Sigma-Compactness is Preserved under Continuous Surjection

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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 $T_A$ is $\sigma$-compact, then $T_B$ is also $\sigma$-compact.


Proof

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.

$\blacksquare$


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