Lindelöf's lemma

In mathematics, Lindelöf's lemma is a simple but useful lemma in topology on the real line, named for the Finnish mathematician Ernst Leonard Lindelöf.

Statement of the lemma

Let the real line have its standard topology. Then every open subset of the real line is a countable union of open intervals.

Generalization

Lindelöf's lemma is also known as the statement that every open cover in a second-countable space has a countable subcover (Kelley 1955:49). This means that every second-countable space is also a Lindelöf space.

In order to see this, let ${\mathcal {U}}$ be an open cover of a topological space ${\mathcal {T}}$ which has a countable basis ${\mathcal {B}}$ . Then, ${\textstyle U=\bigcup \{B\in {\mathcal {B}}\mid B\subseteq U\}}$ for all $U\in {\mathcal {U}}$ . Especially, the system ${\textstyle {\mathcal {B}}_{\mathcal {U}}\colon =\bigcup _{U\in {\mathcal {U}}}\{B\in {\mathcal {B}}\mid B\subseteq U\}}$ is non-empty. As ${\mathcal {U}}$ covers ${\mathcal {T}}$ and as ${\textstyle \bigcup {\mathcal {U}}=\bigcup _{U\in {\mathcal {U}}}\left(\bigcup \{B\in {\mathcal {B}}\mid B\subseteq U\}\right)=\bigcup \left(\bigcup _{U\in {\mathcal {U}}}\{B\in {\mathcal {B}}\mid B\subseteq U\}\right)=\bigcup {\mathcal {B}}_{\mathcal {U}}}$ , the system ${\mathcal {B}}_{\mathcal {U}}$ is a cover of ${\mathcal {T}}$ as well. Moreover, ${\mathcal {B}}_{\mathcal {U}}\subseteq {\mathcal {B}}$ implies that ${\mathcal {B}}_{\mathcal {U}}$ is countable. Since for every $B\in {\mathcal {B}}_{\mathcal {U}}$ the set $\{U\in {\mathcal {U}}\mid B\subseteq U\}$ is non-empty by definition of ${\mathcal {B}}_{\mathcal {U}}$ , there exists ${\textstyle u\in \prod _{B\in {\mathcal {B}}_{\mathcal {U}}}\{U\in {\mathcal {U}}\mid B\subseteq U\}}$ by the axiom of (at least countable) choice. The system ${\textstyle {\mathcal {V}}\colon =u({\mathcal {B}}_{\mathcal {U}})=\{u(B)\mid B\in {\mathcal {B}}_{\mathcal {U}}\}}$ is a subsystem of ${\mathcal {U}}$ by construction. It is countable because ${\mathcal {B}}_{\mathcal {U}}$ is. And since ${\mathcal {B}}_{\mathcal {U}}$ covers ${\mathcal {T}}$ and since ${\textstyle \bigcup {\mathcal {B}}_{\mathcal {U}}\subseteq \bigcup {\mathcal {V}}}$ , the system ${\mathcal {V}}$ is also a cover of ${\mathcal {T}}$ . Hence, ${\mathcal {U}}$ has a countable subcover.

References

J.L. Kelley (1955), General Topology, van Nostrand.

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