Talk:Separable Metacompact Space is Lindelöf/Proof 1
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"A refinement is not necessarily a subcover."
According to the definition here: Definition:Open Refinement it is, which is what this proof has been crafted from. --prime mover (talk) 07:49, 9 May 2013 (UTC)
- No, you misread. If $\mathcal U$ and $\mathcal V$ are both open covers then:
- $\mathcal V$ is an open refinement of $\mathcal U$ iff each element of $\mathcal V$ is a subset of some element of $\mathcal U$.
- $\mathcal V$ is a subcover of $\mathcal U$ iff $\mathcal V \subseteq \mathcal U$.
- Thus every open subcover is an open refinement, but not every open refinement is a subcover. --Dfeuer (talk) 13:44, 9 May 2013 (UTC)
- By the definition given here: Definition:Open Refinement, and the one by which this proof is crafted, an open refinement of a cover is itself a cover.
- This is because, every Definition:Refinement of Cover is also a cover by definition of refinement.
- What definition are you using for "refinement"? --prime mover (talk) 18:21, 9 May 2013 (UTC)
- Counterexample? Sources? --prime mover (talk) 19:29, 9 May 2013 (UTC)
- You are joking, right? Let $X = \{ 1, 2 \}$. Let $\mathcal D$ be the discrete topology on $X$. Then $\mathcal U = \{ \{ 1, 2 \} \}$ is an open cover of $X$. $\mathcal V = \{ \{1\}, \{2\} \}$ is an open refinement of $\mathcal U$. However, it is not a subcover of $\mathcal U$ because $\{1\}$ is in $\mathcal V$ but not in $\mathcal U$. The problem is not with the definitions on ProofWiki (in this case), but with how you are reading them. --Dfeuer (talk) 19:50, 9 May 2013 (UTC)
Yeah that's a bit of a bugger that ... the proof I gave is effectively the one given in Steen and Seebach. Seems they fell into the same trap that you very smugly avoided.
It's useful when you notice stuff like this to explain where it's gone wrong rather than just delete it or just give a two-word put-down. The latter behaviour is prickish. --prime mover (talk) 21:07, 9 May 2013 (UTC)