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Let $(X,\mathcal{T})$ be a topological space. Show that if $(X, \mathcal{T})$ has countable base, it is separable (a) and Lindelöf (b)

My attempt:

Let $\mathcal{B}$ be a countable basis of the given topology.

(a) For every $B \in \mathcal{B}\setminus\{\emptyset\}$, choose $x_B \in B$ (here we use AC) and consider $D:= \{x_B \mid B \in \mathcal{B}\}$. It is clear that this is at most countable. Let $x \in X$ and $V \in \mathcal{V(x)}$. Then, there exists $G \in \mathcal{T}$ such that $x \in G \subseteq V$. But, there exists $\mathcal{A} \subseteq \mathcal{B}$ such that $G = \bigcup\mathcal{A}$, since $\mathcal{B}$ is a basis. Take $A \in \mathcal{A}$ with $x \in A \in \mathcal{B}$. Then $A$ is non empty, and hence $x_A \in A \cap D \subseteq V \cap D$, so $x \in \overline{D}$ and hence $D$ is dense in $X$.

(b) Let $\mathcal{G}$ be an open cover of $X$.

For every $G$ in this cover, there exists $\mathcal{A}_G \subseteq \mathcal{B}$ such that $G= \bigcup \mathcal{A_G}$. Consider the collection $\mathcal{C}:= \bigcup_{G \in \mathcal{G}}\mathcal{A}_G \subseteq \mathcal{B}$.

If $C \in \mathcal{C}$, then there is $G_C \in \mathcal{G}$ such that $C \in \mathcal{A}_{G_C}$. Thus, $C \subseteq G_C$. Consider the countable set ($\mathcal{C}$ countable because $\mathcal{B}$ countable)

$$\{G_C \mid C \in \mathcal{C}\}$$

Then:

$$X \supseteq\bigcup_{C \in \mathcal{C}}G_C \supseteq \bigcup_{C \in \mathcal{C}} C = \bigcup_{G \in \mathcal{G}} \bigcup_{H \in \mathcal{A}_G}H = \bigcup_{G \in \mathcal{G}} \bigcup\mathcal{A}_G = \bigcup_{G \in \mathcal{G}}G = \bigcup\mathcal{G} = X$$

Hence, $\{G_C \mid C \in \mathcal{C}\}$ is a countable subcover of $\mathcal{G}$, and hence $X$ is Lindelöf.

Is this correct?

2 Answers2

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Globally it is correct.

You don't need the axiom of choice, though.

José Carlos Santos
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  • When I pick an element out of every set $B \in \mathcal{B}$, don't I use choice? –  Feb 26 '18 at 13:49
  • And why is my $\mathcal{C}$ a subset of $X$? –  Feb 26 '18 at 13:53
  • Statement about AC is incorrect (unless you meant strictly uncountable choice). Actually, assuming every second countable space has a separable subset is equivalent to countable AC, see: https://math.stackexchange.com/questions/309313/second-countable-implies-separable-axiom-countable-choice – j3M Feb 26 '18 at 13:53
  • @Math_QED But you don't do that. You define $\mathcal C$ as the union of all $\mathcal{A}_G$. Where is a choice being made here? – José Carlos Santos Feb 26 '18 at 13:54
  • @Math_QED Please note that Ive edited my answer. – José Carlos Santos Feb 26 '18 at 13:55
  • So the definition of $\mathcal{C}$ was correct? I don't use AC here. I wrote it in my OP but just before you posted your answer I realised I don't need it, so I removed it. Probably, this is the reason of the confusion here. –  Feb 26 '18 at 13:57
  • @Math_QED Good. So, as it stands now, your proofs are correct. – José Carlos Santos Feb 26 '18 at 13:59
  • Great! Thanks for your answer. –  Feb 26 '18 at 14:00
  • @Math_QED countable choice is definitely needed for this to hold! – Henno Brandsma Feb 26 '18 at 18:37
  • Thanks for your comment. I know this. @JoséCarlosSantos means that the seconds implication doesn't need choice. The link that j3M provides shows that the first implication is equivalent with countable AC. –  Feb 26 '18 at 18:53
  • this question and Asaf's answer show some choice is required. – Henno Brandsma Feb 26 '18 at 19:15
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The separable part I quite agree and we need countable choice there.

The other part also needs choice, I'll make my use of it explicit:

Let $\{B_n: n \in \mathbb{N}\}$ be a countable base for $X$. Let $\mathcal{O}$ be an open cover of $X$. Define for each $n \in \mathbb{N}$:

$$\mathcal{O}_n = \{O \in \mathcal{O}: B_n \subseteq O\}$$

For a given $n$ this set could be empty, finite or infinite. You cannot say beforehand. But let $N'$ be the set of $n$ for which $\mathcal{O}_n$ is non-empty, and for each $n \in N'$ we pick (yes, AC is used, and I'm not ashamed of it) some $O_n \in \mathcal{O}_n$.

Now I claim that $\{O_n: n \in N'\}$ is a countable (obvious, as $N'$ is countable as a subset of $\mathbb{N}$) subcover of $X$: Let $x \in X$. Then $x$ is covered by some $O_x \in \mathcal{O}$, and because the $B_n$ form a base, we have some $n_x \in \mathbb{N}$ such that $x \in B_{n(x)} \subseteq O_x$. We conclude that $O_x \in \mathcal{O}_{n_x}$ by definition and so as we have $O_{n_x} \in \mathcal{O}_{n_x}$ (the choice we made) and $x \in B_{n_x} \subseteq O_{n_x}$, $x$ is indeed covered by the countable subcover.

Henno Brandsma
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