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[parent] continuous image of a compact set is compact (Theorem)
Theorem   The continuous image of a compact set is also compact.
Proof. Let $ X$ and $ Y$ be topological spaces, $ f \colon X \to Y$ be continuous, $ A$ be a compact subset of $ X$, $ I$ be an indexing set, and $ \{V_\alpha\}_{\alpha\in I}$ be an open cover of $ f(A)$. Thus, $ \displaystyle f(A)\subseteq \bigcup_{\alpha\in I} V_\alpha$. Therefore, $ \displaystyle A\subseteq f^{-1}\bigg( f(A) \bigg) \subseteq f^{-1} \left( \bigcup_{\alpha\in I} V_{\alpha} \right)=\bigcup_{\alpha\in I} f^{-1} (V_\alpha)$.

Since $ f$ is continuous, each $ f^{-1}(V_\alpha)$ is an open subset of $ X$. Since $ \displaystyle A\subseteq \bigcup_{\alpha\in I} f^{-1} (V_\alpha)$ and $ A$ is compact, there exists % latex2html id marker 235 $ n \in \mathbb{N}$ with $ \displaystyle A\subseteq \bigcup_{j=1}^n f^{-1} \left( V_{\alpha_j} \right)$ for some $ \alpha_1, \dots , \alpha_n \in I$. Hence, $ \displaystyle f(A)\subseteq f \left( \bigcup_{j=1}^n f^{-1} (V_{\alpha_j}) \ri... ...cup_{j=1}^n V_{\alpha_j} \right) \right) \subseteq \bigcup_{j=1}^n V_{\alpha_j}$. It follows that $ f(A)$ is compact. $ \qedsymbol$



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See Also: compactness is preserved under a continuous map


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Cross-references: open subset, open cover, indexing set, compact subset, topological spaces, compact, compact set, image, continuous
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This is version 13 of continuous image of a compact set is compact, born on 2006-04-30, modified 2007-05-31.
Object id is 7888, canonical name is ContinuousImageOfACompactSetIsCompact.
Accessed 2311 times total.

Classification:
AMS MSC: 54D30 (General topology :: Fairly general properties :: Compactness)
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