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Let $f:[a,b] \rightarrow \mathbb{R}$ be a Lipschitz function. I want to show that it carries $F_\sigma$ sets to $F_\sigma$ sets.

I'm not sure how to demonstrate this. Specifically I'm not sure what property of continuity or Lipschitz would preserve the $F_\sigma$ property. I do know that this is true: $f(\bigcup_{i} A_i)=\bigcup_{i}f(A_i)$.

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What is an $F_{\sigma}$ set? –  Giovanni De Gaetano Oct 4 '12 at 10:47
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A countable union of closed sets. –  abet Oct 4 '12 at 10:48
    
Thank you! And for editing as well! –  Giovanni De Gaetano Oct 4 '12 at 10:49

1 Answer 1

up vote 6 down vote accepted

Hint: A closed set $K \subseteq [a,b]$ is compact (as $[a,b]$ is). Hence its image $f[K]$ under the continuous function $f$ is compact also.

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Does it just immediately follow that $f(\bigcup_{i} K_i)=\bigcup_{i}f(K_i)$ since subsets of compact sets are compact? –  abet Oct 4 '12 at 11:00
    
You said you know about this. It holds $x \in f[\bigcup_i K_i] \iff \exists y \in \bigcup_i K_i. f(y) = x $ $\iff \exists i\exists y \in K_i. f(y) = x \iff \exists i. x \in f[K_i] \iff x\in \bigcup_i f[K_i]$. –  martini Oct 4 '12 at 11:08
    
I think it was just stated as a definition in Principles of Analysis by Rudin. I didn't know (or most likely cannot recall) that it works for compactness. –  abet Oct 4 '12 at 11:14
    
The statement $f[\bigcup_i K_i] = \bigcup_i f[K_i]$ has nothing to do with the compactness of the $K_i$. It just follows from the definition of union and image. –  martini Oct 4 '12 at 11:24
    
I'm sorry...I don't follow. Compact sets are closed and bounded and preserved by a continuous function f. However, $F_\sigma$ just have the property of being a countable union of closed sets. –  abet Oct 4 '12 at 11:32

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