Nested descreasing intersection of nonempty covering compact sets is nonempty (check proof)












0












$begingroup$



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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















0












$begingroup$



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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$



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







share|cite|improve this question











$endgroup$





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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 28 '18 at 22:15









José Carlos Santos

176k24135244




176k24135244










asked Dec 28 '18 at 21:49









Lucas CorrêaLucas Corrêa

1,5761421




1,5761421












  • $begingroup$
    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










  • $begingroup$
    @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










1 Answer
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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$$A_{alpha_1}capcdotscap A_{alpha_m}=A_{max{alpha_1,ldots,alpha_m}}neqemptyset.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for the suggestion, José. I'll justify this!
    $endgroup$
    – Lucas Corrêa
    Dec 28 '18 at 22:13














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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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$$A_{alpha_1}capcdotscap A_{alpha_m}=A_{max{alpha_1,ldots,alpha_m}}neqemptyset.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for the suggestion, José. I'll justify this!
    $endgroup$
    – Lucas Corrêa
    Dec 28 '18 at 22:13


















1












$begingroup$

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$$A_{alpha_1}capcdotscap A_{alpha_m}=A_{max{alpha_1,ldots,alpha_m}}neqemptyset.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for the suggestion, José. I'll justify this!
    $endgroup$
    – Lucas Corrêa
    Dec 28 '18 at 22:13
















1












1








1





$begingroup$

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$$A_{alpha_1}capcdotscap A_{alpha_m}=A_{max{alpha_1,ldots,alpha_m}}neqemptyset.$$






share|cite|improve this answer









$endgroup$



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$$A_{alpha_1}capcdotscap A_{alpha_m}=A_{max{alpha_1,ldots,alpha_m}}neqemptyset.$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 28 '18 at 22:09









José Carlos SantosJosé Carlos Santos

176k24135244




176k24135244












  • $begingroup$
    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






$begingroup$
Thanks for the suggestion, José. I'll justify this!
$endgroup$
– Lucas Corrêa
Dec 28 '18 at 22:13




















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