Lindelöf Theorem

Lindelöf and Countably Compact Topological Spaces

If $X$ is a topological space and $A\subseteq X$ , then $A$ is said to be compact in $X$ if every open cover of $A$ has a finite subcover.

Moreover, we said that $X$ is a compact topological space if every open cover of $X$ has a finite subcover.

We will now look at two similar definitions.

Definition: A topological space $X$ is said to be Lindelöf if every open cover of $X$ has a countable subcover. $X$ is said to be Countably Compact if every countable open cover of $X$ has a finite subcover.

It should be noted that the Lindelöf and countably compact property are weaker than the compactness property. If $X$ is a compact topological space then $X$ is also Lindelöf and countably compact.

For example, since $[0,1]$ is a compact topological space (with the subspace topology from the usual topology on $\mathbb {R}$ ), then by extension, $[0,1]$ is both Lindelöf and countably compact.

Of course, there exists topological spaces which are not compact but are still Lindelöf or countably compact.

For another example, consider the set of natural numbers $\mathbb {N}$ with the discrete topology, i.e., every subset of $\mathbb {N}$ is open. Then $\mathbb {N}$ is not compact, because of the following open cover of $\mathbb {N}$ :

{\mathcal {F}}=\{\{n\}:n\in \mathbb {N} \}=\{\{1\},\{2\},...,\{n\},...\}

Clearly there does not exist any subcollection ${\mathcal {F}}^{*}\subseteq {\mathcal {F}}$ that is finite and still covers $\mathbb {N}$ ! So $\mathbb {N}$ is not compact.

However, $\mathbb {N}$ is Lindelöf. To show this, let ${\mathcal {F}}$ be any open cover of $\mathbb {N}$ . Then we can choose a countable collection of sets from ${\mathcal {F}}$ which also cover $\mathbb {N}$ since each subset of ${\mathcal {F}}$ has cardinality greater than or equal to $1$ .

Unfortunately, $\mathbb {N}$ is not countably compact if we use the example open cover which showed that $\mathbb {N}$ was not compact.

So as we can see, the concept of compactness, Lindelöfness, and countable compactness are different properties.

The Lindelöf Lemma

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 <a href="/second-countable-topological-spaces">Second Countable Topological Spaces</a> 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:

${\begin{aligned}\quad A=\bigcup _{B\in {\mathcal {B}}_{A}}B\end{aligned}}$

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

Failed to parse (unknown function "\begin{align}"): {\displaystyle \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: