Nested descreasing intersection of nonempty covering compact sets is nonempty (check proof)
$begingroup$
Proof that the nested descreasing intersection of nonempty covering compact sets is nonempty.
Let
$$A_{1} supset A_{2} supset cdots supset A_{n} supset cdots$$
be a nested decreasing sequence of compacts. Suppose that $displaystyle bigcap A_{n} = emptyset$. Take $U_{n} = A_{n}^{c}$, then
$$bigcup U_{n} = bigcup A_{n}^{c} = left(bigcap A_{n}right)^{c} = A_{1}.$$
Here, I'm thinking of $A_{1}$ as the main metric space. Since ${U_{n}}$ is an open covering of $A_{1}$, we can extract a finite subcovering, that is,
$$A_{alpha_{1}}^{c}cup A_{alpha_{2}}^{c} cup cdots cup A_{alpha_{m}}^{c} supset A_{1}$$
or
$$(A_{1}setminus A_{alpha_{1}})cup(A_{1}setminus A_{alpha_{2}})cupcdotscup(A_{1}setminus A_{alpha_{m}}) supset A_{1}.$$
But, this is true only if $A_{alpha_{i}} = emptyset$ for some $i$, a contradiction.
Is correct?
proof-verification metric-spaces compactness
edited Dec 28 '18 at 22:15
José Carlos Santos
164k22131234
asked Dec 28 '18 at 21:49
Lucas CorrêaLucas Corrêa
1,6201321
$endgroup$
add a comment |
$begingroup$
Proof that the nested descreasing intersection of nonempty covering compact sets is nonempty.
Let
$$A_{1} supset A_{2} supset cdots supset A_{n} supset cdots$$
be a nested decreasing sequence of compacts. Suppose that $displaystyle bigcap A_{n} = emptyset$. Take $U_{n} = A_{n}^{c}$, then
$$bigcup U_{n} = bigcup A_{n}^{c} = left(bigcap A_{n}right)^{c} = A_{1}.$$
Here, I'm thinking of $A_{1}$ as the main metric space. Since ${U_{n}}$ is an open covering of $A_{1}$, we can extract a finite subcovering, that is,
$$A_{alpha_{1}}^{c}cup A_{alpha_{2}}^{c} cup cdots cup A_{alpha_{m}}^{c} supset A_{1}$$
or
$$(A_{1}setminus A_{alpha_{1}})cup(A_{1}setminus A_{alpha_{2}})cupcdotscup(A_{1}setminus A_{alpha_{m}}) supset A_{1}.$$
But, this is true only if $A_{alpha_{i}} = emptyset$ for some $i$, a contradiction.
Is correct?
proof-verification metric-spaces compactness
edited Dec 28 '18 at 22:15
José Carlos Santos
164k22131234
asked Dec 28 '18 at 21:49
Lucas CorrêaLucas Corrêa
1,6201321
$endgroup$
$begingroup$
Why is ${U_n}$ an open covering? Compact subsets of topological spaces are not necessarily closed. See $X={0,1}$ with the trivial topology, and the compact subset ${0}$.
$endgroup$
– SmileyCraft
Dec 28 '18 at 22:10
$begingroup$
@SmileyCraft I'm pretty sure that, at least in metric spaces (see the tag), compact subsets are closed.
$endgroup$
– Noble Mushtak
Dec 28 '18 at 22:11
$begingroup$
@Noble Mushtak I saw the tag, but what confuses me is that the definition of compactness Lucas uses is very topological. I've never seen it in a real analysis course.
$endgroup$
– SmileyCraft
Dec 28 '18 at 22:12
$begingroup$
@SmileyCraft I'm reading Rudin's Analysis book right now, and they define a compact subset to be such that any open cover of the set has a finite subcover. This is the same definition Lucas seems to be using as well.
$endgroup$
– Noble Mushtak
Dec 28 '18 at 22:14
$begingroup$
@SmileyCraft, thats true! In fact, the book Im using define first compact as sequentially compact. But after, the author present the covering compact definition and in this question is requested to use the second definition.
$endgroup$
– Lucas Corrêa
Dec 28 '18 at 22:16
add a comment |
$begingroup$
Proof that the nested descreasing intersection of nonempty covering compact sets is nonempty.
Let
$$A_{1} supset A_{2} supset cdots supset A_{n} supset cdots$$
be a nested decreasing sequence of compacts. Suppose that $displaystyle bigcap A_{n} = emptyset$. Take $U_{n} = A_{n}^{c}$, then
$$bigcup U_{n} = bigcup A_{n}^{c} = left(bigcap A_{n}right)^{c} = A_{1}.$$
Here, I'm thinking of $A_{1}$ as the main metric space. Since ${U_{n}}$ is an open covering of $A_{1}$, we can extract a finite subcovering, that is,
$$A_{alpha_{1}}^{c}cup A_{alpha_{2}}^{c} cup cdots cup A_{alpha_{m}}^{c} supset A_{1}$$
or
$$(A_{1}setminus A_{alpha_{1}})cup(A_{1}setminus A_{alpha_{2}})cupcdotscup(A_{1}setminus A_{alpha_{m}}) supset A_{1}.$$
But, this is true only if $A_{alpha_{i}} = emptyset$ for some $i$, a contradiction.
Is correct?
proof-verification metric-spaces compactness
edited Dec 28 '18 at 22:15
José Carlos Santos
164k22131234
asked Dec 28 '18 at 21:49
Lucas CorrêaLucas Corrêa
1,6201321
$endgroup$
Proof that the nested descreasing intersection of nonempty covering compact sets is nonempty.
Let
$$A_{1} supset A_{2} supset cdots supset A_{n} supset cdots$$
be a nested decreasing sequence of compacts. Suppose that $displaystyle bigcap A_{n} = emptyset$. Take $U_{n} = A_{n}^{c}$, then
$$bigcup U_{n} = bigcup A_{n}^{c} = left(bigcap A_{n}right)^{c} = A_{1}.$$
Here, I'm thinking of $A_{1}$ as the main metric space. Since ${U_{n}}$ is an open covering of $A_{1}$, we can extract a finite subcovering, that is,
$$A_{alpha_{1}}^{c}cup A_{alpha_{2}}^{c} cup cdots cup A_{alpha_{m}}^{c} supset A_{1}$$
or
$$(A_{1}setminus A_{alpha_{1}})cup(A_{1}setminus A_{alpha_{2}})cupcdotscup(A_{1}setminus A_{alpha_{m}}) supset A_{1}.$$
But, this is true only if $A_{alpha_{i}} = emptyset$ for some $i$, a contradiction.
Is correct?
proof-verification metric-spaces compactness
proof-verification metric-spaces compactness
edited Dec 28 '18 at 22:15
José Carlos Santos
164k22131234
asked Dec 28 '18 at 21:49
Lucas CorrêaLucas Corrêa
1,6201321
edited Dec 28 '18 at 22:15
José Carlos Santos
164k22131234
asked Dec 28 '18 at 21:49
Lucas CorrêaLucas Corrêa
1,6201321
edited Dec 28 '18 at 22:15
José Carlos Santos
164k22131234
edited Dec 28 '18 at 22:15
José Carlos Santos
164k22131234
edited Dec 28 '18 at 22:15
José Carlos Santos
164k22131234
164k22131234
asked Dec 28 '18 at 21:49
Lucas CorrêaLucas Corrêa
1,6201321
asked Dec 28 '18 at 21:49
Lucas CorrêaLucas Corrêa
1,6201321
asked Dec 28 '18 at 21:49
Lucas CorrêaLucas Corrêa
1,6201321
1,6201321
$begingroup$
Why is ${U_n}$ an open covering? Compact subsets of topological spaces are not necessarily closed. See $X={0,1}$ with the trivial topology, and the compact subset ${0}$.
$endgroup$
– SmileyCraft
Dec 28 '18 at 22:10
$begingroup$
@SmileyCraft I'm pretty sure that, at least in metric spaces (see the tag), compact subsets are closed.
$endgroup$
– Noble Mushtak
Dec 28 '18 at 22:11
$begingroup$
@Noble Mushtak I saw the tag, but what confuses me is that the definition of compactness Lucas uses is very topological. I've never seen it in a real analysis course.
$endgroup$
– SmileyCraft
Dec 28 '18 at 22:12
$begingroup$
@SmileyCraft I'm reading Rudin's Analysis book right now, and they define a compact subset to be such that any open cover of the set has a finite subcover. This is the same definition Lucas seems to be using as well.
$endgroup$
– Noble Mushtak
Dec 28 '18 at 22:14
$begingroup$
@SmileyCraft, thats true! In fact, the book Im using define first compact as sequentially compact. But after, the author present the covering compact definition and in this question is requested to use the second definition.
$endgroup$
– Lucas Corrêa
Dec 28 '18 at 22:16
add a comment |
$begingroup$
Why is ${U_n}$ an open covering? Compact subsets of topological spaces are not necessarily closed. See $X={0,1}$ with the trivial topology, and the compact subset ${0}$.
$endgroup$
– SmileyCraft
Dec 28 '18 at 22:10
$begingroup$
@SmileyCraft I'm pretty sure that, at least in metric spaces (see the tag), compact subsets are closed.
$endgroup$
– Noble Mushtak
Dec 28 '18 at 22:11
$begingroup$
@Noble Mushtak I saw the tag, but what confuses me is that the definition of compactness Lucas uses is very topological. I've never seen it in a real analysis course.
$endgroup$
– SmileyCraft
Dec 28 '18 at 22:12
$begingroup$
@SmileyCraft I'm reading Rudin's Analysis book right now, and they define a compact subset to be such that any open cover of the set has a finite subcover. This is the same definition Lucas seems to be using as well.
$endgroup$
– Noble Mushtak
Dec 28 '18 at 22:14
$begingroup$
@SmileyCraft, thats true! In fact, the book Im using define first compact as sequentially compact. But after, the author present the covering compact definition and in this question is requested to use the second definition.
$endgroup$
– Lucas Corrêa
Dec 28 '18 at 22:16
$begingroup$
Why is ${U_n}$ an open covering? Compact subsets of topological spaces are not necessarily closed. See $X={0,1}$ with the trivial topology, and the compact subset ${0}$.
$endgroup$
– SmileyCraft
Dec 28 '18 at 22:10
$begingroup$
Why is ${U_n}$ an open covering? Compact subsets of topological spaces are not necessarily closed. See $X={0,1}$ with the trivial topology, and the compact subset ${0}$.
$endgroup$
– SmileyCraft
Dec 28 '18 at 22:10
$begingroup$
@SmileyCraft I'm pretty sure that, at least in metric spaces (see the tag), compact subsets are closed.
$endgroup$
– Noble Mushtak
Dec 28 '18 at 22:11
$begingroup$
@SmileyCraft I'm pretty sure that, at least in metric spaces (see the tag), compact subsets are closed.
$endgroup$
– Noble Mushtak
Dec 28 '18 at 22:11
$begingroup$
@Noble Mushtak I saw the tag, but what confuses me is that the definition of compactness Lucas uses is very topological. I've never seen it in a real analysis course.
$endgroup$
– SmileyCraft
Dec 28 '18 at 22:12
$begingroup$
@Noble Mushtak I saw the tag, but what confuses me is that the definition of compactness Lucas uses is very topological. I've never seen it in a real analysis course.
$endgroup$
– SmileyCraft
Dec 28 '18 at 22:12
$begingroup$
@SmileyCraft I'm reading Rudin's Analysis book right now, and they define a compact subset to be such that any open cover of the set has a finite subcover. This is the same definition Lucas seems to be using as well.
$endgroup$
– Noble Mushtak
Dec 28 '18 at 22:14
$begingroup$
@SmileyCraft I'm reading Rudin's Analysis book right now, and they define a compact subset to be such that any open cover of the set has a finite subcover. This is the same definition Lucas seems to be using as well.
$endgroup$
– Noble Mushtak
Dec 28 '18 at 22:14
$begingroup$
@SmileyCraft, thats true! In fact, the book Im using define first compact as sequentially compact. But after, the author present the covering compact definition and in this question is requested to use the second definition.
$endgroup$
– Lucas Corrêa
Dec 28 '18 at 22:16
$begingroup$
@SmileyCraft, thats true! In fact, the book Im using define first compact as sequentially compact. But after, the author present the covering compact definition and in this question is requested to use the second definition.
$endgroup$
– Lucas Corrêa
Dec 28 '18 at 22:16
add a comment |
1 Answer
1
active
oldest
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In my opinion, you should add a justification of the assertion according to which that$$(A_1setminus A_{alpha_1})cup(A_1setminus A_{alpha_2})cupcdotscup(A_1setminus A_{alpha_m})supset A_1tag1$$can only occur if $A_1=emptyset$. This happens because$$(1)iff A_{alpha_1}capcdotscap A_{alpha_m}=emptyset,$$but$$A_{alpha_1}capcdotscap A_{alpha_m}=A_{max{alpha_1,ldots,alpha_m}}neqemptyset.$$
answered Dec 28 '18 at 22:09
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
$begingroup$
Thanks for the suggestion, José. I'll justify this!
$endgroup$
– Lucas Corrêa
Dec 28 '18 at 22:13
add a comment |
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1 Answer
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$begingroup$
In my opinion, you should add a justification of the assertion according to which that$$(A_1setminus A_{alpha_1})cup(A_1setminus A_{alpha_2})cupcdotscup(A_1setminus A_{alpha_m})supset A_1tag1$$can only occur if $A_1=emptyset$. This happens because$$(1)iff A_{alpha_1}capcdotscap A_{alpha_m}=emptyset,$$but$$A_{alpha_1}capcdotscap A_{alpha_m}=A_{max{alpha_1,ldots,alpha_m}}neqemptyset.$$
answered Dec 28 '18 at 22:09
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
$begingroup$
Thanks for the suggestion, José. I'll justify this!
$endgroup$
– Lucas Corrêa
Dec 28 '18 at 22:13
add a comment |
$begingroup$
In my opinion, you should add a justification of the assertion according to which that$$(A_1setminus A_{alpha_1})cup(A_1setminus A_{alpha_2})cupcdotscup(A_1setminus A_{alpha_m})supset A_1tag1$$can only occur if $A_1=emptyset$. This happens because$$(1)iff A_{alpha_1}capcdotscap A_{alpha_m}=emptyset,$$but$$A_{alpha_1}capcdotscap A_{alpha_m}=A_{max{alpha_1,ldots,alpha_m}}neqemptyset.$$
answered Dec 28 '18 at 22:09
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
$begingroup$
Thanks for the suggestion, José. I'll justify this!
$endgroup$
– Lucas Corrêa
Dec 28 '18 at 22:13
add a comment |
$begingroup$
In my opinion, you should add a justification of the assertion according to which that$$(A_1setminus A_{alpha_1})cup(A_1setminus A_{alpha_2})cupcdotscup(A_1setminus A_{alpha_m})supset A_1tag1$$can only occur if $A_1=emptyset$. This happens because$$(1)iff A_{alpha_1}capcdotscap A_{alpha_m}=emptyset,$$but$$A_{alpha_1}capcdotscap A_{alpha_m}=A_{max{alpha_1,ldots,alpha_m}}neqemptyset.$$
answered Dec 28 '18 at 22:09
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
In my opinion, you should add a justification of the assertion according to which that$$(A_1setminus A_{alpha_1})cup(A_1setminus A_{alpha_2})cupcdotscup(A_1setminus A_{alpha_m})supset A_1tag1$$can only occur if $A_1=emptyset$. This happens because$$(1)iff A_{alpha_1}capcdotscap A_{alpha_m}=emptyset,$$but$$A_{alpha_1}capcdotscap A_{alpha_m}=A_{max{alpha_1,ldots,alpha_m}}neqemptyset.$$
answered Dec 28 '18 at 22:09
José Carlos SantosJosé Carlos Santos
164k22131234
answered Dec 28 '18 at 22:09
José Carlos SantosJosé Carlos Santos
164k22131234
answered Dec 28 '18 at 22:09
José Carlos SantosJosé Carlos Santos
164k22131234
answered Dec 28 '18 at 22:09
José Carlos SantosJosé Carlos Santos
164k22131234
164k22131234
$begingroup$
Thanks for the suggestion, José. I'll justify this!
$endgroup$
– Lucas Corrêa
Dec 28 '18 at 22:13
add a comment |
$begingroup$
Thanks for the suggestion, José. I'll justify this!
$endgroup$
– Lucas Corrêa
Dec 28 '18 at 22:13
$begingroup$
Thanks for the suggestion, José. I'll justify this!
$endgroup$
– Lucas Corrêa
Dec 28 '18 at 22:13
$begingroup$
Thanks for the suggestion, José. I'll justify this!
$endgroup$
– Lucas Corrêa
Dec 28 '18 at 22:13
add a comment |
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$begingroup$
Why is ${U_n}$ an open covering? Compact subsets of topological spaces are not necessarily closed. See $X={0,1}$ with the trivial topology, and the compact subset ${0}$.
$endgroup$
– SmileyCraft
Dec 28 '18 at 22:10
$begingroup$
@SmileyCraft I'm pretty sure that, at least in metric spaces (see the tag), compact subsets are closed.
$endgroup$
– Noble Mushtak
Dec 28 '18 at 22:11
$begingroup$
@Noble Mushtak I saw the tag, but what confuses me is that the definition of compactness Lucas uses is very topological. I've never seen it in a real analysis course.
$endgroup$
– SmileyCraft
Dec 28 '18 at 22:12
$begingroup$
@SmileyCraft I'm reading Rudin's Analysis book right now, and they define a compact subset to be such that any open cover of the set has a finite subcover. This is the same definition Lucas seems to be using as well.
$endgroup$
– Noble Mushtak
Dec 28 '18 at 22:14
$begingroup$
@SmileyCraft, thats true! In fact, the book Im using define first compact as sequentially compact. But after, the author present the covering compact definition and in this question is requested to use the second definition.
$endgroup$
– Lucas Corrêa
Dec 28 '18 at 22:16