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If every basic open cover has a countable subcover

  1. Jul 13, 2012 #1
    Dan Ma's proof that the Sorgenfrey line is Lindelöf (part A) seems to rely on the assumption that if every basic open cover has a countable subcover, then every open cover has a countable subcover. In trying to prove this, I got a stronger result, namely that if every basic open cover has a countable subcover, then every open cover is countable. If this correct? Here is my proof:

    Every element [itex]C_{\lambda}[/itex] of an arbitrary open cover [itex]\cal{C}=\left \{ C_\lambda | \lambda\in \Lambda \right \}[/itex] is a union of basic open sets, so we can choose a countable cover [itex]\cal{B}[/itex], each of whose elements is a subset of some [itex]C_{\lambda}\in\cal{C}[/itex]. By the axiom of choice, for each [itex]C_\lambda\in\cal{C}[/itex] we can choose one [itex]B_{C_\lambda} \in \cal{B}[/itex]. The set [itex]\beta = \left \{ B_{C_\lambda} | C_\lambda\in\cal{C} \right \}[/itex] of these so-chosen [itex]B_{C_\lambda}[/itex] is a subset of [itex]\cal{B}[/itex], and [itex]\cal{B}[/itex] is countable, therefore [itex]\beta[/itex] is countable. There is a natural bijection [itex]f:\beta\rightarrow\cal{C}[/itex], specified by [itex]f(B_{C_\lambda})=C_\lambda[/itex]. Therefore the arbitrary open cover [itex]\cal{C}[/itex] is countable. [itex]\blacksquare[/itex]
     
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  3. Jul 13, 2012 #2

    micromass

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    No, this is not correct. The mistake lies in the fact that for two different [itex]C_\lambda[/itex], I might choose the same [itex]B_{C_\lambda}[/itex]. So your map f is not necessarily well-defined.
     
  4. Jul 13, 2012 #3
    Ah, I see. Thanks, micromass. Here's my revised proof of "If there exists a basis such that every basic open cover has a countable subcover, then every open cover has a countable subcover."

    Suppose every basic open cover has a countable subcover and that we have an open cover [itex]\frak{C}[/itex]. Each [itex]C\in\frak{C}[/itex] is a union of basic open sets,

    [tex]\bigcup_{\lambda\in S_C}B_\lambda,[/tex]

    so we have a cover

    [tex]\bigcup_{C\in\frak{C}}\bigcup_{\lambda\in S_C}B_\lambda,[/tex]

    whence we can select a countable subcover [itex]\frak{B}=\left \{ B_n | n\in\mathbb{N} \right \}[/itex]. That is to say,

    [tex](\forall x\in X)(\exists C\in\frak{C})(\exists B_n\in\frak{B})[ x \in B_n\subseteq C].[/tex]

    So we can choose a [itex]C_n[/itex], not necessarily unique, from [itex]\frak{C}[/itex] for each [itex]B_n\in\frak{B}[/itex] such that [itex]B_n\subseteq C_n[/itex]. Then [itex]\left \{ C_n | n\in\mathbb{N} \right \}[/itex] is a countable subcover of [itex]\frak{C}. \blacksquare[/itex]
     
  5. Jul 13, 2012 #4

    micromass

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    That seems fine!
     
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