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Let $X$ be a compact metric space and $f \colon X \to X$ be continuous and onto. If every closed, invariant proper subset of $X$ has empty interior, then show that $f$ is topologically transitive.

My attempt:

Suppose $f$ is not topologically transitive. Then there exist non-empty open sets $U$ and $V$ such that $$f^n(U)\cap V=\emptyset\tag{1}$$ for all $n \in \mathbb{N}.$ Let $$G=\bigcup_{n \in \mathbb N}f^n(U)$$ and $$C=\overline{G}$$

Then $C$ is closed, proper(by $(1)$) and invariant. Then by given condition, $$C^{\mathrm{o}}=\emptyset$$

I'm not able to get proceed from here. Any hints?

Sahiba Arora
  • 10,847
  • Maybe consider $\bigcup_n f^{-n} (V)$? That's open (since $f$ is continuous) and disjoint from $U$. Then show that its adherence is invariant (compactness will be useful) and still disjoint from $U$. Conclude. – D. Thomine Sep 22 '17 at 16:07
  • @D.Thomine I know of that proof. I want to know if what I'm doing is right. Or maybe slightly tweaking my choice of $G$ helps. – Sahiba Arora Sep 25 '17 at 17:37
  • Please disregard my previous comment, it is a mistake. Your choice of $C$ works. In addition, $U \subset G \subset C$, so $C$ has non-empty interior, and you can conclude. – D. Thomine Sep 25 '17 at 23:08
  • @D.Thomine So, I should include $U$ in my $G$, right? Basically take $G=\bigcup_{n\geq 0} f^n(U)$? – Sahiba Arora Sep 26 '17 at 14:16
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    For me, $\mathbb{N} = {0, 1, \ldots}$, so I thought it already was the case. Yes, you should include the case $n=0$ (because it's possible that $f^n (U)$ has empty interior for $n > 0$). – D. Thomine Sep 26 '17 at 18:27
  • @D.Thomine If I include $U$ in $G$, then I won't be able to conclude that $G$ is not dense and hence say that $C$ is proper. Since it may happen that $U \cap V \neq \emptyset.$ – Sahiba Arora Dec 02 '17 at 22:56
  • AFAIK, topological transitivity says that for all $U$, $V$ non-empty and open, there exists $n \geq 0$ such that $f^n (U) \cap V \neq \emptyset$. Taking the negation, you can assume that $f^n (U) \cap V = \emptyset$ for all $n \geq 0$, and in particular $U \cap V = \emptyset$. – D. Thomine Dec 03 '17 at 00:17
  • A last comment: even is your sketch of a proof is sound, you can improve it. You begin by trying to prove the contrapositive, and then switch at the last moment to proof ad absurdum. Your proof will be clearer if you conclude directly with the contrapositive ("...and thus there exists a closed invariant set with non empty interior."), or even try to prove the statement directly. – D. Thomine Dec 03 '17 at 00:20
  • The definition I'm following requires the existence of $n >0$ for transitivity. – Sahiba Arora Dec 03 '17 at 00:24

1 Answers1

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We first show that every non-empty, proper, negative $f-$invariant set is dense in $X.$

Let $U$ non-empty, proper, negative $f-$invariant set in $X.$ So $X\setminus U$ is non-empty, proper, closed and invariant. By hypothesis, it has empty interior. It follows that $U$ is dense in $X.$

Now, let $U$ and $V$ be any non-empty, open subsets of $X.$

Then $\bigcup_{k \geq 1} f^{-k}(V)$ is non-empty, proper, negative $f-$invariant set and hence, dense in $X.$ Thus, there exists $n \geq 1$ such that $$f^n(U)\cap V \neq \emptyset.$$

Thus, $f$ is topologically transitive.

Sahiba Arora
  • 10,847