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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.

Additional Question:

(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 $\overline{U_i}\neq X$.

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  • $\begingroup$ B: the rationals will give an example. $\endgroup$ – Henno Brandsma May 14 at 5:13
  • $\begingroup$ @HennoBrandsma How can rationals open $\endgroup$ – Lau May 14 at 8:06
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    $\begingroup$ In $X=\mathbb{Q}$ all sets of the form $\mathbb{Q} \setminus \{q\}$ are open and dense. $\endgroup$ – Henno Brandsma May 14 at 8:09
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    $\begingroup$ They're dense as no singleton $\{q\}$ in $\Bbb Q$ is isolated. $\endgroup$ – Henno Brandsma May 14 at 9:02
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    $\begingroup$ 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). $\endgroup$ – Henno Brandsma May 14 at 11:50
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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 $X$.

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  • $\begingroup$ Why Orb$_f(x_0)$ is dense in $X$? I think $\{1\}$ is isolated because $\frac{n}{1+n^2}\leq\frac{1}{2}$ $\endgroup$ – Lau May 14 at 10:54
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    $\begingroup$ @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. $\endgroup$ – Henno Brandsma May 14 at 11:59
  • $\begingroup$ many thanks again $\endgroup$ – Lau May 14 at 12:00
  • $\begingroup$ 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 $n \to -\infty$. $\endgroup$ – Paul Frost May 14 at 12:00

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