in the unit ball of $c_0(Z)$. If $x_\lambda \to x$ weakly, then $x_\lambda \to x$ pointwise because evaluating at each point in $Z$ is a bounded linear functional. Conversely, assume $x_\lambda \to x$ pointwise. The dual space of $c_0(Z)$ is $\ell_1(Z)$ through the pairing,

$$((a_z)_{z \in Z}, (b_z)_{z \in Z}) = \sum_{z \in Z} a_zb_z$$

Then, for any $y \in \ell_1(Z)$ that's finitely supported, we get $(x_\lambda, y) \to (x, y)$ as the pairing just gives a finite linear combination of pointwise coefficients. We now use a $3\epsilon$ argument. Using density of finitely supported sequences in $\ell_1(Z)$, we can, for any $y \in \ell_1(Z)$, pick $y' \in \ell_1(Z)$ finitely supported s.t. $\|y - y'\| < \epsilon$. But then, as $\|x_\lambda\|, \|x\| \leq 1$, we have,

$$|(x_\lambda, y) - (x_\lambda, y')| < \epsilon, |(x, y) - (x, y')| < \epsilon$$

But as $y'$ is finitely supported, for large $\lambda$, we have $|(x_\lambda, y') - (x, y')| < \epsilon$. Hence, by triangle inequality, we get, for large $\lambda$, $|(x_\lambda, y) - (x, y)| < 3\epsilon$, i.e., $(x_\lambda, y) \to (x, y)$ so $x_\lambda \to x$ weakly.

This can also easily be generalized to show that the $c_0(\Gamma)$ characterization of Eberlein compact spaces implies the Banach space characterization. Indeed, if $X \subset c_0(\Gamma)$ is compact under the topology of pointwise convergence, then its projection onto each coordinate is a compact interval. Thus, by rescaling - possibly by different factors on different indices, and only scaling by a factor less than or equal to $1$, which is necessary to ensure the rescaled $X$ still lies inside $c_0(\Gamma)$ - we see that $X$ is homeomorphic to a subset of the unit ball of $c_0(\Gamma)$ compact under the topology of pointwise convergence. Now just apply the fact proved in previous paragraphs. (It's not immediately obvious to me how to prove the converse, though, except in the case where $X$ is homeomorphic to a weakly compact subset of a separable Banach space.)

David Gao
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  • For the converse, that is if $A\subseteq c_0(\Gamma)$ is weakly compact, then its pointwise compact, you want to use uniform boundedness principle I believe. This implies that weakly compact sets are bounded – Jakobian May 30 '25 at 23:57
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    The converse, namely that a weakly compact subset of a Banach space is homeomorphic to a compact subset of some $c_0(\Gamma)$, is a non-trivial result due to Amir & Lindenstrauss (jstor), see Theorem 1. I have not read the details. – PatrickR May 30 '25 at 23:59
  • @PatrickR I would avoid the phrase "compact subset of $c_0(\Gamma)$". In functional analysis, I would say "compact" defaults to "norm-compact" where we treat $c_0(\Gamma)$ as a Banach space. Instead, its better to say this subset is weakly compact. – Jakobian May 31 '25 at 00:03
  • @Jakobian Good point. I should have said: "compact subset of $c_0(\Gamma)$ in the topology of pointwise convergence", which is what I meant. (But as shown by David, we know it's the same as weakly compact in $c_0(\Gamma)$ as a Banach space in this case) – PatrickR May 31 '25 at 00:18
  • @PatrickR well actually that's what my comment above is for. David Gao have shown that both topologies are the same in a unit ball. But that a weakly compact set can be contained in such ball, needs to be shown using uniform boundedness principle. What David Gao haven't shown, is that pointwise and weak topology on $c_0(\Gamma)$ coincide. – Jakobian May 31 '25 at 00:24
  • I think the converse David Gao was referring to at the end of the post is not about that, but rather about showing than a weakly compact subset of an arbitrary Banach space can be embedded in some $c_0(\Gamma)$ ? – PatrickR May 31 '25 at 00:29
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    @PatrickR It doesn't matter what exactly they were referring to. Its still an important detail to point out, that wasn't addressed anywhere. – Jakobian May 31 '25 at 00:30
  • @Jakobian The fact that weakly compact subsets of $c_0(\Gamma)$ are pointwise compact does not need uniform boundedness principle. The weak topology is stronger than the pointwise topology, and a set compact under a stronger topology is also compact under a weaker topology. – David Gao May 31 '25 at 01:49
  • @PatrickR Ah, not a surprise that this is a nontrivial result. The only case I can prove is the separable case, and indeed the paper mentions that the result is trivial in that case. – David Gao May 31 '25 at 01:52
  • @Jakobian I suppose, just for clarification’s sake: The weak topology and the pointwise topology do not coincide on the entirety of $c_0(\Gamma)$. In fact, ${(a_n)\in c_0:\sum_n\frac{a_n}{2^n}=0}$ is a linear subspace that’s weakly closed but not pointwise closed. They only coincide on uniformly norm-bounded subsets. A weakly compact subset is always pointwise compact. This just follows from the weak topology being stronger than the pointwise topology, no need to use the uniform boundedness principle. But you can use the uniform boundedness principle to show every weakly compact set is… – David Gao May 31 '25 at 02:01
  • … uniformly norm-bounded. As for the converse, it’s not true that every pointwise compact subset is weakly compact. Say, in $c_0$, ${2^ne_n:n\in\mathbb{N}}\cup{0}$ is pointwise compact but not weakly compact. Indeed, a pointwise compact subset is weakly compact iff it is uniformly norm-bounded. (This is why rescaling is needed.) And, finally, the difficulty in the converse is really about how to construct a map from the Banach space that $X$ embeds into to $c_0(\Gamma)$. I didn’t really talk about this much because I don’t know how to prove it, nor was I aiming to prove it in my answer. – David Gao May 31 '25 at 02:06
  • I don’t really think these are needed in the answer, since it’s irrelevant for the proof that a pointwise compact subset of $c_0(\Gamma)$ is homeomorphic to a weakly compact subset of $c_0(\Gamma)$, which is what I was demonstrating in my answer. But I suppose these are very common points of confusion that it is worth addressing in the comments at least. – David Gao May 31 '25 at 02:11