A set is compact in complement topology iff closed in standard topology
$begingroup$
Let $mathbb{R^n}$ be a topological space where we define topology $tau$ as follows: a subset $A subset mathbb{R^n}$ is closed if it is compact in the standard (Euclidean) topology, or if $A = mathbb{R^n}$
(This means that open subsets in $tau$ are those whose complement is compact in standard topology.)
Show that $F subset mathbb{R^n}$ is compact in $tau $ $Leftrightarrow $ $F$ is closed in standard topology.
My proof:
($Rightarrow$)
Let $F$ be compact in $tau$. Let $U $ be a open (in $tau$) cover of $F$. Because $F$ is compact there are open sets $U_1, U_2, ldots , U_n$ so that $F = U_1 cup U_2 cup ldots cup U_n $.
For each $U_i$ is $U_i^c$ compact in standard topology. A union of a finite number of compact sets is compact. So $F$ is compact in standard topology. Because $mathbb{R}$ is a metric space, a compact set $F$ is closed. (In standard topology.)
($Leftarrow$)
Let $F$ be closed in standard topology.
Here I am a bit lost as for how to continue. We know that $F^c$ is open (in s.t.)
Let $U$ be an open cover for $F$ in $tau$. Can we find a finite number of open sets that cover $F$? Those need to have compact complements in standard topology. Can we find a finite number of compact sets that make up a whole $F$?
There was a similar question asked a while ago, but I did not completely understand the answer.
general-topology
edited Dec 22 '18 at 11:19
SvanN
2,0661422
asked Dec 20 '18 at 17:06
CoupeauCoupeau
1296
$endgroup$
add a comment |
$begingroup$
Let $mathbb{R^n}$ be a topological space where we define topology $tau$ as follows: a subset $A subset mathbb{R^n}$ is closed if it is compact in the standard (Euclidean) topology, or if $A = mathbb{R^n}$
(This means that open subsets in $tau$ are those whose complement is compact in standard topology.)
Show that $F subset mathbb{R^n}$ is compact in $tau $ $Leftrightarrow $ $F$ is closed in standard topology.
My proof:
($Rightarrow$)
Let $F$ be compact in $tau$. Let $U $ be a open (in $tau$) cover of $F$. Because $F$ is compact there are open sets $U_1, U_2, ldots , U_n$ so that $F = U_1 cup U_2 cup ldots cup U_n $.
For each $U_i$ is $U_i^c$ compact in standard topology. A union of a finite number of compact sets is compact. So $F$ is compact in standard topology. Because $mathbb{R}$ is a metric space, a compact set $F$ is closed. (In standard topology.)
($Leftarrow$)
Let $F$ be closed in standard topology.
Here I am a bit lost as for how to continue. We know that $F^c$ is open (in s.t.)
Let $U$ be an open cover for $F$ in $tau$. Can we find a finite number of open sets that cover $F$? Those need to have compact complements in standard topology. Can we find a finite number of compact sets that make up a whole $F$?
There was a similar question asked a while ago, but I did not completely understand the answer.
general-topology
edited Dec 22 '18 at 11:19
SvanN
2,0661422
asked Dec 20 '18 at 17:06
CoupeauCoupeau
1296
$endgroup$
1
$begingroup$
The set $mathbb{R}^n$ is open in $mathbb{R}^n$, so the distinction in the definition is unnecessary. Also, are you sure this even defines a topology? The infinite intersection of open sets is not open, so in this definition, an infinite intersection of closed sets need not be closed; but this must hold in any topological space (pretty much by definition).
$endgroup$
– SvanN
Dec 20 '18 at 17:52
$begingroup$
It is true that $mathbb{R^n}$ is open, but as I am defining topology using closed sets I need to make sure that the whole space is also in the topology. And I couldn't cover that case without explicitly saying that $A$ is closed if it is equal to $mathbb{R^n}$. Because $mathbb{R^n}$ is not compact in Euclidean topology. By definition of topology finite union of closed and infinite intersection of closeds need to be closed. This is true in this case.
$endgroup$
– Coupeau
Dec 21 '18 at 14:39
1
$begingroup$
I think you're mixing up the terms 'closed' and 'compact'. No, you do not need the distinction: $mathbb{R}^n$ is open in the usual topology on $mathbb{R}^n$, hence it is closed in your new "topology". But as I indicated, and as jgon explained below, you do not have a topology.
$endgroup$
– SvanN
Dec 21 '18 at 15:45
$begingroup$
OOHH SORRY! This is totally my mistake. I Wrote the definition of topology wrong. It should say$mathbb{R^n}$ is closed in $tau$ if it is COMPACT in standard topology or $A = mathbb{R^n}$
$endgroup$
– Coupeau
Dec 22 '18 at 11:08
add a comment |
$begingroup$
Let $mathbb{R^n}$ be a topological space where we define topology $tau$ as follows: a subset $A subset mathbb{R^n}$ is closed if it is compact in the standard (Euclidean) topology, or if $A = mathbb{R^n}$
(This means that open subsets in $tau$ are those whose complement is compact in standard topology.)
Show that $F subset mathbb{R^n}$ is compact in $tau $ $Leftrightarrow $ $F$ is closed in standard topology.
My proof:
($Rightarrow$)
Let $F$ be compact in $tau$. Let $U $ be a open (in $tau$) cover of $F$. Because $F$ is compact there are open sets $U_1, U_2, ldots , U_n$ so that $F = U_1 cup U_2 cup ldots cup U_n $.
For each $U_i$ is $U_i^c$ compact in standard topology. A union of a finite number of compact sets is compact. So $F$ is compact in standard topology. Because $mathbb{R}$ is a metric space, a compact set $F$ is closed. (In standard topology.)
($Leftarrow$)
Let $F$ be closed in standard topology.
Here I am a bit lost as for how to continue. We know that $F^c$ is open (in s.t.)
Let $U$ be an open cover for $F$ in $tau$. Can we find a finite number of open sets that cover $F$? Those need to have compact complements in standard topology. Can we find a finite number of compact sets that make up a whole $F$?
There was a similar question asked a while ago, but I did not completely understand the answer.
general-topology
edited Dec 22 '18 at 11:19
SvanN
2,0661422
asked Dec 20 '18 at 17:06
CoupeauCoupeau
1296
$endgroup$
Let $mathbb{R^n}$ be a topological space where we define topology $tau$ as follows: a subset $A subset mathbb{R^n}$ is closed if it is compact in the standard (Euclidean) topology, or if $A = mathbb{R^n}$
(This means that open subsets in $tau$ are those whose complement is compact in standard topology.)
Show that $F subset mathbb{R^n}$ is compact in $tau $ $Leftrightarrow $ $F$ is closed in standard topology.
My proof:
($Rightarrow$)
Let $F$ be compact in $tau$. Let $U $ be a open (in $tau$) cover of $F$. Because $F$ is compact there are open sets $U_1, U_2, ldots , U_n$ so that $F = U_1 cup U_2 cup ldots cup U_n $.
For each $U_i$ is $U_i^c$ compact in standard topology. A union of a finite number of compact sets is compact. So $F$ is compact in standard topology. Because $mathbb{R}$ is a metric space, a compact set $F$ is closed. (In standard topology.)
($Leftarrow$)
Let $F$ be closed in standard topology.
Here I am a bit lost as for how to continue. We know that $F^c$ is open (in s.t.)
Let $U$ be an open cover for $F$ in $tau$. Can we find a finite number of open sets that cover $F$? Those need to have compact complements in standard topology. Can we find a finite number of compact sets that make up a whole $F$?
There was a similar question asked a while ago, but I did not completely understand the answer.
general-topology
general-topology
edited Dec 22 '18 at 11:19
SvanN
2,0661422
asked Dec 20 '18 at 17:06
CoupeauCoupeau
1296
edited Dec 22 '18 at 11:19
SvanN
2,0661422
asked Dec 20 '18 at 17:06
CoupeauCoupeau
1296
edited Dec 22 '18 at 11:19
SvanN
2,0661422
edited Dec 22 '18 at 11:19
SvanN
2,0661422
edited Dec 22 '18 at 11:19
SvanN
2,0661422
2,0661422
asked Dec 20 '18 at 17:06
CoupeauCoupeau
1296
asked Dec 20 '18 at 17:06
CoupeauCoupeau
1296
asked Dec 20 '18 at 17:06
CoupeauCoupeau
1296
1296
1
$begingroup$
The set $mathbb{R}^n$ is open in $mathbb{R}^n$, so the distinction in the definition is unnecessary. Also, are you sure this even defines a topology? The infinite intersection of open sets is not open, so in this definition, an infinite intersection of closed sets need not be closed; but this must hold in any topological space (pretty much by definition).
$endgroup$
– SvanN
Dec 20 '18 at 17:52
$begingroup$
It is true that $mathbb{R^n}$ is open, but as I am defining topology using closed sets I need to make sure that the whole space is also in the topology. And I couldn't cover that case without explicitly saying that $A$ is closed if it is equal to $mathbb{R^n}$. Because $mathbb{R^n}$ is not compact in Euclidean topology. By definition of topology finite union of closed and infinite intersection of closeds need to be closed. This is true in this case.
$endgroup$
– Coupeau
Dec 21 '18 at 14:39
1
$begingroup$
I think you're mixing up the terms 'closed' and 'compact'. No, you do not need the distinction: $mathbb{R}^n$ is open in the usual topology on $mathbb{R}^n$, hence it is closed in your new "topology". But as I indicated, and as jgon explained below, you do not have a topology.
$endgroup$
– SvanN
Dec 21 '18 at 15:45
$begingroup$
OOHH SORRY! This is totally my mistake. I Wrote the definition of topology wrong. It should say$mathbb{R^n}$ is closed in $tau$ if it is COMPACT in standard topology or $A = mathbb{R^n}$
$endgroup$
– Coupeau
Dec 22 '18 at 11:08
add a comment |
1
$begingroup$
The set $mathbb{R}^n$ is open in $mathbb{R}^n$, so the distinction in the definition is unnecessary. Also, are you sure this even defines a topology? The infinite intersection of open sets is not open, so in this definition, an infinite intersection of closed sets need not be closed; but this must hold in any topological space (pretty much by definition).
$endgroup$
– SvanN
Dec 20 '18 at 17:52
$begingroup$
It is true that $mathbb{R^n}$ is open, but as I am defining topology using closed sets I need to make sure that the whole space is also in the topology. And I couldn't cover that case without explicitly saying that $A$ is closed if it is equal to $mathbb{R^n}$. Because $mathbb{R^n}$ is not compact in Euclidean topology. By definition of topology finite union of closed and infinite intersection of closeds need to be closed. This is true in this case.
$endgroup$
– Coupeau
Dec 21 '18 at 14:39
1
$begingroup$
I think you're mixing up the terms 'closed' and 'compact'. No, you do not need the distinction: $mathbb{R}^n$ is open in the usual topology on $mathbb{R}^n$, hence it is closed in your new "topology". But as I indicated, and as jgon explained below, you do not have a topology.
$endgroup$
– SvanN
Dec 21 '18 at 15:45
$begingroup$
OOHH SORRY! This is totally my mistake. I Wrote the definition of topology wrong. It should say$mathbb{R^n}$ is closed in $tau$ if it is COMPACT in standard topology or $A = mathbb{R^n}$
$endgroup$
– Coupeau
Dec 22 '18 at 11:08
1
1
$begingroup$
The set $mathbb{R}^n$ is open in $mathbb{R}^n$, so the distinction in the definition is unnecessary. Also, are you sure this even defines a topology? The infinite intersection of open sets is not open, so in this definition, an infinite intersection of closed sets need not be closed; but this must hold in any topological space (pretty much by definition).
$endgroup$
– SvanN
Dec 20 '18 at 17:52
$begingroup$
The set $mathbb{R}^n$ is open in $mathbb{R}^n$, so the distinction in the definition is unnecessary. Also, are you sure this even defines a topology? The infinite intersection of open sets is not open, so in this definition, an infinite intersection of closed sets need not be closed; but this must hold in any topological space (pretty much by definition).
$endgroup$
– SvanN
Dec 20 '18 at 17:52
$begingroup$
It is true that $mathbb{R^n}$ is open, but as I am defining topology using closed sets I need to make sure that the whole space is also in the topology. And I couldn't cover that case without explicitly saying that $A$ is closed if it is equal to $mathbb{R^n}$. Because $mathbb{R^n}$ is not compact in Euclidean topology. By definition of topology finite union of closed and infinite intersection of closeds need to be closed. This is true in this case.
$endgroup$
– Coupeau
Dec 21 '18 at 14:39
$begingroup$
It is true that $mathbb{R^n}$ is open, but as I am defining topology using closed sets I need to make sure that the whole space is also in the topology. And I couldn't cover that case without explicitly saying that $A$ is closed if it is equal to $mathbb{R^n}$. Because $mathbb{R^n}$ is not compact in Euclidean topology. By definition of topology finite union of closed and infinite intersection of closeds need to be closed. This is true in this case.
$endgroup$
– Coupeau
Dec 21 '18 at 14:39
1
1
$begingroup$
I think you're mixing up the terms 'closed' and 'compact'. No, you do not need the distinction: $mathbb{R}^n$ is open in the usual topology on $mathbb{R}^n$, hence it is closed in your new "topology". But as I indicated, and as jgon explained below, you do not have a topology.
$endgroup$
– SvanN
Dec 21 '18 at 15:45
$begingroup$
I think you're mixing up the terms 'closed' and 'compact'. No, you do not need the distinction: $mathbb{R}^n$ is open in the usual topology on $mathbb{R}^n$, hence it is closed in your new "topology". But as I indicated, and as jgon explained below, you do not have a topology.
$endgroup$
– SvanN
Dec 21 '18 at 15:45
$begingroup$
OOHH SORRY! This is totally my mistake. I Wrote the definition of topology wrong. It should say$mathbb{R^n}$ is closed in $tau$ if it is COMPACT in standard topology or $A = mathbb{R^n}$
$endgroup$
– Coupeau
Dec 22 '18 at 11:08
$begingroup$
OOHH SORRY! This is totally my mistake. I Wrote the definition of topology wrong. It should say$mathbb{R^n}$ is closed in $tau$ if it is COMPACT in standard topology or $A = mathbb{R^n}$
$endgroup$
– Coupeau
Dec 22 '18 at 11:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As SvanN says in the comments, your "topology" doesn't form a topology.
The open subsets of $Bbb{R}^n$ are not closed under arbitrary intersections:
$$bigcap_{n=1}^infty left(frac{-1}{n},frac{1}{n}right) = {0},$$
so the open subsets of $Bbb{R}^n$ cannot be the closed sets of a topology.
Instead, the question you've linked suggests you should be defining your topology as
$U$ is open in $tau$ if $U^C$ is compact in the usual topology, plus the null set of course.
Then the question you need to answer is why this definition of $tau$ gives a topology.
After that, you can try to reread the answer to the linked question, and it should hopefully make more sense.
answered Dec 21 '18 at 14:50
jgonjgon
14.9k32042
$endgroup$
$begingroup$
Thank you! After reading this and the comments I saw that I made a mistake forming the question. Yes, it should say that a set $A$ is closed if it is COMPACT in the usual topology. Complementing this we get the definition of topology with compact complements, which is indeed, a topology.
$endgroup$
– Coupeau
Dec 22 '18 at 11:15
$begingroup$
The question is now corrected.
$endgroup$
– Coupeau
Dec 22 '18 at 11:23
add a comment |
Your Answer
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1 Answer
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$begingroup$
As SvanN says in the comments, your "topology" doesn't form a topology.
The open subsets of $Bbb{R}^n$ are not closed under arbitrary intersections:
$$bigcap_{n=1}^infty left(frac{-1}{n},frac{1}{n}right) = {0},$$
so the open subsets of $Bbb{R}^n$ cannot be the closed sets of a topology.
Instead, the question you've linked suggests you should be defining your topology as
$U$ is open in $tau$ if $U^C$ is compact in the usual topology, plus the null set of course.
Then the question you need to answer is why this definition of $tau$ gives a topology.
After that, you can try to reread the answer to the linked question, and it should hopefully make more sense.
answered Dec 21 '18 at 14:50
jgonjgon
14.9k32042
$endgroup$
$begingroup$
Thank you! After reading this and the comments I saw that I made a mistake forming the question. Yes, it should say that a set $A$ is closed if it is COMPACT in the usual topology. Complementing this we get the definition of topology with compact complements, which is indeed, a topology.
$endgroup$
– Coupeau
Dec 22 '18 at 11:15
$begingroup$
The question is now corrected.
$endgroup$
– Coupeau
Dec 22 '18 at 11:23
add a comment |
$begingroup$
As SvanN says in the comments, your "topology" doesn't form a topology.
The open subsets of $Bbb{R}^n$ are not closed under arbitrary intersections:
$$bigcap_{n=1}^infty left(frac{-1}{n},frac{1}{n}right) = {0},$$
so the open subsets of $Bbb{R}^n$ cannot be the closed sets of a topology.
Instead, the question you've linked suggests you should be defining your topology as
$U$ is open in $tau$ if $U^C$ is compact in the usual topology, plus the null set of course.
Then the question you need to answer is why this definition of $tau$ gives a topology.
After that, you can try to reread the answer to the linked question, and it should hopefully make more sense.
answered Dec 21 '18 at 14:50
jgonjgon
14.9k32042
$endgroup$
$begingroup$
Thank you! After reading this and the comments I saw that I made a mistake forming the question. Yes, it should say that a set $A$ is closed if it is COMPACT in the usual topology. Complementing this we get the definition of topology with compact complements, which is indeed, a topology.
$endgroup$
– Coupeau
Dec 22 '18 at 11:15
$begingroup$
The question is now corrected.
$endgroup$
– Coupeau
Dec 22 '18 at 11:23
add a comment |
$begingroup$
As SvanN says in the comments, your "topology" doesn't form a topology.
The open subsets of $Bbb{R}^n$ are not closed under arbitrary intersections:
$$bigcap_{n=1}^infty left(frac{-1}{n},frac{1}{n}right) = {0},$$
so the open subsets of $Bbb{R}^n$ cannot be the closed sets of a topology.
Instead, the question you've linked suggests you should be defining your topology as
$U$ is open in $tau$ if $U^C$ is compact in the usual topology, plus the null set of course.
Then the question you need to answer is why this definition of $tau$ gives a topology.
After that, you can try to reread the answer to the linked question, and it should hopefully make more sense.
answered Dec 21 '18 at 14:50
jgonjgon
14.9k32042
$endgroup$
As SvanN says in the comments, your "topology" doesn't form a topology.
The open subsets of $Bbb{R}^n$ are not closed under arbitrary intersections:
$$bigcap_{n=1}^infty left(frac{-1}{n},frac{1}{n}right) = {0},$$
so the open subsets of $Bbb{R}^n$ cannot be the closed sets of a topology.
Instead, the question you've linked suggests you should be defining your topology as
$U$ is open in $tau$ if $U^C$ is compact in the usual topology, plus the null set of course.
Then the question you need to answer is why this definition of $tau$ gives a topology.
After that, you can try to reread the answer to the linked question, and it should hopefully make more sense.
answered Dec 21 '18 at 14:50
jgonjgon
14.9k32042
answered Dec 21 '18 at 14:50
jgonjgon
14.9k32042
answered Dec 21 '18 at 14:50
jgonjgon
14.9k32042
answered Dec 21 '18 at 14:50
jgonjgon
14.9k32042
14.9k32042
$begingroup$
Thank you! After reading this and the comments I saw that I made a mistake forming the question. Yes, it should say that a set $A$ is closed if it is COMPACT in the usual topology. Complementing this we get the definition of topology with compact complements, which is indeed, a topology.
$endgroup$
– Coupeau
Dec 22 '18 at 11:15
$begingroup$
The question is now corrected.
$endgroup$
– Coupeau
Dec 22 '18 at 11:23
add a comment |
$begingroup$
Thank you! After reading this and the comments I saw that I made a mistake forming the question. Yes, it should say that a set $A$ is closed if it is COMPACT in the usual topology. Complementing this we get the definition of topology with compact complements, which is indeed, a topology.
$endgroup$
– Coupeau
Dec 22 '18 at 11:15
$begingroup$
The question is now corrected.
$endgroup$
– Coupeau
Dec 22 '18 at 11:23
$begingroup$
Thank you! After reading this and the comments I saw that I made a mistake forming the question. Yes, it should say that a set $A$ is closed if it is COMPACT in the usual topology. Complementing this we get the definition of topology with compact complements, which is indeed, a topology.
$endgroup$
– Coupeau
Dec 22 '18 at 11:15
$begingroup$
Thank you! After reading this and the comments I saw that I made a mistake forming the question. Yes, it should say that a set $A$ is closed if it is COMPACT in the usual topology. Complementing this we get the definition of topology with compact complements, which is indeed, a topology.
$endgroup$
– Coupeau
Dec 22 '18 at 11:15
$begingroup$
The question is now corrected.
$endgroup$
– Coupeau
Dec 22 '18 at 11:23
$begingroup$
The question is now corrected.
$endgroup$
– Coupeau
Dec 22 '18 at 11:23
add a comment |
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1
$begingroup$
The set $mathbb{R}^n$ is open in $mathbb{R}^n$, so the distinction in the definition is unnecessary. Also, are you sure this even defines a topology? The infinite intersection of open sets is not open, so in this definition, an infinite intersection of closed sets need not be closed; but this must hold in any topological space (pretty much by definition).
$endgroup$
– SvanN
Dec 20 '18 at 17:52
$begingroup$
It is true that $mathbb{R^n}$ is open, but as I am defining topology using closed sets I need to make sure that the whole space is also in the topology. And I couldn't cover that case without explicitly saying that $A$ is closed if it is equal to $mathbb{R^n}$. Because $mathbb{R^n}$ is not compact in Euclidean topology. By definition of topology finite union of closed and infinite intersection of closeds need to be closed. This is true in this case.
$endgroup$
– Coupeau
Dec 21 '18 at 14:39
1
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I think you're mixing up the terms 'closed' and 'compact'. No, you do not need the distinction: $mathbb{R}^n$ is open in the usual topology on $mathbb{R}^n$, hence it is closed in your new "topology". But as I indicated, and as jgon explained below, you do not have a topology.
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– SvanN
Dec 21 '18 at 15:45
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OOHH SORRY! This is totally my mistake. I Wrote the definition of topology wrong. It should say$mathbb{R^n}$ is closed in $tau$ if it is COMPACT in standard topology or $A = mathbb{R^n}$
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– Coupeau
Dec 22 '18 at 11:08