Sequential Compactness is Preserved under Continuous Surjection

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

Proof

Let $T_A$ be a sequentially compact space.

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

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

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

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

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

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

Thus, $\sequence {x_n}$ has a subsequence that converges.

By definition, $T_B$ is sequentially compact.

$\blacksquare$