Sequentially Compact Metric Space is Lindelöf
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Theorem
Let $M = \struct {X, d}$ be a sequentially compact metric space.
Then $M$ is also a Lindelöf space.
That is, from every open cover of $M$, it is possible to extract a countable subcover.
Proof
Take any open cover $C$ of $M$.
We need to find a countable subset of $C$ which still covers $X$.
We have that a Sequentially Compact Metric Space is Second-Countable.
Thus, by definition, the topology of $M$ has a countable basis.
Let $\BB$ be a countable basis for the topology induced by $d$ of $M$.
Let $x \in X$.
As $C$ covers $X$:
- $\exists U_x \in C: x \in U_x$
As $\BB$ is a basis:
- $\exists B_x \in \BB: x \in B_x \subseteq U_x$
Thus:
- $(1): \quad \forall x \in X: \exists B_x \in \BB: x \in B_x$
Consider the set $\Sigma := \set {B_x: x \in X}$.
$\Sigma$ is a subset of $\BB$.
Hence $\Sigma$ is countable.
As $\Sigma$ contains every $x \in X$ from $(1)$, $\Sigma$ covers $X$.
By construction of $\Sigma$, every open set in $\Sigma$ is contained in some $U \in C$.
For each open set $B \in \Sigma$, choose one $U_B \in C$ such that $B \subseteq U_B$.
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Let $\UU$ be the set defined as:
- $\UU = \set {U_B: B \in \Sigma}$
As $\UU$ does not have more sets than $\Sigma$, $\UU$ is countable.
We have that:
- $\forall B \in \Sigma: B \subseteq U_B$
Thus it follows that $\UU$ covers $X$.
As:
- $\forall U_B \in \UU: U_B \in C$
$U$ is a countable subcover of $C$.
Thus a countable subcover has been obtained from $C$.
As $C$ is arbitrary, it follows that $M$ is a Lindelöf space.
$\blacksquare$