The continuous image of a compact space is compact

Let $𝑋, 𝑌 \in 𝖳 𝗈 𝗉$ and $𝑓 \in 𝖳 𝗈 𝗉 (𝑋, 𝑌)$ . If $𝑋$ is compact, so is $𝑓 (𝑋)$ . #m/thm/topology

Proof

Without loss of generality, consider a continuous surjection $𝑓 : 𝑋 ↠ 𝑌$ . Let ${𝑈 𝛼} 𝛼 \in 𝐼$ be an open cover of $𝑌$ . It follows that ${𝑓 - 1 𝑈 𝛼} 𝛼 \in 𝐼$ is an open cover of $𝑋$ with a finite subcover ${𝑓 - 1 𝑈 𝛼 𝑛} 𝑚 𝑛 = 1$ . Therefore ${𝑓 𝑓 - 1 𝑈 𝛼 𝑛} 𝑚 𝑛 = 1 = {𝑈 𝛼 𝑛} 𝑚 𝑛 = 1$ is a finite subcover of $𝑌$ .

It follows that compactness is a Topological property.

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