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Let $X$ and $Y$ be topological vector spaces and $f : X \to Y$ a linear mapping. If $f$ if continuous, then $f$ is Borel (by definition). Is it still true in general ? I believe it is not but I'm not finding any counterexample.

Paul Tristant
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    You can think of finite dimensional Banach spaces as topological rational vector spaces. Then you can use results about automatic continuity of measurable additive maps between finite dimensional real vector spaces, say here. – Moishe Kohan Jul 24 '24 at 15:52
  • @MoisheKohan Any linear map defined on a finite-dimensional Banach space is automatically continuous, so I don’t think this helps much. – David Gao Jul 24 '24 at 19:22
  • There are obvious counterexamples if we don’t impose the condition that the topologies are Hausdorff. I don’t know what happens after adding the Hausdorff assumption, but a counterexample would follow from the existence of a non-Borel linear subspace. – David Gao Jul 24 '24 at 19:34
  • @DavidGao: Note that in my example it is only $\mathbb Q$-linear. Such maps are typically discontinuous. – Moishe Kohan Jul 24 '24 at 21:02
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    @MoisheKohan I was under the impression the OP meant real/complex TVS, but I suppose if rational TVS is allowed, that is a perfectly good counterexample. I’m not sure about the suggestion on tensoring with $\mathbb{R}$ in your answer. I don’t really know how tensoring is supposed to work when dealing with rational TVS. (Algebraically, sure, but the topology part is beyond me.) – David Gao Jul 25 '24 at 05:05
  • @DavidGao I was indeed thinking about real/complex TVS. – Paul Tristant Jul 25 '24 at 17:33

2 Answers2

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Here is a non-constructive proof of existence of a non-Borel linear map: Let $X$ be a separable infinite-dimensional Banach space. Then it has a Hamel basis $\{x_i\}_{i \in I}$ of cardinality continuum $\mathfrak{c}$. See this paper for a proof of this fact. Thus, it has $2^\mathfrak{c}$ linear subspaces, as any subset of the basis spans a different subspace. However, as $X$ is a Polish space, $X$ only has $\mathfrak{c}$ many Borel subsets, a standard fact that can be proved via transfinite induction. Thus, there is a subspace $K \subset X$ which is not Borel. Let $Y = X/K$ be equipped with any Hausdorff topology (say, choose a Hamel basis of $Y$ and then equip $Y$ with the $\infty$-norm associated to this basis). Then the natural quotient map $\pi: X \to X/K = Y$ is linear but not Borel, as $\pi^{-1}(\{0\}) = K \subset X$ is not Borel.

David Gao
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  • Nice one ! Is it possible to go further and suppose $X$ AND $Y$ are Polish ? – Paul Tristant Jul 25 '24 at 17:07
  • @PaulTristant Yes. $X$ is already Polish in my answer. $X/K$, as a linear space, has Hamel dimension $\leq \mathfrak{c}$ (since $\dim(X) = \mathfrak{c}$). Thus, you can linearly embed $X/K$ into any separable infinite-dimensional Banach space $Y$ and $\pi$ can then be the composition of the quotient map $X \to X/K$ and the embedding $X/K \to Y$. – David Gao Jul 25 '24 at 17:32
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    Also, you can even get $Y = \mathbb{C}$ from this. Again, take $\pi: X \to X/K$ as above and embed $X/K$ into a separable Banach space $E$. IIRC, it is known that, as $E$ is separable, $\pi: X \to E$ is Borel iff $\phi \circ \pi: X \to \mathbb{C}$ is Borel for all $\phi \in E^\ast$. Thus, you can get a non-Borel linear functional on any separable infinite-dimensional Banach space. – David Gao Jul 25 '24 at 17:41
  • If we assume $X,Y$ to be Polish then any example must be non-constructive (requiring axiom of choice). This is because it is consistent with $\mathsf{ZF+DC}$ that any subset of any Polish space has the property of Baire, and a Baire measurable homomorphism between Polish group must be continuous (see the last paragraph on page 4 of this survey). Is choice needed if we allow arbitrary TVS? – n901 Aug 02 '24 at 00:07
  • @n901 If non-Hausdorff topologies are allowed, then there are obvious trivial examples. For Hausdorff TVSs, I suspect that $c_{00}(I)$ (finitely-supported sequences) is a non-Borel subspace of $c_0(I)$ for an uncountable $I$. I don’t know how to prove it though (especially without AC), and even if it is true I don’t know if one can prove $c_0(I)/c_{00}(I)$ admits the structure of a Hausdorff TVS without using AC. – David Gao Aug 02 '24 at 02:02
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Let $X, Y$ be nonzero finite-dimensional Banach spaces. We regard them as $\mathbb Q$-vector spaces $X', Y'$; we topologize them using the original Banach space topologies (which are, thus, Hausdorff). Hence, $X', Y'$ are rational Hausdorff topological vector spaces. Let $f: X'\to Y'$ be a discontinuous $\mathbb Q$-linear map (there are many of these). Then $f$ is not a measurable map, see e.g. this question. Tensoring $X', Y'$ with $\mathbb R$ one probably will get an example of non-Borel linear map of Hausdorff real TVSs, but I did not check this.

Moishe Kohan
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  • Thanks for your answer. I think you're right but I was working on Polish spaces on $\mathbf{R}$ or $\mathbf{C}$. I should have specified it – Paul Tristant Jul 25 '24 at 17:10