# Showing that $f$ is topologically transitive if every closed invariant subset has empty interior

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?

• 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
• 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

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.