Is this set perfect?
$begingroup$
Let $X$ be a compact and Hausdorff space and $f:Xto X$ is a homeomorphism.
Let for all $ineq j$ and $pin X$, we have $f^i(p)neq f^j(p)$. I need to know that whether $overline{{f^n(p)}_{ninmathbb{Z}}}$ is uncountable or not. I think that it is sufficient to $overline{{f^{n}(p)}_{ninmathbb{Z}}}$ is an infinite and closed without isolated point i.e. it is a perfect set. It is clear that it is infinite and close. Let $p$ be an isolated point, this implies that there is an open set $U$ such that $U={p}$, hence ${f^n(U)}_{ninmathbb{Z}}$ is an open cover for compact set $overline{{f^{n}(p)}_{ninmathbb{Z}}}$. Thus there is $m>0$ with $overline{{f^{n}(p)}_{ninmathbb{Z}}}subseteq cup_{n=0}^{m}f^n(U)$, that is a contradiction.
Is my proof true?
Please help me
general-topology dynamical-systems
$endgroup$
add a comment |
$begingroup$
Let $X$ be a compact and Hausdorff space and $f:Xto X$ is a homeomorphism.
Let for all $ineq j$ and $pin X$, we have $f^i(p)neq f^j(p)$. I need to know that whether $overline{{f^n(p)}_{ninmathbb{Z}}}$ is uncountable or not. I think that it is sufficient to $overline{{f^{n}(p)}_{ninmathbb{Z}}}$ is an infinite and closed without isolated point i.e. it is a perfect set. It is clear that it is infinite and close. Let $p$ be an isolated point, this implies that there is an open set $U$ such that $U={p}$, hence ${f^n(U)}_{ninmathbb{Z}}$ is an open cover for compact set $overline{{f^{n}(p)}_{ninmathbb{Z}}}$. Thus there is $m>0$ with $overline{{f^{n}(p)}_{ninmathbb{Z}}}subseteq cup_{n=0}^{m}f^n(U)$, that is a contradiction.
Is my proof true?
Please help me
general-topology dynamical-systems
$endgroup$
add a comment |
$begingroup$
Let $X$ be a compact and Hausdorff space and $f:Xto X$ is a homeomorphism.
Let for all $ineq j$ and $pin X$, we have $f^i(p)neq f^j(p)$. I need to know that whether $overline{{f^n(p)}_{ninmathbb{Z}}}$ is uncountable or not. I think that it is sufficient to $overline{{f^{n}(p)}_{ninmathbb{Z}}}$ is an infinite and closed without isolated point i.e. it is a perfect set. It is clear that it is infinite and close. Let $p$ be an isolated point, this implies that there is an open set $U$ such that $U={p}$, hence ${f^n(U)}_{ninmathbb{Z}}$ is an open cover for compact set $overline{{f^{n}(p)}_{ninmathbb{Z}}}$. Thus there is $m>0$ with $overline{{f^{n}(p)}_{ninmathbb{Z}}}subseteq cup_{n=0}^{m}f^n(U)$, that is a contradiction.
Is my proof true?
Please help me
general-topology dynamical-systems
$endgroup$
Let $X$ be a compact and Hausdorff space and $f:Xto X$ is a homeomorphism.
Let for all $ineq j$ and $pin X$, we have $f^i(p)neq f^j(p)$. I need to know that whether $overline{{f^n(p)}_{ninmathbb{Z}}}$ is uncountable or not. I think that it is sufficient to $overline{{f^{n}(p)}_{ninmathbb{Z}}}$ is an infinite and closed without isolated point i.e. it is a perfect set. It is clear that it is infinite and close. Let $p$ be an isolated point, this implies that there is an open set $U$ such that $U={p}$, hence ${f^n(U)}_{ninmathbb{Z}}$ is an open cover for compact set $overline{{f^{n}(p)}_{ninmathbb{Z}}}$. Thus there is $m>0$ with $overline{{f^{n}(p)}_{ninmathbb{Z}}}subseteq cup_{n=0}^{m}f^n(U)$, that is a contradiction.
Is my proof true?
Please help me
general-topology dynamical-systems
general-topology dynamical-systems
edited Dec 15 '18 at 9:29
Jean Marie
30.5k42154
30.5k42154
asked Dec 15 '18 at 9:05
user479859user479859
887
887
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There are several problems with your proof:
- You wrote “Let $p$ be an isolated point”, but $p$ here already has another meaning.
- Then you write “this implies that there is an open set $U$ such that $U={p}$”. This is wrong. What it implies is that there is an open subset $U$ of $X$ such that $overline{{f^n(p),|,ninmathbb{Z}}}cap U={p}$.
$endgroup$
$begingroup$
@ José Carlos Santos. Thanks. I think that $overline{{f^n(p)}_{ninmathbb{Z}}}$ is an uncountable set. In above proof $f$ is a homeomorphism, hence it is an open map and $f(overline{U})= overline{f(U)}$.
$endgroup$
– user479859
Dec 15 '18 at 9:28
$begingroup$
I've deleted my third remark.
$endgroup$
– José Carlos Santos
Dec 15 '18 at 9:30
$begingroup$
Let $overline{{f^n(p)| ninmathbb{Z}}}cap U={p}$.
$endgroup$
– user479859
Dec 15 '18 at 9:31
$begingroup$
If $overline{f^n(p)| ninmathbb{Z}}}cap U={ f^k(p)}$, in above proof we work with $f^k(p)$.
$endgroup$
– user479859
Dec 15 '18 at 9:32
$begingroup$
What the means of "$ overline{{f^n(p)|ninmathbb{Z}}}$ has an isolated point"?
$endgroup$
– user479859
Dec 15 '18 at 9:35
add a comment |
Your Answer
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There are several problems with your proof:
- You wrote “Let $p$ be an isolated point”, but $p$ here already has another meaning.
- Then you write “this implies that there is an open set $U$ such that $U={p}$”. This is wrong. What it implies is that there is an open subset $U$ of $X$ such that $overline{{f^n(p),|,ninmathbb{Z}}}cap U={p}$.
$endgroup$
$begingroup$
@ José Carlos Santos. Thanks. I think that $overline{{f^n(p)}_{ninmathbb{Z}}}$ is an uncountable set. In above proof $f$ is a homeomorphism, hence it is an open map and $f(overline{U})= overline{f(U)}$.
$endgroup$
– user479859
Dec 15 '18 at 9:28
$begingroup$
I've deleted my third remark.
$endgroup$
– José Carlos Santos
Dec 15 '18 at 9:30
$begingroup$
Let $overline{{f^n(p)| ninmathbb{Z}}}cap U={p}$.
$endgroup$
– user479859
Dec 15 '18 at 9:31
$begingroup$
If $overline{f^n(p)| ninmathbb{Z}}}cap U={ f^k(p)}$, in above proof we work with $f^k(p)$.
$endgroup$
– user479859
Dec 15 '18 at 9:32
$begingroup$
What the means of "$ overline{{f^n(p)|ninmathbb{Z}}}$ has an isolated point"?
$endgroup$
– user479859
Dec 15 '18 at 9:35
add a comment |
$begingroup$
There are several problems with your proof:
- You wrote “Let $p$ be an isolated point”, but $p$ here already has another meaning.
- Then you write “this implies that there is an open set $U$ such that $U={p}$”. This is wrong. What it implies is that there is an open subset $U$ of $X$ such that $overline{{f^n(p),|,ninmathbb{Z}}}cap U={p}$.
$endgroup$
$begingroup$
@ José Carlos Santos. Thanks. I think that $overline{{f^n(p)}_{ninmathbb{Z}}}$ is an uncountable set. In above proof $f$ is a homeomorphism, hence it is an open map and $f(overline{U})= overline{f(U)}$.
$endgroup$
– user479859
Dec 15 '18 at 9:28
$begingroup$
I've deleted my third remark.
$endgroup$
– José Carlos Santos
Dec 15 '18 at 9:30
$begingroup$
Let $overline{{f^n(p)| ninmathbb{Z}}}cap U={p}$.
$endgroup$
– user479859
Dec 15 '18 at 9:31
$begingroup$
If $overline{f^n(p)| ninmathbb{Z}}}cap U={ f^k(p)}$, in above proof we work with $f^k(p)$.
$endgroup$
– user479859
Dec 15 '18 at 9:32
$begingroup$
What the means of "$ overline{{f^n(p)|ninmathbb{Z}}}$ has an isolated point"?
$endgroup$
– user479859
Dec 15 '18 at 9:35
add a comment |
$begingroup$
There are several problems with your proof:
- You wrote “Let $p$ be an isolated point”, but $p$ here already has another meaning.
- Then you write “this implies that there is an open set $U$ such that $U={p}$”. This is wrong. What it implies is that there is an open subset $U$ of $X$ such that $overline{{f^n(p),|,ninmathbb{Z}}}cap U={p}$.
$endgroup$
There are several problems with your proof:
- You wrote “Let $p$ be an isolated point”, but $p$ here already has another meaning.
- Then you write “this implies that there is an open set $U$ such that $U={p}$”. This is wrong. What it implies is that there is an open subset $U$ of $X$ such that $overline{{f^n(p),|,ninmathbb{Z}}}cap U={p}$.
edited Dec 15 '18 at 9:30
answered Dec 15 '18 at 9:20
José Carlos SantosJosé Carlos Santos
165k22132235
165k22132235
$begingroup$
@ José Carlos Santos. Thanks. I think that $overline{{f^n(p)}_{ninmathbb{Z}}}$ is an uncountable set. In above proof $f$ is a homeomorphism, hence it is an open map and $f(overline{U})= overline{f(U)}$.
$endgroup$
– user479859
Dec 15 '18 at 9:28
$begingroup$
I've deleted my third remark.
$endgroup$
– José Carlos Santos
Dec 15 '18 at 9:30
$begingroup$
Let $overline{{f^n(p)| ninmathbb{Z}}}cap U={p}$.
$endgroup$
– user479859
Dec 15 '18 at 9:31
$begingroup$
If $overline{f^n(p)| ninmathbb{Z}}}cap U={ f^k(p)}$, in above proof we work with $f^k(p)$.
$endgroup$
– user479859
Dec 15 '18 at 9:32
$begingroup$
What the means of "$ overline{{f^n(p)|ninmathbb{Z}}}$ has an isolated point"?
$endgroup$
– user479859
Dec 15 '18 at 9:35
add a comment |
$begingroup$
@ José Carlos Santos. Thanks. I think that $overline{{f^n(p)}_{ninmathbb{Z}}}$ is an uncountable set. In above proof $f$ is a homeomorphism, hence it is an open map and $f(overline{U})= overline{f(U)}$.
$endgroup$
– user479859
Dec 15 '18 at 9:28
$begingroup$
I've deleted my third remark.
$endgroup$
– José Carlos Santos
Dec 15 '18 at 9:30
$begingroup$
Let $overline{{f^n(p)| ninmathbb{Z}}}cap U={p}$.
$endgroup$
– user479859
Dec 15 '18 at 9:31
$begingroup$
If $overline{f^n(p)| ninmathbb{Z}}}cap U={ f^k(p)}$, in above proof we work with $f^k(p)$.
$endgroup$
– user479859
Dec 15 '18 at 9:32
$begingroup$
What the means of "$ overline{{f^n(p)|ninmathbb{Z}}}$ has an isolated point"?
$endgroup$
– user479859
Dec 15 '18 at 9:35
$begingroup$
@ José Carlos Santos. Thanks. I think that $overline{{f^n(p)}_{ninmathbb{Z}}}$ is an uncountable set. In above proof $f$ is a homeomorphism, hence it is an open map and $f(overline{U})= overline{f(U)}$.
$endgroup$
– user479859
Dec 15 '18 at 9:28
$begingroup$
@ José Carlos Santos. Thanks. I think that $overline{{f^n(p)}_{ninmathbb{Z}}}$ is an uncountable set. In above proof $f$ is a homeomorphism, hence it is an open map and $f(overline{U})= overline{f(U)}$.
$endgroup$
– user479859
Dec 15 '18 at 9:28
$begingroup$
I've deleted my third remark.
$endgroup$
– José Carlos Santos
Dec 15 '18 at 9:30
$begingroup$
I've deleted my third remark.
$endgroup$
– José Carlos Santos
Dec 15 '18 at 9:30
$begingroup$
Let $overline{{f^n(p)| ninmathbb{Z}}}cap U={p}$.
$endgroup$
– user479859
Dec 15 '18 at 9:31
$begingroup$
Let $overline{{f^n(p)| ninmathbb{Z}}}cap U={p}$.
$endgroup$
– user479859
Dec 15 '18 at 9:31
$begingroup$
If $overline{f^n(p)| ninmathbb{Z}}}cap U={ f^k(p)}$, in above proof we work with $f^k(p)$.
$endgroup$
– user479859
Dec 15 '18 at 9:32
$begingroup$
If $overline{f^n(p)| ninmathbb{Z}}}cap U={ f^k(p)}$, in above proof we work with $f^k(p)$.
$endgroup$
– user479859
Dec 15 '18 at 9:32
$begingroup$
What the means of "$ overline{{f^n(p)|ninmathbb{Z}}}$ has an isolated point"?
$endgroup$
– user479859
Dec 15 '18 at 9:35
$begingroup$
What the means of "$ overline{{f^n(p)|ninmathbb{Z}}}$ has an isolated point"?
$endgroup$
– user479859
Dec 15 '18 at 9:35
add a comment |
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