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Let $X$ be a topological space. $A\subseteq X$ is Lindelöf $\iff$ Every cover of $A$ by open subsets of $X$ has a countable subcover.

My attempt:

Let $(U_i)_{i \in I}$ be a cover for $A$ by open subsets of $X$. So $A\subseteq \bigcup_{i\in I}U_i$. Then $A\subseteq A \cap $ $\bigcup_{i\in I}U_i$ $=$ $\bigcup_{i\in I } A\cap U_i$ ,which is a cover for $A$ by sets open relative to $A$. By assumption, there exists a countable subcover, indexed by $J\subseteq \mathbb{N}$ such that $A\subseteq \bigcup_{j\in J}A\cap U_j$. This implies that $A\subseteq \bigcup_{j\in J}U_j$, which means that every cover of $A$ by open subsets of $X$ has a countable subcover. For the converse, let $(U_i)_{i\in I}$ be sets open relative to $A$ that cover $A$. So $A\subseteq \bigcup_{i\in I}U_i$, where each $U_i \in \tau_A$. Hence, $U_i=U_i' \cap A$ for some $U_{i}' \in \tau_X$. Hence $A\subseteq$ $\bigcup_{i\in I}(U_i' \cap A)$ $=$ $A \cap \bigcup_{i\in I}U_i'$ $\implies$ $A\subseteq$ $\bigcup_{i\in I}U_i'$, and so by assumption there exists an enumeration $U_1',U_2',U_3'...$ by an index set $J\subseteq \mathbb{N}$ such that their union covers $A$, hence $A\subseteq A\cap (U_1' \cup U_2' \cup U_3' \ldots)$ $=$ $\bigcup_{j \in J} (A\cap U_j')$, and so $A$ is Lindelöf.

Is it correct?

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    $\begingroup$ The second sentence of the proof should assert containment, not equality. Likewise for the fifth sentence. $\endgroup$
    – MPW
    Commented Dec 24, 2019 at 13:51

1 Answer 1

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$J$ need not be a subset of $\Bbb N$. It's just a countable subset of $I$, and just say so.

Moreover, we can state $$A = A \cap \bigcup_{i \in I} U_i$$ etc. with equality, so that $A$ is exactly covered by the relatively open cover $\{U_i \cap A: i \in I\}$, which has a countable subcover $\{U_i \cap A: i \in J\}$ where $J \subseteq I$ countable, and then $$A = \bigcup_{i \in J} (U_i \cap A) = A \cap \bigcup_{i \in J} U_i$$ implying $A \subseteq \bigcup_{i \in J} U_i$ as required.

The other direction has an equality instead of an inclusion too in $A = \bigcup_{i \in I} U_i$ as well, when the $U_i$ are open in $A$ (which implies already they are all a subset of $A$). Same remark about $J$ just being countable subset of $I$, not of $\Bbb N$.

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  • $\begingroup$ You're right I see your point, but since they're countable, they can be indexed by the natural numbers, am I right? $\endgroup$
    – user643073
    Commented Dec 24, 2019 at 15:20
  • $\begingroup$ @topologicalmagician No, that's not how it works. There is a bijection between $\Bbb N$ and $J$ but $J$ stays a subset of $I$ if we use this $U_i$ notation throughout. Otherwise you'd have to say $U_{i_n}$, for $n \in \Bbb N$, which is slightly uglier IMHO. Then $J = \{i_n: n \in \Bbb N\}$ explicitly. I try to avoid double indexing if it can be avoided. $\endgroup$ Commented Dec 24, 2019 at 15:23
  • $\begingroup$ Thanks! That makes sense. $\endgroup$
    – user643073
    Commented Dec 24, 2019 at 15:24

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