Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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: $f(\bigcup_{i} A_i)=\bigcup_{i}f(A_i)$.

share|improve this question
What is an $F_{\sigma}$ set? –  Giovanni De Gaetano Oct 4 '12 at 10:47
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.

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.