If every basic open cover has a countable subcover

[Rasalhague]

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

 

[micromass]

No, this is not correct. The mistake lies in the fact that for two different C_\lambda, I might choose the same B_{C_\lambda}. So your map f is not necessarily well-defined.

 

[Rasalhague]

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 \frak{C}. Each C\in\frak{C} is a union of basic open sets,

\bigcup_{\lambda\in S_C}B_\lambda,

so we have a cover

\bigcup_{C\in\frak{C}}\bigcup_{\lambda\in S_C}B_\lambda,

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

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

So we can choose a C_n, not necessarily unique, from \frak{C} for each B_n\in\frak{B} such that B_n\subseteq C_n. Then \left \{ C_n | n\in\mathbb{N} \right \} is a countable subcover of \frak{C}. \blacksquare

 

[micromass]

That seems fine!