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:(adsbygoogle = window.adsbygoogle || []).push({});

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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# If every basic open cover has a countable subcover

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