# Compactness Properties Preserved under Continuous Surjection

## Contents

## Theorem

Let $T_A = \left({S_A, \tau_A}\right)$ and $T_B = \left({S_B, \tau_B}\right)$ 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 $\mathcal U$ of $T_B$.

From Preimage of Cover is Cover, $\left\{{\phi^{-1} \left({U}\right): U \in \mathcal U}\right\}$ 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 $\forall U \in \mathcal U: \phi^{-1} \left({U}\right)$ is open in $T_A$.

So $\left\{{\phi^{-1} \left({U}\right):\ U \in \mathcal U}\right\}$ is an open cover of $T_A$.

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

- $\left\{{\phi^{-1} \left({U_1}\right), \ldots, \phi^{-1} \left({U_n}\right)}\right\}$

We have that $\phi$ is surjective.

So from Surjection iff Right Inverse, $\phi \left({\phi^{-1} \left({A}\right)}\right) = A$

So:

- $\left\{{\phi\left({\phi^{-1}\left({U_1}\right)}\right), \ldots, \phi \left({\phi^{-1} \left({U_n}\right)}\right)}\right\} = \left\{{U_1, \ldots, U_n}\right\} \subseteq \mathcal U$

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

$\blacksquare$

### Proof for $\sigma$-Compactness

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

Then:

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

where $S_i \subseteq S_A$ are compact.

Since $\phi$ is surjective:

- $\displaystyle \phi \left({S_A}\right) = S_B = \phi\left({\bigcup_{i \mathop = 1}^\infty S_i}\right) = \bigcup_{i \mathop = 1}^\infty \phi(S_i)$ using the proven theorem Image of Union under Relation.

From Compactness is Preserved under Continuous Surjection, we have that $\phi \left({S_i}\right)$ 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 $\mathcal U$ of $T_B$.

From Preimage of Cover is Cover, $\mathcal V := \left\{{\phi^{-1} \left({U}\right): U \in \mathcal U}\right\}$ is a cover of $S_A$.

$\mathcal V$ is a countable cover because there it is bijective with $\mathcal U$.

By hypothesis, $\phi$ is continuous.

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

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

So $\mathcal V$ is a countable open cover of $T_A$.

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

- $\left\{{\phi^{-1} \left({U_1}\right), \ldots, \phi^{-1} \left({U_n}\right)}\right\}$

We have that $\phi$ is surjective.

So from Surjection iff Right Inverse:

- $\phi \left({\phi^{-1} \left({A}\right)}\right) = A$

So:

- $\left\{{\phi\left({\phi^{-1}\left({U_1}\right)}\right), \ldots, \phi \left({\phi^{-1} \left({U_n}\right)}\right)}\right\} = \left\{{U_1, \ldots, U_n}\right\} \subseteq \mathcal U$

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

{{explain|It is not clear that $\left\{{U_1, \ldots, U_n}\right\}$ covers $T_B$.}}

$\blacksquare$

### Proof for Sequential Compactness

Let $T_A$ be a sequentially compact space.

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

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

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

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

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

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

Thus, $\left\{{x_n}\right\}$ 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 $\mathcal U$ 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 $T_B$.

$\blacksquare$

## Sources

- 1970: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*... (previous) ... (next): $\text{I}: \ \S 3$: Invariance Properties