# Counterexamples in metric spaces about density

I came across those questions on my school's topological dynamical-system textbook and I had totally no idea about them.

(a) There exist a compact metric space $$X$$, a homeomorphism $$f:X\rightarrow X$$ and a point $$x_0\in X$$, such that $$\text{Orb}_f(x_0)={\{f^k(x_0):k\in\mathbb{Z}\}}$$ is dense in $$X$$ while $${\{f^k(x):k\in\mathbb{Z}^+\}}$$ and $${\{f^k(x):k\in\mathbb{Z}^-\}}$$ are not dense in $$X$$ for any $$x\in X$$.

(b) There exist a metric space $$X$$ with countable many open dense subset $$\{U_i\}$$, s.t. $$\bigcap_{i}U_i=\varnothing$$.

Thanks to Henno Brandsma, now we have (b):

Consider $$X=\mathbb{Q}(=\{r_i\}_{-\infty}^{+\infty})$$ as the subspace of $$\mathbb{E}^1$$ and $$U_i=X-\{r_i\}$$. Clearly, every $$U_i$$ is open since $$U^c_i=\{r_i\}$$ is a point set which is closed. To see the density of $$U_i$$, it is sufficient to show that $$U_i$$ is not closed. In fact, sequence $$\{r_i+\frac{1}{n}\}\subset X$$ converges to $$\{r_i\}\notin U_i$$, which implies that $$U_i$$ is not closed.

EDITED! Sorry I've given the (a) incorrectly and now it is clarified. The first version is wrong because that $$\bigcup_{x\in X}{\{f^k(x):k\in\mathbb{Z}^-\}}=X$$ and $$\bigcup_{x\in X}\text{Orb}_f(x)=X$$ always holds.

(b)$$^\prime$$ Determine whether it is possible that there exist a metric space $$X$$ with countable many open dense subset $$\{U_i\}$$, such that for any $$i\neq j$$, $$U_i\cap U_j=\varnothing$$ holds.

Just now I figured out (b)$$^\prime$$:

If for any $$i\neq j$$, $$U_i\cap U_j=\varnothing$$, we will have $$U_i\subset X-U_j$$. Then $$\overline{U_i}\subset X-U_j$$ because $$X-U_j$$ is closed. Clearly $$X-U_j$$ is a proper subset of $$X$$, which implies $$\overline{U_i}\neq X$$.

• B: the rationals will give an example. – Henno Brandsma May 14 at 5:13
• @HennoBrandsma How can rationals open – Lau May 14 at 8:06
• In $X=\mathbb{Q}$ all sets of the form $\mathbb{Q} \setminus \{q\}$ are open and dense. – Henno Brandsma May 14 at 8:09
• They're dense as no singleton $\{q\}$ in $\Bbb Q$ is isolated. – Henno Brandsma May 14 at 9:02
• As to the new (b)': if $U_i$ is (open and) dense and $U_j$ is open (non-empty) we must have $U_i \cap U_j \neq \emptyset$. So any pair of non-empty open dense sets intersects (and the intersection is again dense and open). So even 2 sets fails, and this holds in any topological space (no metric needed). – Henno Brandsma May 14 at 11:50

Let $$X = \{-1, 1 \} \cup \{ \frac{n}{1+ \lvert n \rvert} : n \in \mathbb Z \}$$ with the subspace topology inherited from $$\mathbb R$$. Let $$x_n = \frac{n}{1+\lvert n \rvert}$$. Define $$f : X \to X, f(-1) = -1, f(1) = 1, f(x_n) = x_{n+1}$$ for $$n \in \mathbb Z$$. This is a homeomorphism. You have $$f^k(x_0) = x_k$$. Thus $$\text{Orb}_f(x_0) = \{ \frac{n}{1+\lvert n \rvert} : n \in \mathbb Z \}$$ which is dense in $$X$$ and $${\{f^k(x):k\in\mathbb{Z}^\pm\}} = \{ \frac{n}{1+\lvert n \rvert} : n \in \mathbb Z^\pm \}$$ which are not dense in $$X$$.
• Why Orb$_f(x_0)$ is dense in $X$? I think $\{1\}$ is isolated because $\frac{n}{1+n^2}\leq\frac{1}{2}$ – Lau May 14 at 10:54
• @Lau the example has been corrected. And we need isolated points anyway (all the $x_n$ are) because if $X$ is compact metric and without isolated points and there is a point with a dense full orbit, there is also a point with a dense forward orbit. – Henno Brandsma May 14 at 11:59
• You are right, I had to correct the definition so that $x_n \to 1$ as $n \to \infty$ and $x_n \to -1$ as $n \to -\infty$. – Paul Frost May 14 at 12:00