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Revision difference : continuous image of a compact set is compact |
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| $\mathbf{Theorem:}$ The continuous image of a compact set is also compact. |
$\mathbf{Theorem:}$ The continuous image of a compact set is also compact. |
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| {\sl 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 $f(A)$. Since $\displaystyle f(A)\sub \bigcup_{\alpha\in I} V_\alpha$, then $\displaystyle A\sub f^{-1}\bigg( f(A) \bigg) \sub f^{-1} \left( \bigcup_{\alpha\in I} V_{\alpha} \right)=\bigcup_{\alpha\in I} f^{-1} (V_\alpha)$. |
{\sl 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 $f(A)$. Since $\displaystyle f(A)\sub \bigcup_{\alpha\in I} V_\alpha$, then $\displaystyle A\sub f^{-1}\bigg( f(A) \bigg) \sub f^{-1} \left( \bigcup_{\alpha\in I} V_{\alpha} \right)=\bigcup_{\alpha\in I} f^{-1} (V_\alpha)$. |
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| Since $f$ is continuous, each $f^{-1}(V_\alpha)$ is an open subset of $X$. Since $\displaystyle A\sub \bigcup_{\alpha\in I} f^{-1} (V_\alpha)$ and $A$ is compact, there exists $n \in \mathbb{N}$ with $\displaystyle A\sub\bigcup_{j=1}^n f^{-1} \left( V_{\alpha_j} \right)$ for some $\alpha_1, \dots , \alpha_n \in I$. Hence, $\displaystyle f(A)\sub f \left( \bigcup_{j=1}^n f^{-1} (V_{\alpha_j}) \right)=f\left( f^{-1} \left( \bigcup_{j=1}^n V_{\alpha_j} \right) \right) \sub \bigcup_{j=1}^n V_{\alpha_j}$. It follows that $f(A)$ is compact, QED. |
Since $f$ is continuous, each $f^{-1}(V_\alpha)$ is an open subset of $X$. Since $\displaystyle A\sub \bigcup_{\alpha\in I} f^{-1} (V_\alpha)$ and $A$ is compact, there exists $n \in \mathbb{N}$ with $\displaystyle A\sub\bigcup_{j=1}^n f^{-1} \left( V_{\alpha_j} \right)$ for some $\alpha_1, \dots , \alpha_n \in I$. Hence, $\displaystyle f(A)\sub f \left( \bigcup_{j=1}^n f^{-1} (V_{\alpha_j}) \right)=f\left( f^{-1} \left( \bigcup_{j=1}^n V_{\alpha_j} \right) \right) \sub \bigcup_{j=1}^n V_{\alpha_j}$. It follows that $f(A)$ is compact, QED. |
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