Continuous Image of Compact Space is Compact/Proof 2
Theorem
Let $T_1$ and $T_2$ be topological spaces.
Let $f: T_1 \to T_2$ be a continuous mapping.
If $T_1$ is compact then so is its image $f \sqbrk {T_1}$ under $f$.
That is, compactness is a continuous invariant.
Proof
Suppose $\UU$ is an open cover of $f \sqbrk {T_1}$ by sets open in $T_2$.
Because $f$ is continuous, it follows that $f^{-1} \sqbrk U$ is open in $T_1$ for all $U \in \UU$.
The set $\set {f^{-1} \sqbrk U: U \in \UU}$ is an open cover of $T_1$, because for any $x \in T_1$, it follows that $\map f x$ must be in some $U \in \UU$.
Because $T_1$ is compact, it has a finite subcover $\set {f^{-1} \sqbrk {U_1}, f^{-1} \sqbrk {U_2}, \ldots, f^{-1} \sqbrk {U_r} }$.
It follows that $\set {U_1, U_2, \ldots, U_r}$ is a finite subcover of $f \sqbrk {T_1}$.
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $5$: Compact spaces: $5.5$: Continuous maps on compact spaces: Proposition $5.5.1$
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- 1953: Walter Rudin: Principles of Mathematical Analysis ... (previous) ... (next): $4.14$