In a proof of a specific theorem, the following is stated: ($\Omega$ is assumed to be a compact metric space)

"Let $H \subset C(\Omega)$ be a countable vector space over $\mathbb{Q}$ which is closed under the operations $\vee, \wedge$ contains the function $1$ and is dense in C(K)"

I can't see how to construct such a set $H$ or stated differently: Why can I assume the existence of such a set?


Well the passage is from a book of Dellacherie on Stochastic Processes. It is used in a proof of the Disintegration property of Measures and it truly is used as a fact and not as an assumption. That is, the proof of the theorem begins by stating "Let H...." and this is not part of any assumption underlying the statement of the theorem. (i don't think it is helpful - but the theorem states that on a compact metric space with its Borel sigma field the disintegration property holds)

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    $\begingroup$ What are the operations $\vee ,\wedge$? $\endgroup$
    – user235708
    Commented Jun 30, 2015 at 8:31
  • $\begingroup$ Probably $\min(f,g)$ and $max(f,g)$, but I may be wrong $\endgroup$
    – Tryss
    Commented Jun 30, 2015 at 8:34

1 Answer 1


In the quoted passage no assertion is made that such a set actually exists. Read "let x be Y" as "assume for the sake of argument that some object called x satisfying Y exists".

In general, this can go several places. It can lead to a contradiction, proving that such an x fails to exist. Or it could show some property that all such objects must have, even if their existence may still be left uncertain. Example: "Let a + bi be a complex root of a real polynomial p ... (chain of reasoning) ... Then a - bi is also a root of p." No assertion was made that all real polynomials have complex roots, only that when those roots exist, their conjugates are also roots.

I don't know if a vector space of functions such as described exists for all compact metric spaces, but without more context from that passage, there's no reason to assume it must in order to follow the chain of reasoning.

  • $\begingroup$ To address the "remark", it doesn't matter whether the passage is part of the statement of the theorem, or its proof. "Let x be Y" should still be read as I described. In fact, the distinction between "theorem" and "intermediate result used to prove a theorem" is pretty fuzzy. That said, if you could provide more context on exactly how the vector space H is used, that might aid the discussion. $\endgroup$ Commented Jun 30, 2015 at 14:48

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