Is there a nonHausdorff topology on $mathbb{R}$ such that we still have the closedness/boundedness...
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Using the standard topology of $mathbb{R}$ is sufficient to ensure that a compact subset of $mathbb{R}$ is closed and bounded and vice versa. A standard topology of $mathbb{R}$ is Hausdorff. I wonder if there is a nonHausdorff topology on $mathbb{R}$ such that with respect to which a subset of $mathbb{R}$ is compact if and only if it is closed and bounded.
Note:
Thanks to the contributors' works. I made a mistake. What I am after is a nonHausdorff topology $tau$ on $mathbb{R}$ such that, if $tau'$ is the usual topology on $mathbb{R}$, a subset of $mathbb{R}$ is $tau$-compact if and only if it is $tau$-closed and $tau'$-bounded. It is great to know the existence of that type of nonHausdorff topology ($tau$-bounded) as given in an answer below too.
real-analysis general-topology real-numbers
asked Nov 21 at 15:58
Gary Moore
17.2k21545
add a comment |
up vote
0
down vote
favorite
Using the standard topology of $mathbb{R}$ is sufficient to ensure that a compact subset of $mathbb{R}$ is closed and bounded and vice versa. A standard topology of $mathbb{R}$ is Hausdorff. I wonder if there is a nonHausdorff topology on $mathbb{R}$ such that with respect to which a subset of $mathbb{R}$ is compact if and only if it is closed and bounded.
Note:
Thanks to the contributors' works. I made a mistake. What I am after is a nonHausdorff topology $tau$ on $mathbb{R}$ such that, if $tau'$ is the usual topology on $mathbb{R}$, a subset of $mathbb{R}$ is $tau$-compact if and only if it is $tau$-closed and $tau'$-bounded. It is great to know the existence of that type of nonHausdorff topology ($tau$-bounded) as given in an answer below too.
real-analysis general-topology real-numbers
asked Nov 21 at 15:58
Gary Moore
17.2k21545
You ask tha compact implies closed and bounded or you ask for an if and only if?
– user126154
Nov 21 at 16:31
@user126154, Hi, thanks. Yes, equivalence is after.
– Gary Moore
Nov 21 at 16:33
Do you mean compactness in the new topology is equivalent to closed and bounded in the old topology, or compactness in the new topology is equivalent to closed and bounded in the new topology?
– Noah Schweber
Nov 21 at 16:44
@NoahSchweber, Hi, thanks. Yes, new topology for both.
– Gary Moore
Nov 21 at 16:45
@NoahSchweber, Hi, sorry. I made a mistake. You are right; I was after the equivalence such that compactness and closedness are relative to the nonHausdorff topology and the boundedness is relative to the usual topology! (thus cofinite example)
– Gary Moore
Nov 21 at 17:08
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Using the standard topology of $mathbb{R}$ is sufficient to ensure that a compact subset of $mathbb{R}$ is closed and bounded and vice versa. A standard topology of $mathbb{R}$ is Hausdorff. I wonder if there is a nonHausdorff topology on $mathbb{R}$ such that with respect to which a subset of $mathbb{R}$ is compact if and only if it is closed and bounded.
Note:
Thanks to the contributors' works. I made a mistake. What I am after is a nonHausdorff topology $tau$ on $mathbb{R}$ such that, if $tau'$ is the usual topology on $mathbb{R}$, a subset of $mathbb{R}$ is $tau$-compact if and only if it is $tau$-closed and $tau'$-bounded. It is great to know the existence of that type of nonHausdorff topology ($tau$-bounded) as given in an answer below too.
real-analysis general-topology real-numbers
asked Nov 21 at 15:58
Gary Moore
17.2k21545
Using the standard topology of $mathbb{R}$ is sufficient to ensure that a compact subset of $mathbb{R}$ is closed and bounded and vice versa. A standard topology of $mathbb{R}$ is Hausdorff. I wonder if there is a nonHausdorff topology on $mathbb{R}$ such that with respect to which a subset of $mathbb{R}$ is compact if and only if it is closed and bounded.
Note:
Thanks to the contributors' works. I made a mistake. What I am after is a nonHausdorff topology $tau$ on $mathbb{R}$ such that, if $tau'$ is the usual topology on $mathbb{R}$, a subset of $mathbb{R}$ is $tau$-compact if and only if it is $tau$-closed and $tau'$-bounded. It is great to know the existence of that type of nonHausdorff topology ($tau$-bounded) as given in an answer below too.
real-analysis general-topology real-numbers
real-analysis general-topology real-numbers
asked Nov 21 at 15:58
Gary Moore
17.2k21545
asked Nov 21 at 15:58
Gary Moore
17.2k21545
edited Nov 21 at 17:12
asked Nov 21 at 15:58
Gary Moore
17.2k21545
asked Nov 21 at 15:58
Gary Moore
17.2k21545
asked Nov 21 at 15:58
Gary Moore
17.2k21545
17.2k21545
You ask tha compact implies closed and bounded or you ask for an if and only if?
– user126154
Nov 21 at 16:31
@user126154, Hi, thanks. Yes, equivalence is after.
– Gary Moore
Nov 21 at 16:33
Do you mean compactness in the new topology is equivalent to closed and bounded in the old topology, or compactness in the new topology is equivalent to closed and bounded in the new topology?
– Noah Schweber
Nov 21 at 16:44
@NoahSchweber, Hi, thanks. Yes, new topology for both.
– Gary Moore
Nov 21 at 16:45
@NoahSchweber, Hi, sorry. I made a mistake. You are right; I was after the equivalence such that compactness and closedness are relative to the nonHausdorff topology and the boundedness is relative to the usual topology! (thus cofinite example)
– Gary Moore
Nov 21 at 17:08
add a comment |
You ask tha compact implies closed and bounded or you ask for an if and only if?
– user126154
Nov 21 at 16:31
@user126154, Hi, thanks. Yes, equivalence is after.
– Gary Moore
Nov 21 at 16:33
Do you mean compactness in the new topology is equivalent to closed and bounded in the old topology, or compactness in the new topology is equivalent to closed and bounded in the new topology?
– Noah Schweber
Nov 21 at 16:44
@NoahSchweber, Hi, thanks. Yes, new topology for both.
– Gary Moore
Nov 21 at 16:45
@NoahSchweber, Hi, sorry. I made a mistake. You are right; I was after the equivalence such that compactness and closedness are relative to the nonHausdorff topology and the boundedness is relative to the usual topology! (thus cofinite example)
– Gary Moore
Nov 21 at 17:08
You ask tha compact implies closed and bounded or you ask for an if and only if?
– user126154
Nov 21 at 16:31
You ask tha compact implies closed and bounded or you ask for an if and only if?
– user126154
Nov 21 at 16:31
@user126154, Hi, thanks. Yes, equivalence is after.
– Gary Moore
Nov 21 at 16:33
@user126154, Hi, thanks. Yes, equivalence is after.
– Gary Moore
Nov 21 at 16:33
Do you mean compactness in the new topology is equivalent to closed and bounded in the old topology, or compactness in the new topology is equivalent to closed and bounded in the new topology?
– Noah Schweber
Nov 21 at 16:44
Do you mean compactness in the new topology is equivalent to closed and bounded in the old topology, or compactness in the new topology is equivalent to closed and bounded in the new topology?
– Noah Schweber
Nov 21 at 16:44
@NoahSchweber, Hi, thanks. Yes, new topology for both.
– Gary Moore
Nov 21 at 16:45
@NoahSchweber, Hi, thanks. Yes, new topology for both.
– Gary Moore
Nov 21 at 16:45
@NoahSchweber, Hi, sorry. I made a mistake. You are right; I was after the equivalence such that compactness and closedness are relative to the nonHausdorff topology and the boundedness is relative to the usual topology! (thus cofinite example)
– Gary Moore
Nov 21 at 17:08
@NoahSchweber, Hi, sorry. I made a mistake. You are right; I was after the equivalence such that compactness and closedness are relative to the nonHausdorff topology and the boundedness is relative to the usual topology! (thus cofinite example)
– Gary Moore
Nov 21 at 17:08
add a comment |
1 Answer
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Consider the equivalence relation generated by $xsim y$ if they both are in $[0,1]$. Then let $tau$ be the topology saturated by such relation (i.e. take the quotient space and then pull back the quotient topology). Explicitly, you declare a set open if it is open in Euclidean topology and if, whenever it contains a point in $[0,1]$ then it contains the whole $[0,1]$.
Such topology in not hausdorff because points in $[0, 1]$ are indistinguishable. But from the fact that the quotient space is homeo to $mathbb R$ you get the desired property.
answered Nov 21 at 16:41
user126154
5,368716
Maybe I misunderstood the question, but if you think the answer is incorrect, please explain so that I can correct. Why downvote?
– user126154
Nov 21 at 16:45
1
+1. Note that this is overkill: you could just fix distinct reals $x,y$ and consider the topology $tau={X:$ $X$ is open in the usual topology and $xin Xiff yin X}$.
– Noah Schweber
Nov 21 at 16:46
Yes, you are right. I thought it was easy to prove required properties by using that the quotient is R.
– user126154
Nov 21 at 16:51
Thank you. The answer makes me wonder further that what if boundedness is required to be with respect to the usual topology (compactness and closedness are with respect to the new topology)? Is it possible to do that?
– Gary Moore
Nov 21 at 16:54
boundedness does not depend on the topology. It depends on the metric or on the order of R. Both induces the standard topology. But in general what does it mean "bounded" for a topology? Many mathematicians use "bounded" with the meaning: "contained in a compact. So the example give works with this terminology.
– user126154
Nov 22 at 12:03
|
show 1 more comment
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1 Answer
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Consider the equivalence relation generated by $xsim y$ if they both are in $[0,1]$. Then let $tau$ be the topology saturated by such relation (i.e. take the quotient space and then pull back the quotient topology). Explicitly, you declare a set open if it is open in Euclidean topology and if, whenever it contains a point in $[0,1]$ then it contains the whole $[0,1]$.
Such topology in not hausdorff because points in $[0, 1]$ are indistinguishable. But from the fact that the quotient space is homeo to $mathbb R$ you get the desired property.
answered Nov 21 at 16:41
user126154
5,368716
Maybe I misunderstood the question, but if you think the answer is incorrect, please explain so that I can correct. Why downvote?
– user126154
Nov 21 at 16:45
1
+1. Note that this is overkill: you could just fix distinct reals $x,y$ and consider the topology $tau={X:$ $X$ is open in the usual topology and $xin Xiff yin X}$.
– Noah Schweber
Nov 21 at 16:46
Yes, you are right. I thought it was easy to prove required properties by using that the quotient is R.
– user126154
Nov 21 at 16:51
Thank you. The answer makes me wonder further that what if boundedness is required to be with respect to the usual topology (compactness and closedness are with respect to the new topology)? Is it possible to do that?
– Gary Moore
Nov 21 at 16:54
boundedness does not depend on the topology. It depends on the metric or on the order of R. Both induces the standard topology. But in general what does it mean "bounded" for a topology? Many mathematicians use "bounded" with the meaning: "contained in a compact. So the example give works with this terminology.
– user126154
Nov 22 at 12:03
|
show 1 more comment
up vote
3
down vote
Consider the equivalence relation generated by $xsim y$ if they both are in $[0,1]$. Then let $tau$ be the topology saturated by such relation (i.e. take the quotient space and then pull back the quotient topology). Explicitly, you declare a set open if it is open in Euclidean topology and if, whenever it contains a point in $[0,1]$ then it contains the whole $[0,1]$.
Such topology in not hausdorff because points in $[0, 1]$ are indistinguishable. But from the fact that the quotient space is homeo to $mathbb R$ you get the desired property.
answered Nov 21 at 16:41
user126154
5,368716
Maybe I misunderstood the question, but if you think the answer is incorrect, please explain so that I can correct. Why downvote?
– user126154
Nov 21 at 16:45
1
+1. Note that this is overkill: you could just fix distinct reals $x,y$ and consider the topology $tau={X:$ $X$ is open in the usual topology and $xin Xiff yin X}$.
– Noah Schweber
Nov 21 at 16:46
Yes, you are right. I thought it was easy to prove required properties by using that the quotient is R.
– user126154
Nov 21 at 16:51
Thank you. The answer makes me wonder further that what if boundedness is required to be with respect to the usual topology (compactness and closedness are with respect to the new topology)? Is it possible to do that?
– Gary Moore
Nov 21 at 16:54
boundedness does not depend on the topology. It depends on the metric or on the order of R. Both induces the standard topology. But in general what does it mean "bounded" for a topology? Many mathematicians use "bounded" with the meaning: "contained in a compact. So the example give works with this terminology.
– user126154
Nov 22 at 12:03
|
show 1 more comment
up vote
3
down vote
up vote
3
down vote
Consider the equivalence relation generated by $xsim y$ if they both are in $[0,1]$. Then let $tau$ be the topology saturated by such relation (i.e. take the quotient space and then pull back the quotient topology). Explicitly, you declare a set open if it is open in Euclidean topology and if, whenever it contains a point in $[0,1]$ then it contains the whole $[0,1]$.
Such topology in not hausdorff because points in $[0, 1]$ are indistinguishable. But from the fact that the quotient space is homeo to $mathbb R$ you get the desired property.
answered Nov 21 at 16:41
user126154
5,368716
Consider the equivalence relation generated by $xsim y$ if they both are in $[0,1]$. Then let $tau$ be the topology saturated by such relation (i.e. take the quotient space and then pull back the quotient topology). Explicitly, you declare a set open if it is open in Euclidean topology and if, whenever it contains a point in $[0,1]$ then it contains the whole $[0,1]$.
Such topology in not hausdorff because points in $[0, 1]$ are indistinguishable. But from the fact that the quotient space is homeo to $mathbb R$ you get the desired property.
answered Nov 21 at 16:41
user126154
5,368716
edited Nov 21 at 16:50
answered Nov 21 at 16:41
user126154
5,368716
answered Nov 21 at 16:41
user126154
5,368716
answered Nov 21 at 16:41
user126154
5,368716
5,368716
Maybe I misunderstood the question, but if you think the answer is incorrect, please explain so that I can correct. Why downvote?
– user126154
Nov 21 at 16:45
1
+1. Note that this is overkill: you could just fix distinct reals $x,y$ and consider the topology $tau={X:$ $X$ is open in the usual topology and $xin Xiff yin X}$.
– Noah Schweber
Nov 21 at 16:46
Yes, you are right. I thought it was easy to prove required properties by using that the quotient is R.
– user126154
Nov 21 at 16:51
Thank you. The answer makes me wonder further that what if boundedness is required to be with respect to the usual topology (compactness and closedness are with respect to the new topology)? Is it possible to do that?
– Gary Moore
Nov 21 at 16:54
boundedness does not depend on the topology. It depends on the metric or on the order of R. Both induces the standard topology. But in general what does it mean "bounded" for a topology? Many mathematicians use "bounded" with the meaning: "contained in a compact. So the example give works with this terminology.
– user126154
Nov 22 at 12:03
|
show 1 more comment
Maybe I misunderstood the question, but if you think the answer is incorrect, please explain so that I can correct. Why downvote?
– user126154
Nov 21 at 16:45
1
+1. Note that this is overkill: you could just fix distinct reals $x,y$ and consider the topology $tau={X:$ $X$ is open in the usual topology and $xin Xiff yin X}$.
– Noah Schweber
Nov 21 at 16:46
Yes, you are right. I thought it was easy to prove required properties by using that the quotient is R.
– user126154
Nov 21 at 16:51
Thank you. The answer makes me wonder further that what if boundedness is required to be with respect to the usual topology (compactness and closedness are with respect to the new topology)? Is it possible to do that?
– Gary Moore
Nov 21 at 16:54
boundedness does not depend on the topology. It depends on the metric or on the order of R. Both induces the standard topology. But in general what does it mean "bounded" for a topology? Many mathematicians use "bounded" with the meaning: "contained in a compact. So the example give works with this terminology.
– user126154
Nov 22 at 12:03
Maybe I misunderstood the question, but if you think the answer is incorrect, please explain so that I can correct. Why downvote?
– user126154
Nov 21 at 16:45
Maybe I misunderstood the question, but if you think the answer is incorrect, please explain so that I can correct. Why downvote?
– user126154
Nov 21 at 16:45
1
1
+1. Note that this is overkill: you could just fix distinct reals $x,y$ and consider the topology $tau={X:$ $X$ is open in the usual topology and $xin Xiff yin X}$.
– Noah Schweber
Nov 21 at 16:46
+1. Note that this is overkill: you could just fix distinct reals $x,y$ and consider the topology $tau={X:$ $X$ is open in the usual topology and $xin Xiff yin X}$.
– Noah Schweber
Nov 21 at 16:46
Yes, you are right. I thought it was easy to prove required properties by using that the quotient is R.
– user126154
Nov 21 at 16:51
Yes, you are right. I thought it was easy to prove required properties by using that the quotient is R.
– user126154
Nov 21 at 16:51
Thank you. The answer makes me wonder further that what if boundedness is required to be with respect to the usual topology (compactness and closedness are with respect to the new topology)? Is it possible to do that?
– Gary Moore
Nov 21 at 16:54
Thank you. The answer makes me wonder further that what if boundedness is required to be with respect to the usual topology (compactness and closedness are with respect to the new topology)? Is it possible to do that?
– Gary Moore
Nov 21 at 16:54
boundedness does not depend on the topology. It depends on the metric or on the order of R. Both induces the standard topology. But in general what does it mean "bounded" for a topology? Many mathematicians use "bounded" with the meaning: "contained in a compact. So the example give works with this terminology.
– user126154
Nov 22 at 12:03
boundedness does not depend on the topology. It depends on the metric or on the order of R. Both induces the standard topology. But in general what does it mean "bounded" for a topology? Many mathematicians use "bounded" with the meaning: "contained in a compact. So the example give works with this terminology.
– user126154
Nov 22 at 12:03
|
show 1 more comment
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You ask tha compact implies closed and bounded or you ask for an if and only if?
– user126154
Nov 21 at 16:31
@user126154, Hi, thanks. Yes, equivalence is after.
– Gary Moore
Nov 21 at 16:33
Do you mean compactness in the new topology is equivalent to closed and bounded in the old topology, or compactness in the new topology is equivalent to closed and bounded in the new topology?
– Noah Schweber
Nov 21 at 16:44
@NoahSchweber, Hi, thanks. Yes, new topology for both.
– Gary Moore
Nov 21 at 16:45
@NoahSchweber, Hi, sorry. I made a mistake. You are right; I was after the equivalence such that compactness and closedness are relative to the nonHausdorff topology and the boundedness is relative to the usual topology! (thus cofinite example)
– Gary Moore
Nov 21 at 17:08