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'continuous image of a compact set is compact'
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| Title of object: |
continuous image of a compact set is compact |
| Canonical Name: |
ContinuousImageOfACompactSetIsCompact |
| Type: |
Theorem |
| Created on: |
2006-04-30 10:48:50 |
| Modified on: |
2007-05-30 03:54:12 |
| Classification: |
msc:54D30 |
Preamble:
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
\def\R{\mathbb{ R}}
\def\N{\mathbb{ N}}
\def\Z{\mathbb{ Z}}
\def\K{\mathbb{ K}}
\def\sub{\subseteq}
\def\bcap{\bigcap}
\def\bcup{\bigcup}
\def\empty{\varnothing}
\def\imp{\rightarrow} |
Content:
$\mathbf{Theorem:}$ The continuous image of a compact set is also compact.
{\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 $f(A)$. Thus, $\displaystyle f(A)\sub \bigcup_{\alpha\in I} V_\alpha$. Therefore, $\displaystyle A\sub f^{-1}\bigg( f(A) \bigg) \sub f^{-1} \left( \bigcup_{\alpha\in I} V_{\alpha} \right)=\bigcup_{\alpha\in I} f^{-1} (V_\alpha)$.
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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