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 compact, then $T_B$ is also compact.


Proof

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 $T_B$.

$\blacksquare$


Also see


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