Sequential 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 sequentially compact, then $T_B$ is also sequentially compact.


Proof

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.

$\blacksquare$


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