Mixing
Product of a compact set and a singleton is compact proof
$begingroup$
It's before we prove that 'Product of two compact sets is compact'. Let $S$ be an open cover of $X times {bullet}$ where $X$ is compact. Then $pi_1(S)$ is an open cover of $X$ so there is a finite subcover then pick one $A_n$ from $S$ corresponding to each $pi_1(A_n)$ but I think it may not cover $X times {bullet}$. How to complete the proof?
Edit
There are topological spaces $X$(which is compact), $Y$ and the product topology on $X times Y$ is given by the subbase $U times V$ where $U$ is open in $X$ and $V$ is open in $Y$. Pick an element $bullet in Y$. Let $S$ be an open cover of $X times {bullet}$. Then $pi_1(S)$ is an open cover of $X$ so there is a finite subcover then pick one $A_n$ from $S$ corresponding to each $pi_1(A_n)$ but I think it may not cover $X times {bullet}$. For example let $Y$ be a $T_1$ space and pick another element $bulletbullet in Y$. There exists an open set $W$ containing $bulletbullet$ but not
$bullet$. Lets construct an open cover of $X times {bullet}$
$$S cup {B times W mid B in pi_1(S)}$$
Now we can apply this open cover to upper proof, and when we are picking $A_n$ from $S$ corresponding to each $pi_1(A_n)$, we may pick all the sets from ${B times W mid B in pi_1(S)}$ so in fact it doesn't cover $X times {bullet}$. Am I misunderstanding something?
general-topology compactness product-space
asked Jan 26 at 13:41
RisRis
397
$endgroup$
add a comment |
$begingroup$
It's before we prove that 'Product of two compact sets is compact'. Let $S$ be an open cover of $X times {bullet}$ where $X$ is compact. Then $pi_1(S)$ is an open cover of $X$ so there is a finite subcover then pick one $A_n$ from $S$ corresponding to each $pi_1(A_n)$ but I think it may not cover $X times {bullet}$. How to complete the proof?
Edit
There are topological spaces $X$(which is compact), $Y$ and the product topology on $X times Y$ is given by the subbase $U times V$ where $U$ is open in $X$ and $V$ is open in $Y$. Pick an element $bullet in Y$. Let $S$ be an open cover of $X times {bullet}$. Then $pi_1(S)$ is an open cover of $X$ so there is a finite subcover then pick one $A_n$ from $S$ corresponding to each $pi_1(A_n)$ but I think it may not cover $X times {bullet}$. For example let $Y$ be a $T_1$ space and pick another element $bulletbullet in Y$. There exists an open set $W$ containing $bulletbullet$ but not
$bullet$. Lets construct an open cover of $X times {bullet}$
$$S cup {B times W mid B in pi_1(S)}$$
Now we can apply this open cover to upper proof, and when we are picking $A_n$ from $S$ corresponding to each $pi_1(A_n)$, we may pick all the sets from ${B times W mid B in pi_1(S)}$ so in fact it doesn't cover $X times {bullet}$. Am I misunderstanding something?
general-topology compactness product-space
asked Jan 26 at 13:41
RisRis
397
$endgroup$
add a comment |
$begingroup$
It's before we prove that 'Product of two compact sets is compact'. Let $S$ be an open cover of $X times {bullet}$ where $X$ is compact. Then $pi_1(S)$ is an open cover of $X$ so there is a finite subcover then pick one $A_n$ from $S$ corresponding to each $pi_1(A_n)$ but I think it may not cover $X times {bullet}$. How to complete the proof?
Edit
There are topological spaces $X$(which is compact), $Y$ and the product topology on $X times Y$ is given by the subbase $U times V$ where $U$ is open in $X$ and $V$ is open in $Y$. Pick an element $bullet in Y$. Let $S$ be an open cover of $X times {bullet}$. Then $pi_1(S)$ is an open cover of $X$ so there is a finite subcover then pick one $A_n$ from $S$ corresponding to each $pi_1(A_n)$ but I think it may not cover $X times {bullet}$. For example let $Y$ be a $T_1$ space and pick another element $bulletbullet in Y$. There exists an open set $W$ containing $bulletbullet$ but not
$bullet$. Lets construct an open cover of $X times {bullet}$
$$S cup {B times W mid B in pi_1(S)}$$
Now we can apply this open cover to upper proof, and when we are picking $A_n$ from $S$ corresponding to each $pi_1(A_n)$, we may pick all the sets from ${B times W mid B in pi_1(S)}$ so in fact it doesn't cover $X times {bullet}$. Am I misunderstanding something?
general-topology compactness product-space
asked Jan 26 at 13:41
RisRis
397
$endgroup$
It's before we prove that 'Product of two compact sets is compact'. Let $S$ be an open cover of $X times {bullet}$ where $X$ is compact. Then $pi_1(S)$ is an open cover of $X$ so there is a finite subcover then pick one $A_n$ from $S$ corresponding to each $pi_1(A_n)$ but I think it may not cover $X times {bullet}$. How to complete the proof?
Edit
There are topological spaces $X$(which is compact), $Y$ and the product topology on $X times Y$ is given by the subbase $U times V$ where $U$ is open in $X$ and $V$ is open in $Y$. Pick an element $bullet in Y$. Let $S$ be an open cover of $X times {bullet}$. Then $pi_1(S)$ is an open cover of $X$ so there is a finite subcover then pick one $A_n$ from $S$ corresponding to each $pi_1(A_n)$ but I think it may not cover $X times {bullet}$. For example let $Y$ be a $T_1$ space and pick another element $bulletbullet in Y$. There exists an open set $W$ containing $bulletbullet$ but not
$bullet$. Lets construct an open cover of $X times {bullet}$
$$S cup {B times W mid B in pi_1(S)}$$
Now we can apply this open cover to upper proof, and when we are picking $A_n$ from $S$ corresponding to each $pi_1(A_n)$, we may pick all the sets from ${B times W mid B in pi_1(S)}$ so in fact it doesn't cover $X times {bullet}$. Am I misunderstanding something?
general-topology compactness product-space
general-topology compactness product-space
asked Jan 26 at 13:41
RisRis
397
asked Jan 26 at 13:41
RisRis
397
edited Jan 26 at 14:35
Ris
asked Jan 26 at 13:41
RisRis
397
asked Jan 26 at 13:41
RisRis
397
asked Jan 26 at 13:41
RisRis
397
397
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It will cover $X times {∙}$.
You have a finite collection $pi_1(A_1),dots,pi_1(A_n)$ of open subsets of $X$ that cover $X$. Choose any $(x,∙) in A_i$. Then $x in X$, so there exists $A_i$ such that $x in pi_1(A_i)$. This means there exists $(y,z) in A_isubseteq X times {∙}$ such that $pi_1(y,z)=x$. But the only $z in {∙}$ is $∙$ itself, and $pi_1(y,z)=y$ by definition. So we have $(x,∙) in A_i$.
However, you should also justify why $pi_1(A_i)$ is open. In general the image of an open set is not necessarily open, but it will hold for projections.
answered Jan 26 at 13:54
kccukccu
10.6k11229
$endgroup$
$begingroup$
I edited the question sorry for short explanations.
$endgroup$
– Ris
Jan 26 at 14:26
$begingroup$
In your edit, is $Y$ compact?
$endgroup$
– kccu
Jan 26 at 14:32
$begingroup$
Not necessarily.
$endgroup$
– Ris
Jan 26 at 14:34
$begingroup$
Then $X times Y$ is not necessarily compact. (Also, you are using $X$ to denote an element of $pi_1(S)$, but $X$ is also the name of your compact space...)
$endgroup$
– kccu
Jan 26 at 14:34
$begingroup$
Oh I edited it to $B$. Then $X times { bullet }$ is also not necessarily compact?
$endgroup$
– Ris
Jan 26 at 14:36
add a comment |
$begingroup$
$pi_1$ is a homeomorphism between X×{•} and X.
answered Jan 26 at 14:02
William ElliotWilliam Elliot
8,7962820
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
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$begingroup$
It will cover $X times {∙}$.
You have a finite collection $pi_1(A_1),dots,pi_1(A_n)$ of open subsets of $X$ that cover $X$. Choose any $(x,∙) in A_i$. Then $x in X$, so there exists $A_i$ such that $x in pi_1(A_i)$. This means there exists $(y,z) in A_isubseteq X times {∙}$ such that $pi_1(y,z)=x$. But the only $z in {∙}$ is $∙$ itself, and $pi_1(y,z)=y$ by definition. So we have $(x,∙) in A_i$.
However, you should also justify why $pi_1(A_i)$ is open. In general the image of an open set is not necessarily open, but it will hold for projections.
answered Jan 26 at 13:54
kccukccu
10.6k11229
$endgroup$
$begingroup$
I edited the question sorry for short explanations.
$endgroup$
– Ris
Jan 26 at 14:26
$begingroup$
In your edit, is $Y$ compact?
$endgroup$
– kccu
Jan 26 at 14:32
$begingroup$
Not necessarily.
$endgroup$
– Ris
Jan 26 at 14:34
$begingroup$
Then $X times Y$ is not necessarily compact. (Also, you are using $X$ to denote an element of $pi_1(S)$, but $X$ is also the name of your compact space...)
$endgroup$
– kccu
Jan 26 at 14:34
$begingroup$
Oh I edited it to $B$. Then $X times { bullet }$ is also not necessarily compact?
$endgroup$
– Ris
Jan 26 at 14:36
add a comment |
$begingroup$
It will cover $X times {∙}$.
You have a finite collection $pi_1(A_1),dots,pi_1(A_n)$ of open subsets of $X$ that cover $X$. Choose any $(x,∙) in A_i$. Then $x in X$, so there exists $A_i$ such that $x in pi_1(A_i)$. This means there exists $(y,z) in A_isubseteq X times {∙}$ such that $pi_1(y,z)=x$. But the only $z in {∙}$ is $∙$ itself, and $pi_1(y,z)=y$ by definition. So we have $(x,∙) in A_i$.
However, you should also justify why $pi_1(A_i)$ is open. In general the image of an open set is not necessarily open, but it will hold for projections.
answered Jan 26 at 13:54
kccukccu
10.6k11229
$endgroup$
$begingroup$
I edited the question sorry for short explanations.
$endgroup$
– Ris
Jan 26 at 14:26
$begingroup$
In your edit, is $Y$ compact?
$endgroup$
– kccu
Jan 26 at 14:32
$begingroup$
Not necessarily.
$endgroup$
– Ris
Jan 26 at 14:34
$begingroup$
Then $X times Y$ is not necessarily compact. (Also, you are using $X$ to denote an element of $pi_1(S)$, but $X$ is also the name of your compact space...)
$endgroup$
– kccu
Jan 26 at 14:34
$begingroup$
Oh I edited it to $B$. Then $X times { bullet }$ is also not necessarily compact?
$endgroup$
– Ris
Jan 26 at 14:36
add a comment |
$begingroup$
It will cover $X times {∙}$.
You have a finite collection $pi_1(A_1),dots,pi_1(A_n)$ of open subsets of $X$ that cover $X$. Choose any $(x,∙) in A_i$. Then $x in X$, so there exists $A_i$ such that $x in pi_1(A_i)$. This means there exists $(y,z) in A_isubseteq X times {∙}$ such that $pi_1(y,z)=x$. But the only $z in {∙}$ is $∙$ itself, and $pi_1(y,z)=y$ by definition. So we have $(x,∙) in A_i$.
However, you should also justify why $pi_1(A_i)$ is open. In general the image of an open set is not necessarily open, but it will hold for projections.
answered Jan 26 at 13:54
kccukccu
10.6k11229
$endgroup$
It will cover $X times {∙}$.
You have a finite collection $pi_1(A_1),dots,pi_1(A_n)$ of open subsets of $X$ that cover $X$. Choose any $(x,∙) in A_i$. Then $x in X$, so there exists $A_i$ such that $x in pi_1(A_i)$. This means there exists $(y,z) in A_isubseteq X times {∙}$ such that $pi_1(y,z)=x$. But the only $z in {∙}$ is $∙$ itself, and $pi_1(y,z)=y$ by definition. So we have $(x,∙) in A_i$.
However, you should also justify why $pi_1(A_i)$ is open. In general the image of an open set is not necessarily open, but it will hold for projections.
answered Jan 26 at 13:54
kccukccu
10.6k11229
answered Jan 26 at 13:54
kccukccu
10.6k11229
answered Jan 26 at 13:54
kccukccu
10.6k11229
answered Jan 26 at 13:54
kccukccu
10.6k11229
10.6k11229
$begingroup$
I edited the question sorry for short explanations.
$endgroup$
– Ris
Jan 26 at 14:26
$begingroup$
In your edit, is $Y$ compact?
$endgroup$
– kccu
Jan 26 at 14:32
$begingroup$
Not necessarily.
$endgroup$
– Ris
Jan 26 at 14:34
$begingroup$
Then $X times Y$ is not necessarily compact. (Also, you are using $X$ to denote an element of $pi_1(S)$, but $X$ is also the name of your compact space...)
$endgroup$
– kccu
Jan 26 at 14:34
$begingroup$
Oh I edited it to $B$. Then $X times { bullet }$ is also not necessarily compact?
$endgroup$
– Ris
Jan 26 at 14:36
add a comment |
$begingroup$
I edited the question sorry for short explanations.
$endgroup$
– Ris
Jan 26 at 14:26
$begingroup$
In your edit, is $Y$ compact?
$endgroup$
– kccu
Jan 26 at 14:32
$begingroup$
Not necessarily.
$endgroup$
– Ris
Jan 26 at 14:34
$begingroup$
Then $X times Y$ is not necessarily compact. (Also, you are using $X$ to denote an element of $pi_1(S)$, but $X$ is also the name of your compact space...)
$endgroup$
– kccu
Jan 26 at 14:34
$begingroup$
Oh I edited it to $B$. Then $X times { bullet }$ is also not necessarily compact?
$endgroup$
– Ris
Jan 26 at 14:36
$begingroup$
I edited the question sorry for short explanations.
$endgroup$
– Ris
Jan 26 at 14:26
$begingroup$
I edited the question sorry for short explanations.
$endgroup$
– Ris
Jan 26 at 14:26
$begingroup$
In your edit, is $Y$ compact?
$endgroup$
– kccu
Jan 26 at 14:32
$begingroup$
In your edit, is $Y$ compact?
$endgroup$
– kccu
Jan 26 at 14:32
$begingroup$
Not necessarily.
$endgroup$
– Ris
Jan 26 at 14:34
$begingroup$
Not necessarily.
$endgroup$
– Ris
Jan 26 at 14:34
$begingroup$
Then $X times Y$ is not necessarily compact. (Also, you are using $X$ to denote an element of $pi_1(S)$, but $X$ is also the name of your compact space...)
$endgroup$
– kccu
Jan 26 at 14:34
$begingroup$
Then $X times Y$ is not necessarily compact. (Also, you are using $X$ to denote an element of $pi_1(S)$, but $X$ is also the name of your compact space...)
$endgroup$
– kccu
Jan 26 at 14:34
$begingroup$
Oh I edited it to $B$. Then $X times { bullet }$ is also not necessarily compact?
$endgroup$
– Ris
Jan 26 at 14:36
$begingroup$
Oh I edited it to $B$. Then $X times { bullet }$ is also not necessarily compact?
$endgroup$
– Ris
Jan 26 at 14:36
add a comment |
$begingroup$
$pi_1$ is a homeomorphism between X×{•} and X.
answered Jan 26 at 14:02
William ElliotWilliam Elliot
8,7962820
$endgroup$
add a comment |
$begingroup$
$pi_1$ is a homeomorphism between X×{•} and X.
answered Jan 26 at 14:02
William ElliotWilliam Elliot
8,7962820
$endgroup$
add a comment |
$begingroup$
$pi_1$ is a homeomorphism between X×{•} and X.
answered Jan 26 at 14:02
William ElliotWilliam Elliot
8,7962820
$endgroup$
$pi_1$ is a homeomorphism between X×{•} and X.
answered Jan 26 at 14:02
William ElliotWilliam Elliot
8,7962820
answered Jan 26 at 14:02
William ElliotWilliam Elliot
8,7962820
answered Jan 26 at 14:02
William ElliotWilliam Elliot
8,7962820
answered Jan 26 at 14:02
William ElliotWilliam Elliot
8,7962820
8,7962820
add a comment |
add a comment |
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