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Prove that every compact metric space $K$ has a countable base, and that $K$ is therefore separable.
$begingroup$
Prove that every compact metric space $K$ has a countable base, and that $K$ is therefore separable.
How does the following look?
Proof:
For each $n in mathbb{N}$, make an open cover of $K$ by neighborhoods of radius $frac{1}{n}$, and we have a finite subcover by compactness, i.e.
$$K subset bigcup_{x in K} N_{frac{1}{n}}(x) Rightarrow exists x_1, ..., x_N in K text{ such that } K subset bigcup_{i=1}^{N} N_{frac{1}{n}} (x_i)$$
Doing this for every $n in mathbb{N}$, we get a countable union of finite collections of sets, so the collection of these sets, call it $S$, is countable.
We claim that $S$ is a countable base for $K$, which is defined as a countable collection of open sets such that for any $x in K$ and any open set $G$ with $x in G$, there is some $V in S$ such that $x in V subset G$.
Let $x in K$ and let $G$ be any open set with $x in G$. Then since $G$ is open, there is some $r > 0$ such that $N_r(x) subset G$. Choose $n in mathbb{N}$ such that $frac{1}{n} < frac{r}{2}$, so that the maximal distance between points in a neighborhood of radius $frac{1}{n}$ is $r$. Then there must be some $i$ such that $x in N_{frac{1}{n}}(x_i) subset N_r(x)$ because any neighborhood of radius $frac{1}{n}$ containing $x$ cannot contain points a distance more than $r$ away. This shows that $S$ is a countable base.
The second part of the question asks us to show that $K$ is separable. Let ${V_n}$ be our countable base for $K$. For each $n in mathbb{N}$, choose $x_n in V_n$, and let $E = { x_n | n in mathbb{n} }$. We claim that $E$ is a countable dense set, which would show that $K$ is separable.
First, note that $E$ is clearly countable. To show that it's dense, we need to show that $overline{E} = K$. This is equivalent to showing that $(overline{E})^c = emptyset$. Now $(overline{E})^c$ is an open set because it's the complement of a closed set, $overline{E}$. If $(overline{E})^c$ is nonempty, then there is some $x in (overline{E})^c$, which is open, so since ${V_n}$ is a base, there is some $n$ such that $x in V_n subset (overline{E})^c$, which implies that $x_n in (overline{E})^c$, a contradiction, because
$x_n in E implies x_n in overline{E} implies x_n notin (overline{E})^c$.
Therefore, $(overline{E})^c = emptyset$, so that $overline{E} = K$.
Q.E.D.
general-topology proof-verification metric-spaces compactness second-countable
edited Dec 28 '16 at 8:48
Martin Sleziak
44.7k10119272
asked Feb 3 '14 at 6:50
PandaManPandaMan
1,17411333
$endgroup$
add a comment |
$begingroup$
Prove that every compact metric space $K$ has a countable base, and that $K$ is therefore separable.
How does the following look?
Proof:
For each $n in mathbb{N}$, make an open cover of $K$ by neighborhoods of radius $frac{1}{n}$, and we have a finite subcover by compactness, i.e.
$$K subset bigcup_{x in K} N_{frac{1}{n}}(x) Rightarrow exists x_1, ..., x_N in K text{ such that } K subset bigcup_{i=1}^{N} N_{frac{1}{n}} (x_i)$$
Doing this for every $n in mathbb{N}$, we get a countable union of finite collections of sets, so the collection of these sets, call it $S$, is countable.
We claim that $S$ is a countable base for $K$, which is defined as a countable collection of open sets such that for any $x in K$ and any open set $G$ with $x in G$, there is some $V in S$ such that $x in V subset G$.
Let $x in K$ and let $G$ be any open set with $x in G$. Then since $G$ is open, there is some $r > 0$ such that $N_r(x) subset G$. Choose $n in mathbb{N}$ such that $frac{1}{n} < frac{r}{2}$, so that the maximal distance between points in a neighborhood of radius $frac{1}{n}$ is $r$. Then there must be some $i$ such that $x in N_{frac{1}{n}}(x_i) subset N_r(x)$ because any neighborhood of radius $frac{1}{n}$ containing $x$ cannot contain points a distance more than $r$ away. This shows that $S$ is a countable base.
The second part of the question asks us to show that $K$ is separable. Let ${V_n}$ be our countable base for $K$. For each $n in mathbb{N}$, choose $x_n in V_n$, and let $E = { x_n | n in mathbb{n} }$. We claim that $E$ is a countable dense set, which would show that $K$ is separable.
First, note that $E$ is clearly countable. To show that it's dense, we need to show that $overline{E} = K$. This is equivalent to showing that $(overline{E})^c = emptyset$. Now $(overline{E})^c$ is an open set because it's the complement of a closed set, $overline{E}$. If $(overline{E})^c$ is nonempty, then there is some $x in (overline{E})^c$, which is open, so since ${V_n}$ is a base, there is some $n$ such that $x in V_n subset (overline{E})^c$, which implies that $x_n in (overline{E})^c$, a contradiction, because
$x_n in E implies x_n in overline{E} implies x_n notin (overline{E})^c$.
Therefore, $(overline{E})^c = emptyset$, so that $overline{E} = K$.
Q.E.D.
general-topology proof-verification metric-spaces compactness second-countable
edited Dec 28 '16 at 8:48
Martin Sleziak
44.7k10119272
asked Feb 3 '14 at 6:50
PandaManPandaMan
1,17411333
$endgroup$
1
$begingroup$
You don't need open covers. Sequential compactness plus a contradiction proof do the trick
$endgroup$
– Gabriel Romon
Feb 3 '14 at 6:54
1
$begingroup$
Looks fine to me.
$endgroup$
– copper.hat
Feb 3 '14 at 6:57
$begingroup$
In general if X is a metric space and Y is a dense subset of X then the set of open balls centered at members of Y with rational radii is a base for X. So if X is separable then X is second-countable.
$endgroup$
– DanielWainfleet
Feb 17 at 22:27
add a comment |
$begingroup$
Prove that every compact metric space $K$ has a countable base, and that $K$ is therefore separable.
How does the following look?
Proof:
For each $n in mathbb{N}$, make an open cover of $K$ by neighborhoods of radius $frac{1}{n}$, and we have a finite subcover by compactness, i.e.
$$K subset bigcup_{x in K} N_{frac{1}{n}}(x) Rightarrow exists x_1, ..., x_N in K text{ such that } K subset bigcup_{i=1}^{N} N_{frac{1}{n}} (x_i)$$
Doing this for every $n in mathbb{N}$, we get a countable union of finite collections of sets, so the collection of these sets, call it $S$, is countable.
We claim that $S$ is a countable base for $K$, which is defined as a countable collection of open sets such that for any $x in K$ and any open set $G$ with $x in G$, there is some $V in S$ such that $x in V subset G$.
Let $x in K$ and let $G$ be any open set with $x in G$. Then since $G$ is open, there is some $r > 0$ such that $N_r(x) subset G$. Choose $n in mathbb{N}$ such that $frac{1}{n} < frac{r}{2}$, so that the maximal distance between points in a neighborhood of radius $frac{1}{n}$ is $r$. Then there must be some $i$ such that $x in N_{frac{1}{n}}(x_i) subset N_r(x)$ because any neighborhood of radius $frac{1}{n}$ containing $x$ cannot contain points a distance more than $r$ away. This shows that $S$ is a countable base.
The second part of the question asks us to show that $K$ is separable. Let ${V_n}$ be our countable base for $K$. For each $n in mathbb{N}$, choose $x_n in V_n$, and let $E = { x_n | n in mathbb{n} }$. We claim that $E$ is a countable dense set, which would show that $K$ is separable.
First, note that $E$ is clearly countable. To show that it's dense, we need to show that $overline{E} = K$. This is equivalent to showing that $(overline{E})^c = emptyset$. Now $(overline{E})^c$ is an open set because it's the complement of a closed set, $overline{E}$. If $(overline{E})^c$ is nonempty, then there is some $x in (overline{E})^c$, which is open, so since ${V_n}$ is a base, there is some $n$ such that $x in V_n subset (overline{E})^c$, which implies that $x_n in (overline{E})^c$, a contradiction, because
$x_n in E implies x_n in overline{E} implies x_n notin (overline{E})^c$.
Therefore, $(overline{E})^c = emptyset$, so that $overline{E} = K$.
Q.E.D.
general-topology proof-verification metric-spaces compactness second-countable
edited Dec 28 '16 at 8:48
Martin Sleziak
44.7k10119272
asked Feb 3 '14 at 6:50
PandaManPandaMan
1,17411333
$endgroup$
Prove that every compact metric space $K$ has a countable base, and that $K$ is therefore separable.
How does the following look?
Proof:
For each $n in mathbb{N}$, make an open cover of $K$ by neighborhoods of radius $frac{1}{n}$, and we have a finite subcover by compactness, i.e.
$$K subset bigcup_{x in K} N_{frac{1}{n}}(x) Rightarrow exists x_1, ..., x_N in K text{ such that } K subset bigcup_{i=1}^{N} N_{frac{1}{n}} (x_i)$$
Doing this for every $n in mathbb{N}$, we get a countable union of finite collections of sets, so the collection of these sets, call it $S$, is countable.
We claim that $S$ is a countable base for $K$, which is defined as a countable collection of open sets such that for any $x in K$ and any open set $G$ with $x in G$, there is some $V in S$ such that $x in V subset G$.
Let $x in K$ and let $G$ be any open set with $x in G$. Then since $G$ is open, there is some $r > 0$ such that $N_r(x) subset G$. Choose $n in mathbb{N}$ such that $frac{1}{n} < frac{r}{2}$, so that the maximal distance between points in a neighborhood of radius $frac{1}{n}$ is $r$. Then there must be some $i$ such that $x in N_{frac{1}{n}}(x_i) subset N_r(x)$ because any neighborhood of radius $frac{1}{n}$ containing $x$ cannot contain points a distance more than $r$ away. This shows that $S$ is a countable base.
The second part of the question asks us to show that $K$ is separable. Let ${V_n}$ be our countable base for $K$. For each $n in mathbb{N}$, choose $x_n in V_n$, and let $E = { x_n | n in mathbb{n} }$. We claim that $E$ is a countable dense set, which would show that $K$ is separable.
First, note that $E$ is clearly countable. To show that it's dense, we need to show that $overline{E} = K$. This is equivalent to showing that $(overline{E})^c = emptyset$. Now $(overline{E})^c$ is an open set because it's the complement of a closed set, $overline{E}$. If $(overline{E})^c$ is nonempty, then there is some $x in (overline{E})^c$, which is open, so since ${V_n}$ is a base, there is some $n$ such that $x in V_n subset (overline{E})^c$, which implies that $x_n in (overline{E})^c$, a contradiction, because
$x_n in E implies x_n in overline{E} implies x_n notin (overline{E})^c$.
Therefore, $(overline{E})^c = emptyset$, so that $overline{E} = K$.
Q.E.D.
general-topology proof-verification metric-spaces compactness second-countable
general-topology proof-verification metric-spaces compactness second-countable
edited Dec 28 '16 at 8:48
Martin Sleziak
44.7k10119272
asked Feb 3 '14 at 6:50
PandaManPandaMan
1,17411333
edited Dec 28 '16 at 8:48
Martin Sleziak
44.7k10119272
asked Feb 3 '14 at 6:50
PandaManPandaMan
1,17411333
edited Dec 28 '16 at 8:48
Martin Sleziak
44.7k10119272
edited Dec 28 '16 at 8:48
Martin Sleziak
44.7k10119272
edited Dec 28 '16 at 8:48
Martin Sleziak
44.7k10119272
44.7k10119272
asked Feb 3 '14 at 6:50
PandaManPandaMan
1,17411333
asked Feb 3 '14 at 6:50
PandaManPandaMan
1,17411333
asked Feb 3 '14 at 6:50
PandaManPandaMan
1,17411333
1,17411333
1
$begingroup$
You don't need open covers. Sequential compactness plus a contradiction proof do the trick
$endgroup$
– Gabriel Romon
Feb 3 '14 at 6:54
1
$begingroup$
Looks fine to me.
$endgroup$
– copper.hat
Feb 3 '14 at 6:57
$begingroup$
In general if X is a metric space and Y is a dense subset of X then the set of open balls centered at members of Y with rational radii is a base for X. So if X is separable then X is second-countable.
$endgroup$
– DanielWainfleet
Feb 17 at 22:27
add a comment |
1
$begingroup$
You don't need open covers. Sequential compactness plus a contradiction proof do the trick
$endgroup$
– Gabriel Romon
Feb 3 '14 at 6:54
1
$begingroup$
Looks fine to me.
$endgroup$
– copper.hat
Feb 3 '14 at 6:57
$begingroup$
In general if X is a metric space and Y is a dense subset of X then the set of open balls centered at members of Y with rational radii is a base for X. So if X is separable then X is second-countable.
$endgroup$
– DanielWainfleet
Feb 17 at 22:27
1
1
$begingroup$
You don't need open covers. Sequential compactness plus a contradiction proof do the trick
$endgroup$
– Gabriel Romon
Feb 3 '14 at 6:54
$begingroup$
You don't need open covers. Sequential compactness plus a contradiction proof do the trick
$endgroup$
– Gabriel Romon
Feb 3 '14 at 6:54
1
1
$begingroup$
Looks fine to me.
$endgroup$
– copper.hat
Feb 3 '14 at 6:57
$begingroup$
Looks fine to me.
$endgroup$
– copper.hat
Feb 3 '14 at 6:57
$begingroup$
In general if X is a metric space and Y is a dense subset of X then the set of open balls centered at members of Y with rational radii is a base for X. So if X is separable then X is second-countable.
$endgroup$
– DanielWainfleet
Feb 17 at 22:27
$begingroup$
In general if X is a metric space and Y is a dense subset of X then the set of open balls centered at members of Y with rational radii is a base for X. So if X is separable then X is second-countable.
$endgroup$
– DanielWainfleet
Feb 17 at 22:27
add a comment |
1 Answer
1
active
oldest
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$begingroup$
I liked it a lot.
But to proof that $E$ is a dense subset of $K$, you can do it in the following way, for each open set $U$ of $K$, there is a $V_{n}subseteq U$ so, you will have $x_{n}in Ucap E$, that complete the proof.
I used an equivalence of density: E is a dense subset of $K$ if and only if for each open subset $U$ of $K$ exists a $xin Ucap E$.
answered Jan 16 at 13:53
Gil AstudilloGil Astudillo
315
$endgroup$
add a comment |
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$begingroup$
I liked it a lot.
But to proof that $E$ is a dense subset of $K$, you can do it in the following way, for each open set $U$ of $K$, there is a $V_{n}subseteq U$ so, you will have $x_{n}in Ucap E$, that complete the proof.
I used an equivalence of density: E is a dense subset of $K$ if and only if for each open subset $U$ of $K$ exists a $xin Ucap E$.
answered Jan 16 at 13:53
Gil AstudilloGil Astudillo
315
$endgroup$
add a comment |
$begingroup$
I liked it a lot.
But to proof that $E$ is a dense subset of $K$, you can do it in the following way, for each open set $U$ of $K$, there is a $V_{n}subseteq U$ so, you will have $x_{n}in Ucap E$, that complete the proof.
I used an equivalence of density: E is a dense subset of $K$ if and only if for each open subset $U$ of $K$ exists a $xin Ucap E$.
answered Jan 16 at 13:53
Gil AstudilloGil Astudillo
315
$endgroup$
add a comment |
$begingroup$
I liked it a lot.
But to proof that $E$ is a dense subset of $K$, you can do it in the following way, for each open set $U$ of $K$, there is a $V_{n}subseteq U$ so, you will have $x_{n}in Ucap E$, that complete the proof.
I used an equivalence of density: E is a dense subset of $K$ if and only if for each open subset $U$ of $K$ exists a $xin Ucap E$.
answered Jan 16 at 13:53
Gil AstudilloGil Astudillo
315
$endgroup$
I liked it a lot.
But to proof that $E$ is a dense subset of $K$, you can do it in the following way, for each open set $U$ of $K$, there is a $V_{n}subseteq U$ so, you will have $x_{n}in Ucap E$, that complete the proof.
I used an equivalence of density: E is a dense subset of $K$ if and only if for each open subset $U$ of $K$ exists a $xin Ucap E$.
answered Jan 16 at 13:53
Gil AstudilloGil Astudillo
315
answered Jan 16 at 13:53
Gil AstudilloGil Astudillo
315
answered Jan 16 at 13:53
Gil AstudilloGil Astudillo
315
answered Jan 16 at 13:53
Gil AstudilloGil Astudillo
315
315
add a comment |
add a comment |
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$begingroup$
You don't need open covers. Sequential compactness plus a contradiction proof do the trick
$endgroup$
– Gabriel Romon
Feb 3 '14 at 6:54
1
$begingroup$
Looks fine to me.
$endgroup$
– copper.hat
Feb 3 '14 at 6:57
$begingroup$
In general if X is a metric space and Y is a dense subset of X then the set of open balls centered at members of Y with rational radii is a base for X. So if X is separable then X is second-countable.
$endgroup$
– DanielWainfleet
Feb 17 at 22:27