Let $X$ be a compact topological space with continuous map $f\colon X\to X$. $Y:=\bigcap_{n\in\mathbb{N}}f^n(X)\subseteq X$ is closed and hence compact itself.

Assume $\overline{A}=Y$, i.e. $A$ is be dense in $Y$.

Does this imply $\overline{A}=X$, i.e. that $A$ is dense in $X$?

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    $\begingroup$ Did you instead mean $Y \subset \overline{A}$? $\endgroup$ – B. Mehta May 11 '18 at 13:05
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    $\begingroup$ You can always consider $A=Y\neq X$. So $A$ is trivially dense in $Y$. If $Y$ is closed then $\bar A=Y\neq X$. $\endgroup$ – user126154 May 11 '18 at 13:24

No it does not. For a counterexample take $Y = [0, 1]$ and $X = [-1, 2]$ as subsets of $\mathbb{R}$ (where $\mathbb{R}$ has the standard topology). Note that $X$ is compact and $Y \subseteq X$ and $Y$ is closed and thus also compact. Then let $A = (0, 1) \subseteq Y \subseteq X$.

Then we have $\overline{A} = Y$ and certainly $\overline{A} = [0, 1] \neq [-1, 2] = X$.

  • $\begingroup$ @Randall Sorry I edited my answer $\endgroup$ – Perturbative May 11 '18 at 13:16
  • $\begingroup$ Please see the edit of my question. $\endgroup$ – Rhjg May 11 '18 at 15:35
  • $\begingroup$ @Rhjg You have COMPLETELY changed the question, voiding everyone's work in answering your first question. $\endgroup$ – Randall May 11 '18 at 17:46
  • $\begingroup$ @Randall if you like you can change the question to its original version again and I will ask this new question in another thread $\endgroup$ – Rhjg May 11 '18 at 17:54
  • $\begingroup$ @Rhjg Please do so $\endgroup$ – Perturbative May 12 '18 at 11:22

If $Y$ is a proper subset of $X$, so $X \neq Y$, then $A$ is certainly not dense in $X$. The closure cannot both be $Y$ and $X$ at the same time...

"$A$ dense in $Y$" implies $A \subseteq Y$ and so $\overline{A} \subseteq \overline{Y} = Y \neq X$.

  • $\begingroup$ Please see the edit of my question. I specified Y. $\endgroup$ – Rhjg May 11 '18 at 15:35
  • $\begingroup$ @Rhjg It doesn't alter my answer. $Y$ is stil closed. What is $A$ precisely, a subset of $Y$? $\endgroup$ – Henno Brandsma May 11 '18 at 16:40
  • $\begingroup$ Let $x\in X$ be fixed. $A$ is the set of all backward orbits $\{x^{-i}: i\geq 0\}$ where $x^{-i}\in f^{-i}(x)$, $x^0:=x$ and $x^{-i}=f(x^{-(i+1)})$. $\endgroup$ – Rhjg May 11 '18 at 17:36

Lemma: Given $A\subseteq Y \subseteq X$, where $X$ is a topological space and $Y$ is endowed with the sub-space topology. Then the closure of $A$ in $Y$ is the intersection with $Y$ of the closure of $A$ in $X$: $$\overline{A}^Y=\overline{A}^X\cap Y$$

From this lemma it follows that we can speak about $\overline{A}$ without confusion.

If $Y\neq X$ is closed in $X$, then $\overline A\subseteq Y\neq X$ so no subset of $Y$ can be dense in $X$.

Proof of Lemma: $\overline{A}^X\cap Y$ is closed in $Y$, hence $\overline{A}^Y\subseteq \overline{A}^X\cap Y$ (because $\overline{A}^Y$ is the smallest closed subet of $Y$ containing $A$).

On the other hand, sinche $\overline{A}^Y$ is closed in $Y$, and $Y$ is endowed with the subspace topology, then there is a closed subset $C$ of $X$ such that $\overline{A}^Y=C\cap Y$. As above, since $C$ is closed in $X$, we have $\overline{A}^X\subseteq C$, whence $\overline{A}^X\cap Y\subseteq C\cap Y=\overline{A}^Y$.

Since $\overline{A}^Y\subseteq \overline{A}^X\cap Y\subseteq \overline{A}^Y$, both inclusions are in fact equalities. $\square$


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