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I'm trying to understand a proof of the following theorem (from Banach Algebra Techniques in Operator Theory by Douglas):

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Here is my question:

Where is the Hausdorff property used in the proof?

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    You don't need Hausdorffness. Let $\sim$ be the smallest equivalence relation on $X$ such that $\sim$ is a closed subset of $X \times X$; then every continuous map $X \to \mathbb{R}$ will factor through the quotient map $X \to X / {\sim}$. But $X / {\sim}$ is compact Hausdorff as soon as $X$ is compact, and this argument shows that $C (X) \cong C (X / {\sim})$. – Zhen Lin Sep 28 '15 at 14:39
  • @ZhenLin True. But it's much simpler to just note that Hausdorff is simply not used at any point in the proof. – David C. Ullrich Sep 28 '15 at 14:40
  • where did you found this proof ? – Surb Sep 28 '15 at 14:41
  • @Surb ??? It's exactly the standard argument - you can find it anywhere the result is proved. – David C. Ullrich Sep 28 '15 at 14:43
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    @DavidC.Ullrich Indeed. But this observation lets us make the same conclusion without inspecting the proof. – Zhen Lin Sep 28 '15 at 14:45
  • I was asking, because may be in the proof they use a result that they proved only on Hausdorff space that would explain why the precise that $X$ is Hausdorff. – Surb Sep 28 '15 at 14:46
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    @ZhenLin True. Also works for readers who are somehow unable to inspect the proof... – David C. Ullrich Sep 28 '15 at 14:52

2 Answers2

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I don't believe that Hausdorff is used; the result is true for compact $K$. They state the result for compact Hausdorff $K$ just because nobody cares about $C(K)$ for non-Hausdorff compact $K$.

David C. Ullrich
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  • Of course it is used. – Surb Sep 28 '15 at 14:30
  • @Surb Where? You say in your answer that it is used to show that a neighborhood of $x_0$ is not just ${x_0}$. That fact is not used in the proof. – David C. Ullrich Sep 28 '15 at 14:32
  • How would you justify that there is an $x\in U\backslash {x_0}$ such that $|f_N(x_0)-f_N(x)|<3\varepsilon$ (7th line of the proof). – Surb Sep 28 '15 at 14:35
  • You don't need to. You just want to prove it for any $x \in U$. – jxnh Sep 28 '15 at 14:35
  • @Surb Why would you imagine that fact is needed? You're familiar with how the empty set works, right? A statement "for every $x\in E\dots$ does not imply or require that $E$ be nonempty. – David C. Ullrich Sep 28 '15 at 14:38
  • @Surb You might also note, come to think of it, that it's not even true that Hausdorff implies that a neighborhood of $x_0$ cannot be ${x_0}$. – David C. Ullrich Sep 28 '15 at 14:39
  • Ok, I thought it was important. Then sorry for the disturbing :-) – Surb Sep 28 '15 at 14:40
  • @Surb I'm about to make an unimportant edit to my post, just so that whoever downvoted it is not stumped by the "cannot change vote until post is edited" rule... – David C. Ullrich Sep 28 '15 at 14:41
  • By the way, I'm the downvoter, and I'm sorry for that. I thought you were wrong. So if you edit again, I will erase my downvote :-) – Surb Sep 28 '15 at 14:43
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I don't believe the Hausdorff property is used.

In fact, as far as I can tell, even the compactness is only needed for the suprema to exist.

So in fact, if you allow a metric to be $\infty$ at some pairs, the same proof shows that $C(X)$ is a complete metric space for any topological space $X$.

Even if you insist that metrics should never assume $\infty$, it still shows that $C_b(X)$ (the space of bounded continuous functions) is always a complete metric space (in this case, you need to notice that an uniform limit of bounded functions is bounded, but that is trivial).

tomasz
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