Mixing
Product of a compact topological space and a singleton in another topological space is compact proof
$begingroup$
It's before we prove that 'Product of two compact sets is compact'. 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 $X times {bullet}$. How to complete the proof? 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 $S cup {B times W mid B in pi_1(S)}$ 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 27 at 1:15
RisRis
397
$endgroup$
add a comment |
$begingroup$
It's before we prove that 'Product of two compact sets is compact'. 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 $X times {bullet}$. How to complete the proof? 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 $S cup {B times W mid B in pi_1(S)}$ 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 27 at 1:15
RisRis
397
$endgroup$
add a comment |
$begingroup$
It's before we prove that 'Product of two compact sets is compact'. 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 $X times {bullet}$. How to complete the proof? 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 $S cup {B times W mid B in pi_1(S)}$ 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 27 at 1:15
RisRis
397
$endgroup$
It's before we prove that 'Product of two compact sets is compact'. 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 $X times {bullet}$. How to complete the proof? 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 $S cup {B times W mid B in pi_1(S)}$ 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 27 at 1:15
RisRis
397
asked Jan 27 at 1:15
RisRis
397
edited Jan 27 at 1:21
Ris
asked Jan 27 at 1:15
RisRis
397
asked Jan 27 at 1:15
RisRis
397
asked Jan 27 at 1:15
RisRis
397
397
add a comment |
add a comment |
1 Answer
1
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$begingroup$
There are two possible interpretations of the sentence "Let $S$ be an open cover of $X times {bullet}$":
Let $S$ be a collection of open subsets of $X times Y$ which cover $X times {bullet}$. In this case, you must be careful how you proceed, for the reason that you have observed.
Let $S$ be a collection of open subsets of $X times {bullet}$ which cover $X times {bullet}$, where $X times {bullet}$ is given the subspace topology of $X times Y$. In this case, you can simply observe that $pi_1: X times {bullet} to X$ is a homeomorphism, and the proof you've described works perfectly.
In the case that you are using version 1, observe that the intersection of an open subset of $X times Y$ with $X times {bullet}$ forms an open subset of $X times {bullet}$, which projects under $pi_1$. onto an open subset of $X$.
answered Jan 27 at 1:37
Dustan LevensteinDustan Levenstein
10k11647
$endgroup$
$begingroup$
Why is it an open subset? Perhaps from compactness of $X$? If we intersect an open disk in $mathbb{R}^2$ with a line $mathbb{R} times {bullet}$ it results in something like an open interval but it is not an open set of course. But $mathbb{R}$ is not compact.
$endgroup$
– Ris
Jan 27 at 4:53
$begingroup$
Actually there exists a closed square centered at the origin containing the open disk so we can replace $mathbb{R}$ with a closed interval $I$.
$endgroup$
– Ris
Jan 27 at 5:04
$begingroup$
@Ris it is literally the definition of the subspace topology that an open subset of $X times Y$ intersected with $X times {bullet}$ is an open subset of $X times {bullet}$.
$endgroup$
– Dustan Levenstein
Jan 27 at 18:25
$begingroup$
If you want to avoid using the subspace topology, just prove that $pi_1(U cap X times {bullet}) subset X$ is open for $U subset X times Y$ open.
$endgroup$
– Dustan Levenstein
Jan 27 at 18:28
$begingroup$
Oh I thought that it meant to be the subset which is open in $X times Y$. Does the term "open subset of $A$" always refer to an open set of the subspace topology on $A$? I was always confused of that.
$endgroup$
– Ris
Jan 28 at 3:59
|
show 1 more comment
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1 Answer
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1 Answer
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active
oldest
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active
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$begingroup$
There are two possible interpretations of the sentence "Let $S$ be an open cover of $X times {bullet}$":
Let $S$ be a collection of open subsets of $X times Y$ which cover $X times {bullet}$. In this case, you must be careful how you proceed, for the reason that you have observed.
Let $S$ be a collection of open subsets of $X times {bullet}$ which cover $X times {bullet}$, where $X times {bullet}$ is given the subspace topology of $X times Y$. In this case, you can simply observe that $pi_1: X times {bullet} to X$ is a homeomorphism, and the proof you've described works perfectly.
In the case that you are using version 1, observe that the intersection of an open subset of $X times Y$ with $X times {bullet}$ forms an open subset of $X times {bullet}$, which projects under $pi_1$. onto an open subset of $X$.
answered Jan 27 at 1:37
Dustan LevensteinDustan Levenstein
10k11647
$endgroup$
$begingroup$
Why is it an open subset? Perhaps from compactness of $X$? If we intersect an open disk in $mathbb{R}^2$ with a line $mathbb{R} times {bullet}$ it results in something like an open interval but it is not an open set of course. But $mathbb{R}$ is not compact.
$endgroup$
– Ris
Jan 27 at 4:53
$begingroup$
Actually there exists a closed square centered at the origin containing the open disk so we can replace $mathbb{R}$ with a closed interval $I$.
$endgroup$
– Ris
Jan 27 at 5:04
$begingroup$
@Ris it is literally the definition of the subspace topology that an open subset of $X times Y$ intersected with $X times {bullet}$ is an open subset of $X times {bullet}$.
$endgroup$
– Dustan Levenstein
Jan 27 at 18:25
$begingroup$
If you want to avoid using the subspace topology, just prove that $pi_1(U cap X times {bullet}) subset X$ is open for $U subset X times Y$ open.
$endgroup$
– Dustan Levenstein
Jan 27 at 18:28
$begingroup$
Oh I thought that it meant to be the subset which is open in $X times Y$. Does the term "open subset of $A$" always refer to an open set of the subspace topology on $A$? I was always confused of that.
$endgroup$
– Ris
Jan 28 at 3:59
|
show 1 more comment
$begingroup$
There are two possible interpretations of the sentence "Let $S$ be an open cover of $X times {bullet}$":
Let $S$ be a collection of open subsets of $X times Y$ which cover $X times {bullet}$. In this case, you must be careful how you proceed, for the reason that you have observed.
Let $S$ be a collection of open subsets of $X times {bullet}$ which cover $X times {bullet}$, where $X times {bullet}$ is given the subspace topology of $X times Y$. In this case, you can simply observe that $pi_1: X times {bullet} to X$ is a homeomorphism, and the proof you've described works perfectly.
In the case that you are using version 1, observe that the intersection of an open subset of $X times Y$ with $X times {bullet}$ forms an open subset of $X times {bullet}$, which projects under $pi_1$. onto an open subset of $X$.
answered Jan 27 at 1:37
Dustan LevensteinDustan Levenstein
10k11647
$endgroup$
$begingroup$
Why is it an open subset? Perhaps from compactness of $X$? If we intersect an open disk in $mathbb{R}^2$ with a line $mathbb{R} times {bullet}$ it results in something like an open interval but it is not an open set of course. But $mathbb{R}$ is not compact.
$endgroup$
– Ris
Jan 27 at 4:53
$begingroup$
Actually there exists a closed square centered at the origin containing the open disk so we can replace $mathbb{R}$ with a closed interval $I$.
$endgroup$
– Ris
Jan 27 at 5:04
$begingroup$
@Ris it is literally the definition of the subspace topology that an open subset of $X times Y$ intersected with $X times {bullet}$ is an open subset of $X times {bullet}$.
$endgroup$
– Dustan Levenstein
Jan 27 at 18:25
$begingroup$
If you want to avoid using the subspace topology, just prove that $pi_1(U cap X times {bullet}) subset X$ is open for $U subset X times Y$ open.
$endgroup$
– Dustan Levenstein
Jan 27 at 18:28
$begingroup$
Oh I thought that it meant to be the subset which is open in $X times Y$. Does the term "open subset of $A$" always refer to an open set of the subspace topology on $A$? I was always confused of that.
$endgroup$
– Ris
Jan 28 at 3:59
|
show 1 more comment
$begingroup$
There are two possible interpretations of the sentence "Let $S$ be an open cover of $X times {bullet}$":
Let $S$ be a collection of open subsets of $X times Y$ which cover $X times {bullet}$. In this case, you must be careful how you proceed, for the reason that you have observed.
Let $S$ be a collection of open subsets of $X times {bullet}$ which cover $X times {bullet}$, where $X times {bullet}$ is given the subspace topology of $X times Y$. In this case, you can simply observe that $pi_1: X times {bullet} to X$ is a homeomorphism, and the proof you've described works perfectly.
In the case that you are using version 1, observe that the intersection of an open subset of $X times Y$ with $X times {bullet}$ forms an open subset of $X times {bullet}$, which projects under $pi_1$. onto an open subset of $X$.
answered Jan 27 at 1:37
Dustan LevensteinDustan Levenstein
10k11647
$endgroup$
There are two possible interpretations of the sentence "Let $S$ be an open cover of $X times {bullet}$":
Let $S$ be a collection of open subsets of $X times Y$ which cover $X times {bullet}$. In this case, you must be careful how you proceed, for the reason that you have observed.
Let $S$ be a collection of open subsets of $X times {bullet}$ which cover $X times {bullet}$, where $X times {bullet}$ is given the subspace topology of $X times Y$. In this case, you can simply observe that $pi_1: X times {bullet} to X$ is a homeomorphism, and the proof you've described works perfectly.
In the case that you are using version 1, observe that the intersection of an open subset of $X times Y$ with $X times {bullet}$ forms an open subset of $X times {bullet}$, which projects under $pi_1$. onto an open subset of $X$.
answered Jan 27 at 1:37
Dustan LevensteinDustan Levenstein
10k11647
answered Jan 27 at 1:37
Dustan LevensteinDustan Levenstein
10k11647
answered Jan 27 at 1:37
Dustan LevensteinDustan Levenstein
10k11647
answered Jan 27 at 1:37
Dustan LevensteinDustan Levenstein
10k11647
10k11647
$begingroup$
Why is it an open subset? Perhaps from compactness of $X$? If we intersect an open disk in $mathbb{R}^2$ with a line $mathbb{R} times {bullet}$ it results in something like an open interval but it is not an open set of course. But $mathbb{R}$ is not compact.
$endgroup$
– Ris
Jan 27 at 4:53
$begingroup$
Actually there exists a closed square centered at the origin containing the open disk so we can replace $mathbb{R}$ with a closed interval $I$.
$endgroup$
– Ris
Jan 27 at 5:04
$begingroup$
@Ris it is literally the definition of the subspace topology that an open subset of $X times Y$ intersected with $X times {bullet}$ is an open subset of $X times {bullet}$.
$endgroup$
– Dustan Levenstein
Jan 27 at 18:25
$begingroup$
If you want to avoid using the subspace topology, just prove that $pi_1(U cap X times {bullet}) subset X$ is open for $U subset X times Y$ open.
$endgroup$
– Dustan Levenstein
Jan 27 at 18:28
$begingroup$
Oh I thought that it meant to be the subset which is open in $X times Y$. Does the term "open subset of $A$" always refer to an open set of the subspace topology on $A$? I was always confused of that.
$endgroup$
– Ris
Jan 28 at 3:59
|
show 1 more comment
$begingroup$
Why is it an open subset? Perhaps from compactness of $X$? If we intersect an open disk in $mathbb{R}^2$ with a line $mathbb{R} times {bullet}$ it results in something like an open interval but it is not an open set of course. But $mathbb{R}$ is not compact.
$endgroup$
– Ris
Jan 27 at 4:53
$begingroup$
Actually there exists a closed square centered at the origin containing the open disk so we can replace $mathbb{R}$ with a closed interval $I$.
$endgroup$
– Ris
Jan 27 at 5:04
$begingroup$
@Ris it is literally the definition of the subspace topology that an open subset of $X times Y$ intersected with $X times {bullet}$ is an open subset of $X times {bullet}$.
$endgroup$
– Dustan Levenstein
Jan 27 at 18:25
$begingroup$
If you want to avoid using the subspace topology, just prove that $pi_1(U cap X times {bullet}) subset X$ is open for $U subset X times Y$ open.
$endgroup$
– Dustan Levenstein
Jan 27 at 18:28
$begingroup$
Oh I thought that it meant to be the subset which is open in $X times Y$. Does the term "open subset of $A$" always refer to an open set of the subspace topology on $A$? I was always confused of that.
$endgroup$
– Ris
Jan 28 at 3:59
$begingroup$
Why is it an open subset? Perhaps from compactness of $X$? If we intersect an open disk in $mathbb{R}^2$ with a line $mathbb{R} times {bullet}$ it results in something like an open interval but it is not an open set of course. But $mathbb{R}$ is not compact.
$endgroup$
– Ris
Jan 27 at 4:53
$begingroup$
Why is it an open subset? Perhaps from compactness of $X$? If we intersect an open disk in $mathbb{R}^2$ with a line $mathbb{R} times {bullet}$ it results in something like an open interval but it is not an open set of course. But $mathbb{R}$ is not compact.
$endgroup$
– Ris
Jan 27 at 4:53
$begingroup$
Actually there exists a closed square centered at the origin containing the open disk so we can replace $mathbb{R}$ with a closed interval $I$.
$endgroup$
– Ris
Jan 27 at 5:04
$begingroup$
Actually there exists a closed square centered at the origin containing the open disk so we can replace $mathbb{R}$ with a closed interval $I$.
$endgroup$
– Ris
Jan 27 at 5:04
$begingroup$
@Ris it is literally the definition of the subspace topology that an open subset of $X times Y$ intersected with $X times {bullet}$ is an open subset of $X times {bullet}$.
$endgroup$
– Dustan Levenstein
Jan 27 at 18:25
$begingroup$
@Ris it is literally the definition of the subspace topology that an open subset of $X times Y$ intersected with $X times {bullet}$ is an open subset of $X times {bullet}$.
$endgroup$
– Dustan Levenstein
Jan 27 at 18:25
$begingroup$
If you want to avoid using the subspace topology, just prove that $pi_1(U cap X times {bullet}) subset X$ is open for $U subset X times Y$ open.
$endgroup$
– Dustan Levenstein
Jan 27 at 18:28
$begingroup$
If you want to avoid using the subspace topology, just prove that $pi_1(U cap X times {bullet}) subset X$ is open for $U subset X times Y$ open.
$endgroup$
– Dustan Levenstein
Jan 27 at 18:28
$begingroup$
Oh I thought that it meant to be the subset which is open in $X times Y$. Does the term "open subset of $A$" always refer to an open set of the subspace topology on $A$? I was always confused of that.
$endgroup$
– Ris
Jan 28 at 3:59
$begingroup$
Oh I thought that it meant to be the subset which is open in $X times Y$. Does the term "open subset of $A$" always refer to an open set of the subspace topology on $A$? I was always confused of that.
$endgroup$
– Ris
Jan 28 at 3:59
|
show 1 more comment
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