The Lindelöf Lemma

# The Lindelöf Lemma

Recall from the Lindelöf and Countably Compact Topological Spaces page that a topological space $X$ is said to be Lindelöf if every open cover of $X$ has a countable subcover.

Furthermore, we said that $X$ is countably compact if every countable open cover of $X$ has a finite subcover.

Awhile back, on the Second Countable Topological Spaces page we said that a space $X$ is second countable if there exists a countable basis for the topology on $X$.

We will now look at a very important connection between Lindelöf spaces and second countable spaces which we state and prove below.

Lemma 1 (The Lindelöf Lemma): If $X$ is a second countable topological space then $X$ is Lindelöf. |

**Proof:**Let $X$ be a second countable topology space. Then there exists a countable basis $\mathcal B$ of the topology $\tau$ on $X$.

- Now, let $\mathcal F$ be any open cover of $X$ so that:

\begin{align} \quad X = \bigcup_{A \in \mathcal F} A \end{align}

- Note that each $A \in \mathcal F$ is an open set since $\mathcal F$ is an open cover of $X$. Therefore, for each $A$ there exists a subcollection $\mathcal B_A \subseteq \mathcal B$ such that:

\begin{align} \quad A = \bigcup_{B \in \mathcal B_A} B \end{align}

- Define an open cover $\mathcal C$ of $X$ as follows:

\begin{align} \quad \mathcal C = \left \{ B \in \mathcal B : A = \bigcup_{B \in \mathcal B_A} B, \: \mathrm{for \: some \:} A \in \mathcal F \right \} \end{align}

- Then $\mathcal C$ is countable since $\mathcal C \subseteq \mathcal B$ and $\mathcal B$ is countable.

- We now define an open subcover of $\mathcal F$. For each element $C \in \mathcal C$, take an element $A \in \mathcal F$ such that $C \subseteq A$. This is possible since $\mathcal C$ is a subcollection of the countable basis $\mathcal B$ and every element $A \in \mathcal F$ is the union of some collection of basis elements. Then $\mathcal F^*$ is countable, and moreover:

\begin{align} \quad X = \bigcup_{A \in \mathcal F^*} A \end{align}

- Hence every open cover $\mathcal F$ of $X$ has a countable subcover $\mathcal F^*$. Therefore, $X$ is Lindelöf. $\blacksquare$