Lindelöf Property is Preserved under Continuous Surjection

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Theorem

Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.

Let $\phi: T_A \to T_B$ be a continuous surjection.


If $T_A$ is a Lindelöf space, then $T_B$ is also a Lindelöf space.


Proof

Let $T_A$ be a Lindelöf space.

Take an open cover $\mathcal U$ of $T_B$.

From Preimage of Cover is Cover, $\set {\phi^{-1} \sqbrk U: U \in \mathcal U}$ is a cover of $S_A$.

But $\phi$ is continuous, and for all $U \in \mathcal U$, $U$ is open in $T_B$.

It follows that $\forall U \in \mathcal U: \phi^{-1} \sqbrk U$ is open in $T_A$.

So $\set {\phi^{-1} \sqbrk U: U \in \mathcal U}$ is an open cover of $T_A$.

$T_A$ is Lindelöf , so we take a countable subcover:

$\set {\phi^{-1} \sqbrk {U_1}, \ldots, \phi^{-1} \sqbrk {U_n}, \ldots}$

We have that $\phi$ is surjective.

So from Surjection iff Right Inverse:

$\phi \sqbrk {\phi^{-1} \sqbrk A} = A$

So:

$\set {\phi \sqbrk {\phi^{-1} \sqbrk {U_1} }, \ldots, \phi \sqbrk {\phi^{-1} \sqbrk {U_n} }, \ldots} = \set {U_1, \ldots, U_n, \ldots} \subseteq \mathcal U$

is a countable subcover of $\mathcal U$ on $T_B$.

$\blacksquare$


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