# Lindelöf space condition

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?

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

$$J$$ 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 $$\Bbb N$$.

• You're right I see your point, but since they're countable, they can be indexed by the natural numbers, am I right?
– user643073
Commented Dec 24, 2019 at 15:20
• @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. Commented Dec 24, 2019 at 15:23
• Thanks! That makes sense.
– user643073
Commented Dec 24, 2019 at 15:24