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!

 

[blue_raver22]

Yes, your proof is correct. By using the axiom of choice, you have shown that every open cover can be reduced to a countable subcover. This is a stronger result than the assumption made in Dan Ma's proof, which only states that every basic open cover has a countable subcover. Your proof shows that this assumption can be extended to all open covers. This is a useful result and can be applied to other topological spaces as well. Great job!