Compactness Properties 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 $T_A$ has one of the following properties, then $T_B$ has the same property:

Compact Space
$\sigma$-Compact Space
Countable Compact Space
Sequential Compact Space
Lindelöf Space


Proof

Proof for Compactness

Let $T_A$ be compact.

Take an open cover $\UU$ of $T_B$.

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

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

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

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


$T_A$ is compact, so we take a finite subcover:

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

We have that $\phi$ is surjective.

So from Surjection iff Right Inverse, $\phi \sqbrk {\phi^{-1} \sqbrk A} = A$

So:

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

is a finite subcover of $\UU$ on $T_B$.

$\blacksquare$


Proof for $\sigma$-Compactness

Let $T_A$ be $\sigma$-compact.

Then:

$\ds S_A = \bigcup_{i \mathop = 1}^\infty S_i$

where $S_i \subseteq S_A$ are compact.

Since $\phi$ is surjective, we have from Image of Union under Relation:

$\ds \phi \sqbrk {S_A} = S_B = \phi \sqbrk {\bigcup_{i \mathop = 1}^\infty S_i} = \bigcup_{i \mathop = 1}^\infty \phi \sqbrk {S_i}$

From Compactness is Preserved under Continuous Surjection, we have that $\phi \sqbrk {S_i}$ is compact for all $i \in \N$.

So $S_B$ is the union of a countable number of compact subsets.

Thus, by definition, $T_B$ is also $\sigma$-compact.

$\blacksquare$


Proof for Countable Compactness

Let $T_A$ be countably compact.

Take a countable open cover $\UU$ of $T_B$.

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

$\VV$ is a countable cover because there it is bijective with $\UU$.


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By hypothesis, $\phi$ is continuous.

For all $U \in \UU$, $U$ is open in $T_B$.

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

So $\VV$ is a countable open cover of $T_A$.

$T_A$ is countably compact, so we take a finite subcover:

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

We have that $\phi$ is surjective.

So from Surjection iff Right Inverse:

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

So:

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

is a finite subcover of $\UU$ on $T_B$.


This article, or a section of it, needs explaining.
In particular: It is not clear that $\set {U_1, \ldots, U_n}$ covers $T_B$.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.
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$\blacksquare$


Proof for Sequential Compactness

Let $T_A$ be a sequentially compact space.

Take an infinite sequence $\sequence {x_n} \subseteq S_B$.

From the surjectivity of $\phi$, there exists another infinite sequence $\sequence {y_n} \subseteq S_A$ such that $\map \phi {y_n} = x_n$.

By the definition of sequential compactness, there exists a subsequence of $\sequence {y_n}$

Let this subsequence be named $\sequence {y_{n_k} }$.

Let $\sequence {y_{n_k} }$ converge to $y \in S_A$ with respect to $T_A$.

From the continuity of $\phi$, it is concluded that $\map \phi {y_{n_k} } = x_{n_k}$ converges to $\map \phi y \in S_B$.

Thus, $\sequence {x_n}$ has a subsequence that converges.

By definition, $T_B$ is sequentially compact.

$\blacksquare$


Proof for Lindelöf Property

Let $T_A$ be a Lindelöf space.

Take an open cover $\UU$ of $T_B$.

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

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

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

So $\set {\phi^{-1} \sqbrk U: U \in \UU}$ 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:

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

is a countable subcover of $\UU$ on $T_B$.

$\blacksquare$


Sources

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