Closed Subspace of Lindelöf Space is Lindelöf Space

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Theorem

Let $T = \struct {S, \tau}$ be a Lindelöf space.

Let $C = \struct {H, \tau_H}$ be a subspace of $T$.

Let $C$ be closed in $T$.

Then $\struct {H, \tau}$ is Lindelöf space.

That is, the property of being Lindelöf is weakly hereditary.

Proof

Let $T$ be a Lindelöf space.

Let $C$ be a closed subspace of $T$.

Let $\UU$ be an open cover of $H$.

We have that $H$ is closed in $T$.

It follows by definition of closed that $H \setminus C$ is open in $T$.

So if we add $S \setminus H$ to $\UU$, we see that $\UU \cup \set {S \setminus H}$ is also an open cover of $S$.

As $T$ is compact, there exists a countable subcover of $\UU \cup \set {S \setminus H}$, say $\VV = \set {U_1, U_2, \ldots}$.

This covers $H$ by the fact that it covers $S$.

Suppose $S \setminus H$ is an element of $\VV$.

Then $S \setminus H$ may be removed from $\VV$, and the rest of $\VV$ still covers $H$.

Thus we have a countable subcover of $\UU$ which covers $H$.

Hence $C$ is a Lindelöf space.

$\blacksquare$

Also see

• Closed Subspace of Compact Space is Compact

Sources

• 1970: Stephen Willard: General Topology: Chapter $4$: Separation and Countability: $\S16$: Countability Properties: Theorem $16.6(\text b)$