Mixing
Intersection of finite number of compact sets is compact?
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Is the the intersection of a finite number of compact sets is compact? If not please give a counter example to demonstrate this is not true.
I said that this is true because the intersection of finite number of compact sets are closed. Which therefore means that it will be bounded because the intersection is contained by every set. I am not sure if this is correct.
Thank you for the help
general-topology examples-counterexamples compactness
edited Jun 1 '13 at 18:37
M.Sina
1,20511130
asked Nov 5 '12 at 16:37
John BuchtaJohn Buchta
153125
$endgroup$
add a comment |
$begingroup$
Is the the intersection of a finite number of compact sets is compact? If not please give a counter example to demonstrate this is not true.
I said that this is true because the intersection of finite number of compact sets are closed. Which therefore means that it will be bounded because the intersection is contained by every set. I am not sure if this is correct.
Thank you for the help
general-topology examples-counterexamples compactness
edited Jun 1 '13 at 18:37
M.Sina
1,20511130
asked Nov 5 '12 at 16:37
John BuchtaJohn Buchta
153125
$endgroup$
6
$begingroup$
What is your definition of compactness?
$endgroup$
– Phira
Nov 5 '12 at 16:41
add a comment |
$begingroup$
Is the the intersection of a finite number of compact sets is compact? If not please give a counter example to demonstrate this is not true.
I said that this is true because the intersection of finite number of compact sets are closed. Which therefore means that it will be bounded because the intersection is contained by every set. I am not sure if this is correct.
Thank you for the help
general-topology examples-counterexamples compactness
edited Jun 1 '13 at 18:37
M.Sina
1,20511130
asked Nov 5 '12 at 16:37
John BuchtaJohn Buchta
153125
$endgroup$
Is the the intersection of a finite number of compact sets is compact? If not please give a counter example to demonstrate this is not true.
I said that this is true because the intersection of finite number of compact sets are closed. Which therefore means that it will be bounded because the intersection is contained by every set. I am not sure if this is correct.
Thank you for the help
general-topology examples-counterexamples compactness
general-topology examples-counterexamples compactness
edited Jun 1 '13 at 18:37
M.Sina
1,20511130
asked Nov 5 '12 at 16:37
John BuchtaJohn Buchta
153125
edited Jun 1 '13 at 18:37
M.Sina
1,20511130
asked Nov 5 '12 at 16:37
John BuchtaJohn Buchta
153125
edited Jun 1 '13 at 18:37
M.Sina
1,20511130
edited Jun 1 '13 at 18:37
M.Sina
1,20511130
edited Jun 1 '13 at 18:37
M.Sina
1,20511130
1,20511130
asked Nov 5 '12 at 16:37
John BuchtaJohn Buchta
153125
asked Nov 5 '12 at 16:37
John BuchtaJohn Buchta
153125
asked Nov 5 '12 at 16:37
John BuchtaJohn Buchta
153125
153125
6
$begingroup$
What is your definition of compactness?
$endgroup$
– Phira
Nov 5 '12 at 16:41
add a comment |
6
$begingroup$
What is your definition of compactness?
$endgroup$
– Phira
Nov 5 '12 at 16:41
6
6
$begingroup$
What is your definition of compactness?
$endgroup$
– Phira
Nov 5 '12 at 16:41
$begingroup$
What is your definition of compactness?
$endgroup$
– Phira
Nov 5 '12 at 16:41
add a comment |
3 Answers
3
active
oldest
votes
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For Hausdorff spaces your statement is true, since compact sets in a Hausdorff space must be closed and a closed subset of a compact set is compact. In fact, in this case, the intersection of any family of compact sets is compact (by the same argument). However, in general it is false.
Take $mathbb{N}$ with the discrete topology and add in two more points $x_1$ and $x_2$. Declare that the only open sets containing $x_i$ to be ${x_i}cup mathbb{N}$ and ${x_1 , x_2}cup mathbb{N}$. (If you can't see it immediately, check this gives a topology on ${x_1 , x_2}cup mathbb{N}$).
Now ${x_i}cup mathbb{N}$ is compact for $i=1,2$, since any open cover must contain ${x_i}cup mathbb{N}$ (it is the only open set containing $x_i$). However, their intersection, $mathbb{N}$, is infinite and discrete, so definitely not compact.
$endgroup$
$begingroup$
How is {xi}∪N compact? What is the finite subcover of the only open cover -- itself? You can't cover N with a finite subcover right?
$endgroup$
– verticese
Apr 7 '16 at 15:56
$begingroup$
As written, that sentence was wrong. There is more than one open cover. However every open cover must contain ${x_i}cup mathbb{N}$, and hence has a finite subcover consisting of a single set. I've edited the final paragraph to remove this inaccuracy.
$endgroup$
– user45861
Apr 7 '16 at 21:30
$begingroup$
I am sorry I am still confused. Say I take the open cover {xi}∪N, how can you find a finite subcover? If you cannot even cover N with a finite subcover, how can you cover {xi}∪N with one? Thanks
$endgroup$
– verticese
Apr 14 '16 at 12:55
$begingroup$
@verticese The open cover ${x_i}cupBbb N$ must contain an open member $U$ with $x_1in U$. But then necessarily $Bbb Nsubset U$ as well.
$endgroup$
– Hagen von Eitzen
Apr 14 '16 at 13:06
1
$begingroup$
I think you are confused about what finite cover means. A cover is a collection of sets. It is finite if this collection contains only finitely many sets. It does not mean that the sets in the cover are finite (indeed, as in this case the sets themselves can be infinite).
$endgroup$
– user45861
Apr 14 '16 at 13:52
|
show 1 more comment
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In a $T_2$ space (Hausdorff), compact subsets are closed. Let be a $T_2$ space and let $F$ be a family of compact subsets of $S$. To avoid trivial cases let $F$ have at least one non-empty member $f$.
Now $G=cap F$ is closed and $G^c=Sbackslash G$ is open. If $V$ is a family of open sets with $cup Vsupset G,$ then $V'=Vcup {G^c}$ is an open cover of $f.$ (Indeed, $cup V'=S.$)
Since $f$ is compact, there exists a finite $Hsubset V'$ with $cup Hsupset f.$ Then $H'=Hbackslash {G^c}$ is a finite subset of $V$, and $cup H'supset G.$
So $G$ is compact.
Note that we needed each member of $F$ to be closed, so that $G$ is closed, so that $G^c$ is open, so that $V'$ is an open family. As an earlier answer shows, if $S$ is not $T_2$ then this can fail.
answered Apr 7 '16 at 22:27
DanielWainfleetDanielWainfleet
35.3k31648
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add a comment |
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I would like to supply another answer to this interesting question. I learnt the following example in a masters thesis of Cecil Eugene Denney (Kansas State University, 1966). Here is a link to the thesis. What follows below is a modified version of Example 3 in page 27 of the thesis.
Example. Let $X={a, b}$ be a topological space whose only open sets are $emptyset$ and $X$ itself (so $X$ has the indiscrete topology). Let $Y=mathbb{R}$ be the set of all real numbers with the standard topology. We will consider the topological space $Xtimes Y$. Consider the following two subsets of $Xtimes Y$:
$$
C = {(a, y): 1leq y < 3} cup {(b, y): 3leq y leq 4}
$$
and
$$
D = {(a, y): 2<y leq 4} cup {(b, y): 1leq y leq 2}.
$$
We claim that $C$ and $D$ are both compact. Indeed, in given any open cover $mathcal{U}$ for $C$, each open set must be of the form $Xtimes U$ where $U$ is an open subset of $Y$ (this is where we use the fact that the only open subsets of $X$ are $emptyset$ and $X$). Using the projection $pi_Y: Xtimes Y to Y$, we note that the collection of all $U$ such that $Xtimes Uinmathcal{U}$ forms an open cover of $pi_Y(C) = [1, 4]$ which is compact. Thus, there are finitely many $U_i$ which cover $[1, 4]$, and consequently finitely many $Xtimes U_i$ which cover $C$. This shows $C$ is compact, and the verification for compactness of $D$ is identical.
However, $Ccap D = {(a, y): 2 < y < 3}$ is homeomorphic, via the projection $pi_Y$, to the open interval $(2, 3)$ in $Y=mathbb{R}$, and it is well-known that $(2, 3)$ is not compact.
answered Jan 19 at 9:28
PrismPrism
5,05731981
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For Hausdorff spaces your statement is true, since compact sets in a Hausdorff space must be closed and a closed subset of a compact set is compact. In fact, in this case, the intersection of any family of compact sets is compact (by the same argument). However, in general it is false.
Take $mathbb{N}$ with the discrete topology and add in two more points $x_1$ and $x_2$. Declare that the only open sets containing $x_i$ to be ${x_i}cup mathbb{N}$ and ${x_1 , x_2}cup mathbb{N}$. (If you can't see it immediately, check this gives a topology on ${x_1 , x_2}cup mathbb{N}$).
Now ${x_i}cup mathbb{N}$ is compact for $i=1,2$, since any open cover must contain ${x_i}cup mathbb{N}$ (it is the only open set containing $x_i$). However, their intersection, $mathbb{N}$, is infinite and discrete, so definitely not compact.
$endgroup$
$begingroup$
How is {xi}∪N compact? What is the finite subcover of the only open cover -- itself? You can't cover N with a finite subcover right?
$endgroup$
– verticese
Apr 7 '16 at 15:56
$begingroup$
As written, that sentence was wrong. There is more than one open cover. However every open cover must contain ${x_i}cup mathbb{N}$, and hence has a finite subcover consisting of a single set. I've edited the final paragraph to remove this inaccuracy.
$endgroup$
– user45861
Apr 7 '16 at 21:30
$begingroup$
I am sorry I am still confused. Say I take the open cover {xi}∪N, how can you find a finite subcover? If you cannot even cover N with a finite subcover, how can you cover {xi}∪N with one? Thanks
$endgroup$
– verticese
Apr 14 '16 at 12:55
$begingroup$
@verticese The open cover ${x_i}cupBbb N$ must contain an open member $U$ with $x_1in U$. But then necessarily $Bbb Nsubset U$ as well.
$endgroup$
– Hagen von Eitzen
Apr 14 '16 at 13:06
1
$begingroup$
I think you are confused about what finite cover means. A cover is a collection of sets. It is finite if this collection contains only finitely many sets. It does not mean that the sets in the cover are finite (indeed, as in this case the sets themselves can be infinite).
$endgroup$
– user45861
Apr 14 '16 at 13:52
|
show 1 more comment
$begingroup$
For Hausdorff spaces your statement is true, since compact sets in a Hausdorff space must be closed and a closed subset of a compact set is compact. In fact, in this case, the intersection of any family of compact sets is compact (by the same argument). However, in general it is false.
Take $mathbb{N}$ with the discrete topology and add in two more points $x_1$ and $x_2$. Declare that the only open sets containing $x_i$ to be ${x_i}cup mathbb{N}$ and ${x_1 , x_2}cup mathbb{N}$. (If you can't see it immediately, check this gives a topology on ${x_1 , x_2}cup mathbb{N}$).
Now ${x_i}cup mathbb{N}$ is compact for $i=1,2$, since any open cover must contain ${x_i}cup mathbb{N}$ (it is the only open set containing $x_i$). However, their intersection, $mathbb{N}$, is infinite and discrete, so definitely not compact.
$endgroup$
$begingroup$
How is {xi}∪N compact? What is the finite subcover of the only open cover -- itself? You can't cover N with a finite subcover right?
$endgroup$
– verticese
Apr 7 '16 at 15:56
$begingroup$
As written, that sentence was wrong. There is more than one open cover. However every open cover must contain ${x_i}cup mathbb{N}$, and hence has a finite subcover consisting of a single set. I've edited the final paragraph to remove this inaccuracy.
$endgroup$
– user45861
Apr 7 '16 at 21:30
$begingroup$
I am sorry I am still confused. Say I take the open cover {xi}∪N, how can you find a finite subcover? If you cannot even cover N with a finite subcover, how can you cover {xi}∪N with one? Thanks
$endgroup$
– verticese
Apr 14 '16 at 12:55
$begingroup$
@verticese The open cover ${x_i}cupBbb N$ must contain an open member $U$ with $x_1in U$. But then necessarily $Bbb Nsubset U$ as well.
$endgroup$
– Hagen von Eitzen
Apr 14 '16 at 13:06
1
$begingroup$
I think you are confused about what finite cover means. A cover is a collection of sets. It is finite if this collection contains only finitely many sets. It does not mean that the sets in the cover are finite (indeed, as in this case the sets themselves can be infinite).
$endgroup$
– user45861
Apr 14 '16 at 13:52
|
show 1 more comment
$begingroup$
For Hausdorff spaces your statement is true, since compact sets in a Hausdorff space must be closed and a closed subset of a compact set is compact. In fact, in this case, the intersection of any family of compact sets is compact (by the same argument). However, in general it is false.
Take $mathbb{N}$ with the discrete topology and add in two more points $x_1$ and $x_2$. Declare that the only open sets containing $x_i$ to be ${x_i}cup mathbb{N}$ and ${x_1 , x_2}cup mathbb{N}$. (If you can't see it immediately, check this gives a topology on ${x_1 , x_2}cup mathbb{N}$).
Now ${x_i}cup mathbb{N}$ is compact for $i=1,2$, since any open cover must contain ${x_i}cup mathbb{N}$ (it is the only open set containing $x_i$). However, their intersection, $mathbb{N}$, is infinite and discrete, so definitely not compact.
$endgroup$
For Hausdorff spaces your statement is true, since compact sets in a Hausdorff space must be closed and a closed subset of a compact set is compact. In fact, in this case, the intersection of any family of compact sets is compact (by the same argument). However, in general it is false.
Take $mathbb{N}$ with the discrete topology and add in two more points $x_1$ and $x_2$. Declare that the only open sets containing $x_i$ to be ${x_i}cup mathbb{N}$ and ${x_1 , x_2}cup mathbb{N}$. (If you can't see it immediately, check this gives a topology on ${x_1 , x_2}cup mathbb{N}$).
Now ${x_i}cup mathbb{N}$ is compact for $i=1,2$, since any open cover must contain ${x_i}cup mathbb{N}$ (it is the only open set containing $x_i$). However, their intersection, $mathbb{N}$, is infinite and discrete, so definitely not compact.
edited Apr 7 '16 at 21:27
answered Nov 5 '12 at 17:13
user45861
$begingroup$
How is {xi}∪N compact? What is the finite subcover of the only open cover -- itself? You can't cover N with a finite subcover right?
$endgroup$
– verticese
Apr 7 '16 at 15:56
$begingroup$
As written, that sentence was wrong. There is more than one open cover. However every open cover must contain ${x_i}cup mathbb{N}$, and hence has a finite subcover consisting of a single set. I've edited the final paragraph to remove this inaccuracy.
$endgroup$
– user45861
Apr 7 '16 at 21:30
$begingroup$
I am sorry I am still confused. Say I take the open cover {xi}∪N, how can you find a finite subcover? If you cannot even cover N with a finite subcover, how can you cover {xi}∪N with one? Thanks
$endgroup$
– verticese
Apr 14 '16 at 12:55
$begingroup$
@verticese The open cover ${x_i}cupBbb N$ must contain an open member $U$ with $x_1in U$. But then necessarily $Bbb Nsubset U$ as well.
$endgroup$
– Hagen von Eitzen
Apr 14 '16 at 13:06
1
$begingroup$
I think you are confused about what finite cover means. A cover is a collection of sets. It is finite if this collection contains only finitely many sets. It does not mean that the sets in the cover are finite (indeed, as in this case the sets themselves can be infinite).
$endgroup$
– user45861
Apr 14 '16 at 13:52
|
show 1 more comment
$begingroup$
How is {xi}∪N compact? What is the finite subcover of the only open cover -- itself? You can't cover N with a finite subcover right?
$endgroup$
– verticese
Apr 7 '16 at 15:56
$begingroup$
As written, that sentence was wrong. There is more than one open cover. However every open cover must contain ${x_i}cup mathbb{N}$, and hence has a finite subcover consisting of a single set. I've edited the final paragraph to remove this inaccuracy.
$endgroup$
– user45861
Apr 7 '16 at 21:30
$begingroup$
I am sorry I am still confused. Say I take the open cover {xi}∪N, how can you find a finite subcover? If you cannot even cover N with a finite subcover, how can you cover {xi}∪N with one? Thanks
$endgroup$
– verticese
Apr 14 '16 at 12:55
$begingroup$
@verticese The open cover ${x_i}cupBbb N$ must contain an open member $U$ with $x_1in U$. But then necessarily $Bbb Nsubset U$ as well.
$endgroup$
– Hagen von Eitzen
Apr 14 '16 at 13:06
1
$begingroup$
I think you are confused about what finite cover means. A cover is a collection of sets. It is finite if this collection contains only finitely many sets. It does not mean that the sets in the cover are finite (indeed, as in this case the sets themselves can be infinite).
$endgroup$
– user45861
Apr 14 '16 at 13:52
$begingroup$
How is {xi}∪N compact? What is the finite subcover of the only open cover -- itself? You can't cover N with a finite subcover right?
$endgroup$
– verticese
Apr 7 '16 at 15:56
$begingroup$
How is {xi}∪N compact? What is the finite subcover of the only open cover -- itself? You can't cover N with a finite subcover right?
$endgroup$
– verticese
Apr 7 '16 at 15:56
$begingroup$
As written, that sentence was wrong. There is more than one open cover. However every open cover must contain ${x_i}cup mathbb{N}$, and hence has a finite subcover consisting of a single set. I've edited the final paragraph to remove this inaccuracy.
$endgroup$
– user45861
Apr 7 '16 at 21:30
$begingroup$
As written, that sentence was wrong. There is more than one open cover. However every open cover must contain ${x_i}cup mathbb{N}$, and hence has a finite subcover consisting of a single set. I've edited the final paragraph to remove this inaccuracy.
$endgroup$
– user45861
Apr 7 '16 at 21:30
$begingroup$
I am sorry I am still confused. Say I take the open cover {xi}∪N, how can you find a finite subcover? If you cannot even cover N with a finite subcover, how can you cover {xi}∪N with one? Thanks
$endgroup$
– verticese
Apr 14 '16 at 12:55
$begingroup$
I am sorry I am still confused. Say I take the open cover {xi}∪N, how can you find a finite subcover? If you cannot even cover N with a finite subcover, how can you cover {xi}∪N with one? Thanks
$endgroup$
– verticese
Apr 14 '16 at 12:55
$begingroup$
@verticese The open cover ${x_i}cupBbb N$ must contain an open member $U$ with $x_1in U$. But then necessarily $Bbb Nsubset U$ as well.
$endgroup$
– Hagen von Eitzen
Apr 14 '16 at 13:06
$begingroup$
@verticese The open cover ${x_i}cupBbb N$ must contain an open member $U$ with $x_1in U$. But then necessarily $Bbb Nsubset U$ as well.
$endgroup$
– Hagen von Eitzen
Apr 14 '16 at 13:06
1
1
$begingroup$
I think you are confused about what finite cover means. A cover is a collection of sets. It is finite if this collection contains only finitely many sets. It does not mean that the sets in the cover are finite (indeed, as in this case the sets themselves can be infinite).
$endgroup$
– user45861
Apr 14 '16 at 13:52
$begingroup$
I think you are confused about what finite cover means. A cover is a collection of sets. It is finite if this collection contains only finitely many sets. It does not mean that the sets in the cover are finite (indeed, as in this case the sets themselves can be infinite).
$endgroup$
– user45861
Apr 14 '16 at 13:52
|
show 1 more comment
$begingroup$
In a $T_2$ space (Hausdorff), compact subsets are closed. Let be a $T_2$ space and let $F$ be a family of compact subsets of $S$. To avoid trivial cases let $F$ have at least one non-empty member $f$.
Now $G=cap F$ is closed and $G^c=Sbackslash G$ is open. If $V$ is a family of open sets with $cup Vsupset G,$ then $V'=Vcup {G^c}$ is an open cover of $f.$ (Indeed, $cup V'=S.$)
Since $f$ is compact, there exists a finite $Hsubset V'$ with $cup Hsupset f.$ Then $H'=Hbackslash {G^c}$ is a finite subset of $V$, and $cup H'supset G.$
So $G$ is compact.
Note that we needed each member of $F$ to be closed, so that $G$ is closed, so that $G^c$ is open, so that $V'$ is an open family. As an earlier answer shows, if $S$ is not $T_2$ then this can fail.
answered Apr 7 '16 at 22:27
DanielWainfleetDanielWainfleet
35.3k31648
$endgroup$
add a comment |
$begingroup$
In a $T_2$ space (Hausdorff), compact subsets are closed. Let be a $T_2$ space and let $F$ be a family of compact subsets of $S$. To avoid trivial cases let $F$ have at least one non-empty member $f$.
Now $G=cap F$ is closed and $G^c=Sbackslash G$ is open. If $V$ is a family of open sets with $cup Vsupset G,$ then $V'=Vcup {G^c}$ is an open cover of $f.$ (Indeed, $cup V'=S.$)
Since $f$ is compact, there exists a finite $Hsubset V'$ with $cup Hsupset f.$ Then $H'=Hbackslash {G^c}$ is a finite subset of $V$, and $cup H'supset G.$
So $G$ is compact.
Note that we needed each member of $F$ to be closed, so that $G$ is closed, so that $G^c$ is open, so that $V'$ is an open family. As an earlier answer shows, if $S$ is not $T_2$ then this can fail.
answered Apr 7 '16 at 22:27
DanielWainfleetDanielWainfleet
35.3k31648
$endgroup$
add a comment |
$begingroup$
In a $T_2$ space (Hausdorff), compact subsets are closed. Let be a $T_2$ space and let $F$ be a family of compact subsets of $S$. To avoid trivial cases let $F$ have at least one non-empty member $f$.
Now $G=cap F$ is closed and $G^c=Sbackslash G$ is open. If $V$ is a family of open sets with $cup Vsupset G,$ then $V'=Vcup {G^c}$ is an open cover of $f.$ (Indeed, $cup V'=S.$)
Since $f$ is compact, there exists a finite $Hsubset V'$ with $cup Hsupset f.$ Then $H'=Hbackslash {G^c}$ is a finite subset of $V$, and $cup H'supset G.$
So $G$ is compact.
Note that we needed each member of $F$ to be closed, so that $G$ is closed, so that $G^c$ is open, so that $V'$ is an open family. As an earlier answer shows, if $S$ is not $T_2$ then this can fail.
answered Apr 7 '16 at 22:27
DanielWainfleetDanielWainfleet
35.3k31648
$endgroup$
In a $T_2$ space (Hausdorff), compact subsets are closed. Let be a $T_2$ space and let $F$ be a family of compact subsets of $S$. To avoid trivial cases let $F$ have at least one non-empty member $f$.
Now $G=cap F$ is closed and $G^c=Sbackslash G$ is open. If $V$ is a family of open sets with $cup Vsupset G,$ then $V'=Vcup {G^c}$ is an open cover of $f.$ (Indeed, $cup V'=S.$)
Since $f$ is compact, there exists a finite $Hsubset V'$ with $cup Hsupset f.$ Then $H'=Hbackslash {G^c}$ is a finite subset of $V$, and $cup H'supset G.$
So $G$ is compact.
Note that we needed each member of $F$ to be closed, so that $G$ is closed, so that $G^c$ is open, so that $V'$ is an open family. As an earlier answer shows, if $S$ is not $T_2$ then this can fail.
answered Apr 7 '16 at 22:27
DanielWainfleetDanielWainfleet
35.3k31648
answered Apr 7 '16 at 22:27
DanielWainfleetDanielWainfleet
35.3k31648
answered Apr 7 '16 at 22:27
DanielWainfleetDanielWainfleet
35.3k31648
answered Apr 7 '16 at 22:27
DanielWainfleetDanielWainfleet
35.3k31648
35.3k31648
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I would like to supply another answer to this interesting question. I learnt the following example in a masters thesis of Cecil Eugene Denney (Kansas State University, 1966). Here is a link to the thesis. What follows below is a modified version of Example 3 in page 27 of the thesis.
Example. Let $X={a, b}$ be a topological space whose only open sets are $emptyset$ and $X$ itself (so $X$ has the indiscrete topology). Let $Y=mathbb{R}$ be the set of all real numbers with the standard topology. We will consider the topological space $Xtimes Y$. Consider the following two subsets of $Xtimes Y$:
$$
C = {(a, y): 1leq y < 3} cup {(b, y): 3leq y leq 4}
$$
and
$$
D = {(a, y): 2<y leq 4} cup {(b, y): 1leq y leq 2}.
$$
We claim that $C$ and $D$ are both compact. Indeed, in given any open cover $mathcal{U}$ for $C$, each open set must be of the form $Xtimes U$ where $U$ is an open subset of $Y$ (this is where we use the fact that the only open subsets of $X$ are $emptyset$ and $X$). Using the projection $pi_Y: Xtimes Y to Y$, we note that the collection of all $U$ such that $Xtimes Uinmathcal{U}$ forms an open cover of $pi_Y(C) = [1, 4]$ which is compact. Thus, there are finitely many $U_i$ which cover $[1, 4]$, and consequently finitely many $Xtimes U_i$ which cover $C$. This shows $C$ is compact, and the verification for compactness of $D$ is identical.
However, $Ccap D = {(a, y): 2 < y < 3}$ is homeomorphic, via the projection $pi_Y$, to the open interval $(2, 3)$ in $Y=mathbb{R}$, and it is well-known that $(2, 3)$ is not compact.
answered Jan 19 at 9:28
PrismPrism
5,05731981
$endgroup$
add a comment |
$begingroup$
I would like to supply another answer to this interesting question. I learnt the following example in a masters thesis of Cecil Eugene Denney (Kansas State University, 1966). Here is a link to the thesis. What follows below is a modified version of Example 3 in page 27 of the thesis.
Example. Let $X={a, b}$ be a topological space whose only open sets are $emptyset$ and $X$ itself (so $X$ has the indiscrete topology). Let $Y=mathbb{R}$ be the set of all real numbers with the standard topology. We will consider the topological space $Xtimes Y$. Consider the following two subsets of $Xtimes Y$:
$$
C = {(a, y): 1leq y < 3} cup {(b, y): 3leq y leq 4}
$$
and
$$
D = {(a, y): 2<y leq 4} cup {(b, y): 1leq y leq 2}.
$$
We claim that $C$ and $D$ are both compact. Indeed, in given any open cover $mathcal{U}$ for $C$, each open set must be of the form $Xtimes U$ where $U$ is an open subset of $Y$ (this is where we use the fact that the only open subsets of $X$ are $emptyset$ and $X$). Using the projection $pi_Y: Xtimes Y to Y$, we note that the collection of all $U$ such that $Xtimes Uinmathcal{U}$ forms an open cover of $pi_Y(C) = [1, 4]$ which is compact. Thus, there are finitely many $U_i$ which cover $[1, 4]$, and consequently finitely many $Xtimes U_i$ which cover $C$. This shows $C$ is compact, and the verification for compactness of $D$ is identical.
However, $Ccap D = {(a, y): 2 < y < 3}$ is homeomorphic, via the projection $pi_Y$, to the open interval $(2, 3)$ in $Y=mathbb{R}$, and it is well-known that $(2, 3)$ is not compact.
answered Jan 19 at 9:28
PrismPrism
5,05731981
$endgroup$
add a comment |
$begingroup$
I would like to supply another answer to this interesting question. I learnt the following example in a masters thesis of Cecil Eugene Denney (Kansas State University, 1966). Here is a link to the thesis. What follows below is a modified version of Example 3 in page 27 of the thesis.
Example. Let $X={a, b}$ be a topological space whose only open sets are $emptyset$ and $X$ itself (so $X$ has the indiscrete topology). Let $Y=mathbb{R}$ be the set of all real numbers with the standard topology. We will consider the topological space $Xtimes Y$. Consider the following two subsets of $Xtimes Y$:
$$
C = {(a, y): 1leq y < 3} cup {(b, y): 3leq y leq 4}
$$
and
$$
D = {(a, y): 2<y leq 4} cup {(b, y): 1leq y leq 2}.
$$
We claim that $C$ and $D$ are both compact. Indeed, in given any open cover $mathcal{U}$ for $C$, each open set must be of the form $Xtimes U$ where $U$ is an open subset of $Y$ (this is where we use the fact that the only open subsets of $X$ are $emptyset$ and $X$). Using the projection $pi_Y: Xtimes Y to Y$, we note that the collection of all $U$ such that $Xtimes Uinmathcal{U}$ forms an open cover of $pi_Y(C) = [1, 4]$ which is compact. Thus, there are finitely many $U_i$ which cover $[1, 4]$, and consequently finitely many $Xtimes U_i$ which cover $C$. This shows $C$ is compact, and the verification for compactness of $D$ is identical.
However, $Ccap D = {(a, y): 2 < y < 3}$ is homeomorphic, via the projection $pi_Y$, to the open interval $(2, 3)$ in $Y=mathbb{R}$, and it is well-known that $(2, 3)$ is not compact.
answered Jan 19 at 9:28
PrismPrism
5,05731981
$endgroup$
I would like to supply another answer to this interesting question. I learnt the following example in a masters thesis of Cecil Eugene Denney (Kansas State University, 1966). Here is a link to the thesis. What follows below is a modified version of Example 3 in page 27 of the thesis.
Example. Let $X={a, b}$ be a topological space whose only open sets are $emptyset$ and $X$ itself (so $X$ has the indiscrete topology). Let $Y=mathbb{R}$ be the set of all real numbers with the standard topology. We will consider the topological space $Xtimes Y$. Consider the following two subsets of $Xtimes Y$:
$$
C = {(a, y): 1leq y < 3} cup {(b, y): 3leq y leq 4}
$$
and
$$
D = {(a, y): 2<y leq 4} cup {(b, y): 1leq y leq 2}.
$$
We claim that $C$ and $D$ are both compact. Indeed, in given any open cover $mathcal{U}$ for $C$, each open set must be of the form $Xtimes U$ where $U$ is an open subset of $Y$ (this is where we use the fact that the only open subsets of $X$ are $emptyset$ and $X$). Using the projection $pi_Y: Xtimes Y to Y$, we note that the collection of all $U$ such that $Xtimes Uinmathcal{U}$ forms an open cover of $pi_Y(C) = [1, 4]$ which is compact. Thus, there are finitely many $U_i$ which cover $[1, 4]$, and consequently finitely many $Xtimes U_i$ which cover $C$. This shows $C$ is compact, and the verification for compactness of $D$ is identical.
However, $Ccap D = {(a, y): 2 < y < 3}$ is homeomorphic, via the projection $pi_Y$, to the open interval $(2, 3)$ in $Y=mathbb{R}$, and it is well-known that $(2, 3)$ is not compact.
answered Jan 19 at 9:28
PrismPrism
5,05731981
edited Jan 19 at 9:44
answered Jan 19 at 9:28
PrismPrism
5,05731981
answered Jan 19 at 9:28
PrismPrism
5,05731981
answered Jan 19 at 9:28
PrismPrism
5,05731981
5,05731981
add a comment |
add a comment |
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What is your definition of compactness?
$endgroup$
– Phira
Nov 5 '12 at 16:41