Sigma-Compactness 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 $\sigma$-compact, then $T_B$ is also $\sigma$-compact.
Proof
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.
$\blacksquare$
Also see
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $3$: Compactness: Invariance Properties