Mixing
Examples of connected door spaces.
$begingroup$
A topological space $X$ is a door space if any subset of $X$ is either open or closed (or both). Naturally, a connected door space is that in which any proper subset is either open or closed, but not both. According to this paper, there are only three types of topologies yielded by connected door spaces: particular point topologies, excluded point topologies, and $T_1$ topologies in which any two non-disjoint open sets have infinite intersection.
Is there any explicit example of a connected door space of the third type? The cofinite topology on any infinite set satisfies the third type but is never a door space, and I can't really think of spaces in which open intersections are infinite...
general-topology connectedness
asked Jan 27 at 0:50
RlosRlos
1437
$endgroup$
add a comment |
$begingroup$
A topological space $X$ is a door space if any subset of $X$ is either open or closed (or both). Naturally, a connected door space is that in which any proper subset is either open or closed, but not both. According to this paper, there are only three types of topologies yielded by connected door spaces: particular point topologies, excluded point topologies, and $T_1$ topologies in which any two non-disjoint open sets have infinite intersection.
Is there any explicit example of a connected door space of the third type? The cofinite topology on any infinite set satisfies the third type but is never a door space, and I can't really think of spaces in which open intersections are infinite...
general-topology connectedness
asked Jan 27 at 0:50
RlosRlos
1437
$endgroup$
$begingroup$
The name 'door space' is so brilliant, I love it!
$endgroup$
– SmileyCraft
Jan 27 at 3:24
$begingroup$
Just saying, the Euclidean topology also satisfies the third type, but is not a door space.
$endgroup$
– SmileyCraft
Jan 27 at 3:30
$begingroup$
Before anyone else wastes their time with this, the straight forward Zorn's lemma approach does not work. There is a chain of connected topologies for which any upper bound is disconnected. In particular, let $mathcal{T}_r$ be the connected topology on $mathbb{R}$ generated by ${(-infty,l):lleq-r}cup{(u,infty):rleq u}cup{{0}}$. Then any upper bound of ${mathcal{T}_r:r>0}$ must have $(-infty,0]$ and $(0,infty)$ both as open sets.
$endgroup$
– SmileyCraft
Jan 27 at 4:12
1
$begingroup$
Yes, I love the concept as well and that's essentially why I became interested in these spaces haha. Unfortunately (or fortunately?) I just found out that a door space is of the third type described above if and only if excluding the empty set from the topology gives a free ultrafilter. To my knowledge, no one has ever constructed an explicit free ultrafilter, so I suppose that that answers my own question...
$endgroup$
– Rlos
Jan 27 at 4:40
$begingroup$
Indeed an ultrafilter is necessary and there can be no "explicit" example of an ultrafilter, as is well-known. But just use a little bit of AC and we can prove one exists...
$endgroup$
– Henno Brandsma
Jan 27 at 10:31
add a comment |
$begingroup$
A topological space $X$ is a door space if any subset of $X$ is either open or closed (or both). Naturally, a connected door space is that in which any proper subset is either open or closed, but not both. According to this paper, there are only three types of topologies yielded by connected door spaces: particular point topologies, excluded point topologies, and $T_1$ topologies in which any two non-disjoint open sets have infinite intersection.
Is there any explicit example of a connected door space of the third type? The cofinite topology on any infinite set satisfies the third type but is never a door space, and I can't really think of spaces in which open intersections are infinite...
general-topology connectedness
asked Jan 27 at 0:50
RlosRlos
1437
$endgroup$
A topological space $X$ is a door space if any subset of $X$ is either open or closed (or both). Naturally, a connected door space is that in which any proper subset is either open or closed, but not both. According to this paper, there are only three types of topologies yielded by connected door spaces: particular point topologies, excluded point topologies, and $T_1$ topologies in which any two non-disjoint open sets have infinite intersection.
Is there any explicit example of a connected door space of the third type? The cofinite topology on any infinite set satisfies the third type but is never a door space, and I can't really think of spaces in which open intersections are infinite...
general-topology connectedness
general-topology connectedness
asked Jan 27 at 0:50
RlosRlos
1437
asked Jan 27 at 0:50
RlosRlos
1437
asked Jan 27 at 0:50
RlosRlos
1437
asked Jan 27 at 0:50
RlosRlos
1437
asked Jan 27 at 0:50
RlosRlos
1437
1437
$begingroup$
The name 'door space' is so brilliant, I love it!
$endgroup$
– SmileyCraft
Jan 27 at 3:24
$begingroup$
Just saying, the Euclidean topology also satisfies the third type, but is not a door space.
$endgroup$
– SmileyCraft
Jan 27 at 3:30
$begingroup$
Before anyone else wastes their time with this, the straight forward Zorn's lemma approach does not work. There is a chain of connected topologies for which any upper bound is disconnected. In particular, let $mathcal{T}_r$ be the connected topology on $mathbb{R}$ generated by ${(-infty,l):lleq-r}cup{(u,infty):rleq u}cup{{0}}$. Then any upper bound of ${mathcal{T}_r:r>0}$ must have $(-infty,0]$ and $(0,infty)$ both as open sets.
$endgroup$
– SmileyCraft
Jan 27 at 4:12
1
$begingroup$
Yes, I love the concept as well and that's essentially why I became interested in these spaces haha. Unfortunately (or fortunately?) I just found out that a door space is of the third type described above if and only if excluding the empty set from the topology gives a free ultrafilter. To my knowledge, no one has ever constructed an explicit free ultrafilter, so I suppose that that answers my own question...
$endgroup$
– Rlos
Jan 27 at 4:40
$begingroup$
Indeed an ultrafilter is necessary and there can be no "explicit" example of an ultrafilter, as is well-known. But just use a little bit of AC and we can prove one exists...
$endgroup$
– Henno Brandsma
Jan 27 at 10:31
add a comment |
$begingroup$
The name 'door space' is so brilliant, I love it!
$endgroup$
– SmileyCraft
Jan 27 at 3:24
$begingroup$
Just saying, the Euclidean topology also satisfies the third type, but is not a door space.
$endgroup$
– SmileyCraft
Jan 27 at 3:30
$begingroup$
Before anyone else wastes their time with this, the straight forward Zorn's lemma approach does not work. There is a chain of connected topologies for which any upper bound is disconnected. In particular, let $mathcal{T}_r$ be the connected topology on $mathbb{R}$ generated by ${(-infty,l):lleq-r}cup{(u,infty):rleq u}cup{{0}}$. Then any upper bound of ${mathcal{T}_r:r>0}$ must have $(-infty,0]$ and $(0,infty)$ both as open sets.
$endgroup$
– SmileyCraft
Jan 27 at 4:12
1
$begingroup$
Yes, I love the concept as well and that's essentially why I became interested in these spaces haha. Unfortunately (or fortunately?) I just found out that a door space is of the third type described above if and only if excluding the empty set from the topology gives a free ultrafilter. To my knowledge, no one has ever constructed an explicit free ultrafilter, so I suppose that that answers my own question...
$endgroup$
– Rlos
Jan 27 at 4:40
$begingroup$
Indeed an ultrafilter is necessary and there can be no "explicit" example of an ultrafilter, as is well-known. But just use a little bit of AC and we can prove one exists...
$endgroup$
– Henno Brandsma
Jan 27 at 10:31
$begingroup$
The name 'door space' is so brilliant, I love it!
$endgroup$
– SmileyCraft
Jan 27 at 3:24
$begingroup$
The name 'door space' is so brilliant, I love it!
$endgroup$
– SmileyCraft
Jan 27 at 3:24
$begingroup$
Just saying, the Euclidean topology also satisfies the third type, but is not a door space.
$endgroup$
– SmileyCraft
Jan 27 at 3:30
$begingroup$
Just saying, the Euclidean topology also satisfies the third type, but is not a door space.
$endgroup$
– SmileyCraft
Jan 27 at 3:30
$begingroup$
Before anyone else wastes their time with this, the straight forward Zorn's lemma approach does not work. There is a chain of connected topologies for which any upper bound is disconnected. In particular, let $mathcal{T}_r$ be the connected topology on $mathbb{R}$ generated by ${(-infty,l):lleq-r}cup{(u,infty):rleq u}cup{{0}}$. Then any upper bound of ${mathcal{T}_r:r>0}$ must have $(-infty,0]$ and $(0,infty)$ both as open sets.
$endgroup$
– SmileyCraft
Jan 27 at 4:12
$begingroup$
Before anyone else wastes their time with this, the straight forward Zorn's lemma approach does not work. There is a chain of connected topologies for which any upper bound is disconnected. In particular, let $mathcal{T}_r$ be the connected topology on $mathbb{R}$ generated by ${(-infty,l):lleq-r}cup{(u,infty):rleq u}cup{{0}}$. Then any upper bound of ${mathcal{T}_r:r>0}$ must have $(-infty,0]$ and $(0,infty)$ both as open sets.
$endgroup$
– SmileyCraft
Jan 27 at 4:12
1
1
$begingroup$
Yes, I love the concept as well and that's essentially why I became interested in these spaces haha. Unfortunately (or fortunately?) I just found out that a door space is of the third type described above if and only if excluding the empty set from the topology gives a free ultrafilter. To my knowledge, no one has ever constructed an explicit free ultrafilter, so I suppose that that answers my own question...
$endgroup$
– Rlos
Jan 27 at 4:40
$begingroup$
Yes, I love the concept as well and that's essentially why I became interested in these spaces haha. Unfortunately (or fortunately?) I just found out that a door space is of the third type described above if and only if excluding the empty set from the topology gives a free ultrafilter. To my knowledge, no one has ever constructed an explicit free ultrafilter, so I suppose that that answers my own question...
$endgroup$
– Rlos
Jan 27 at 4:40
$begingroup$
Indeed an ultrafilter is necessary and there can be no "explicit" example of an ultrafilter, as is well-known. But just use a little bit of AC and we can prove one exists...
$endgroup$
– Henno Brandsma
Jan 27 at 10:31
$begingroup$
Indeed an ultrafilter is necessary and there can be no "explicit" example of an ultrafilter, as is well-known. But just use a little bit of AC and we can prove one exists...
$endgroup$
– Henno Brandsma
Jan 27 at 10:31
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $X$ be an infinite set and (using a mild form of choice) $mathcal{F}$ a free ultrafilter on $X$.
Let $mathcal{T}={emptyset} cup mathcal{F}$, which is a topology on $X$: unions follow from closedness under enlargements, finite intersections follow from the filter axioms as well, and we added the empty set explicitly. That any two non-empty open sets intersect also is a filter axiom ($emptyset notin mathcal{F})$.
It is well known that for all subsets $A$ of $X$ either $A$ or its complement is in $mathcal{F}$ (this is a classic ultrafilter property). This both implies that $X$ is $T_1$ ( as the ultrafilter is free it contains all cofinite sets) and that it is a door space.
This construction gives us a lot of non-homeomorphic spaces, as there are many so-called types of ultrafilters; See set theory for more on this.
answered Jan 27 at 8:16
Henno BrandsmaHenno Brandsma
113k348123
$endgroup$
1
$begingroup$
And this is the most general example: if $X$ is a nonempty door space in which the intersection of any two nonempty open sets is nonempty, then the set of all nonempty open sets is an ultrafilter (a free ultrafilter if $X$ is a $text{T}_1$-space).
$endgroup$
– bof
Jan 27 at 8:55
$begingroup$
This connection between ultrafilters and topologies is most certainly interesting. If such ultrafilter $mathcal{F}$ is principal instead of free, one gets one of the better known door spaces: a particular point space. But not all door spaces come from ultrafilters, as any excluded point topology shows. Henno, do you know if someone has ever investigated this topic any further?
$endgroup$
– Rlos
Jan 27 at 19:20
$begingroup$
Also, in your previous comment you mentioned that textit{there can be no explicit examples} of free ultrafilters. Does this mean that it has been proven that their existence dependes solely on AC?
$endgroup$
– Rlos
Jan 27 at 19:24
$begingroup$
@Rlos I believe there are models of ZF that have no free ultrafilters. E.g. there are models of ZF and dependent choice in which all real subsets are measurable, if we assume some large cardinal (Solovay model). This implies that even countable choice doesn’t give us an ultrafilter on the natural numbers in that case.
$endgroup$
– Henno Brandsma
Jan 27 at 19:29
$begingroup$
@Rlos I believe there aren’t any other notable papers on door spaces that I know of (but I don’t follow that area of topology very closely). I had seen your referenced paper before. Look in MR to find more, e.g.
$endgroup$
– Henno Brandsma
Jan 27 at 19:32
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3088994%2fexamples-of-connected-door-spaces%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $X$ be an infinite set and (using a mild form of choice) $mathcal{F}$ a free ultrafilter on $X$.
Let $mathcal{T}={emptyset} cup mathcal{F}$, which is a topology on $X$: unions follow from closedness under enlargements, finite intersections follow from the filter axioms as well, and we added the empty set explicitly. That any two non-empty open sets intersect also is a filter axiom ($emptyset notin mathcal{F})$.
It is well known that for all subsets $A$ of $X$ either $A$ or its complement is in $mathcal{F}$ (this is a classic ultrafilter property). This both implies that $X$ is $T_1$ ( as the ultrafilter is free it contains all cofinite sets) and that it is a door space.
This construction gives us a lot of non-homeomorphic spaces, as there are many so-called types of ultrafilters; See set theory for more on this.
answered Jan 27 at 8:16
Henno BrandsmaHenno Brandsma
113k348123
$endgroup$
1
$begingroup$
And this is the most general example: if $X$ is a nonempty door space in which the intersection of any two nonempty open sets is nonempty, then the set of all nonempty open sets is an ultrafilter (a free ultrafilter if $X$ is a $text{T}_1$-space).
$endgroup$
– bof
Jan 27 at 8:55
$begingroup$
This connection between ultrafilters and topologies is most certainly interesting. If such ultrafilter $mathcal{F}$ is principal instead of free, one gets one of the better known door spaces: a particular point space. But not all door spaces come from ultrafilters, as any excluded point topology shows. Henno, do you know if someone has ever investigated this topic any further?
$endgroup$
– Rlos
Jan 27 at 19:20
$begingroup$
Also, in your previous comment you mentioned that textit{there can be no explicit examples} of free ultrafilters. Does this mean that it has been proven that their existence dependes solely on AC?
$endgroup$
– Rlos
Jan 27 at 19:24
$begingroup$
@Rlos I believe there are models of ZF that have no free ultrafilters. E.g. there are models of ZF and dependent choice in which all real subsets are measurable, if we assume some large cardinal (Solovay model). This implies that even countable choice doesn’t give us an ultrafilter on the natural numbers in that case.
$endgroup$
– Henno Brandsma
Jan 27 at 19:29
$begingroup$
@Rlos I believe there aren’t any other notable papers on door spaces that I know of (but I don’t follow that area of topology very closely). I had seen your referenced paper before. Look in MR to find more, e.g.
$endgroup$
– Henno Brandsma
Jan 27 at 19:32
add a comment |
$begingroup$
Let $X$ be an infinite set and (using a mild form of choice) $mathcal{F}$ a free ultrafilter on $X$.
Let $mathcal{T}={emptyset} cup mathcal{F}$, which is a topology on $X$: unions follow from closedness under enlargements, finite intersections follow from the filter axioms as well, and we added the empty set explicitly. That any two non-empty open sets intersect also is a filter axiom ($emptyset notin mathcal{F})$.
It is well known that for all subsets $A$ of $X$ either $A$ or its complement is in $mathcal{F}$ (this is a classic ultrafilter property). This both implies that $X$ is $T_1$ ( as the ultrafilter is free it contains all cofinite sets) and that it is a door space.
This construction gives us a lot of non-homeomorphic spaces, as there are many so-called types of ultrafilters; See set theory for more on this.
answered Jan 27 at 8:16
Henno BrandsmaHenno Brandsma
113k348123
$endgroup$
1
$begingroup$
And this is the most general example: if $X$ is a nonempty door space in which the intersection of any two nonempty open sets is nonempty, then the set of all nonempty open sets is an ultrafilter (a free ultrafilter if $X$ is a $text{T}_1$-space).
$endgroup$
– bof
Jan 27 at 8:55
$begingroup$
This connection between ultrafilters and topologies is most certainly interesting. If such ultrafilter $mathcal{F}$ is principal instead of free, one gets one of the better known door spaces: a particular point space. But not all door spaces come from ultrafilters, as any excluded point topology shows. Henno, do you know if someone has ever investigated this topic any further?
$endgroup$
– Rlos
Jan 27 at 19:20
$begingroup$
Also, in your previous comment you mentioned that textit{there can be no explicit examples} of free ultrafilters. Does this mean that it has been proven that their existence dependes solely on AC?
$endgroup$
– Rlos
Jan 27 at 19:24
$begingroup$
@Rlos I believe there are models of ZF that have no free ultrafilters. E.g. there are models of ZF and dependent choice in which all real subsets are measurable, if we assume some large cardinal (Solovay model). This implies that even countable choice doesn’t give us an ultrafilter on the natural numbers in that case.
$endgroup$
– Henno Brandsma
Jan 27 at 19:29
$begingroup$
@Rlos I believe there aren’t any other notable papers on door spaces that I know of (but I don’t follow that area of topology very closely). I had seen your referenced paper before. Look in MR to find more, e.g.
$endgroup$
– Henno Brandsma
Jan 27 at 19:32
add a comment |
$begingroup$
Let $X$ be an infinite set and (using a mild form of choice) $mathcal{F}$ a free ultrafilter on $X$.
Let $mathcal{T}={emptyset} cup mathcal{F}$, which is a topology on $X$: unions follow from closedness under enlargements, finite intersections follow from the filter axioms as well, and we added the empty set explicitly. That any two non-empty open sets intersect also is a filter axiom ($emptyset notin mathcal{F})$.
It is well known that for all subsets $A$ of $X$ either $A$ or its complement is in $mathcal{F}$ (this is a classic ultrafilter property). This both implies that $X$ is $T_1$ ( as the ultrafilter is free it contains all cofinite sets) and that it is a door space.
This construction gives us a lot of non-homeomorphic spaces, as there are many so-called types of ultrafilters; See set theory for more on this.
answered Jan 27 at 8:16
Henno BrandsmaHenno Brandsma
113k348123
$endgroup$
Let $X$ be an infinite set and (using a mild form of choice) $mathcal{F}$ a free ultrafilter on $X$.
Let $mathcal{T}={emptyset} cup mathcal{F}$, which is a topology on $X$: unions follow from closedness under enlargements, finite intersections follow from the filter axioms as well, and we added the empty set explicitly. That any two non-empty open sets intersect also is a filter axiom ($emptyset notin mathcal{F})$.
It is well known that for all subsets $A$ of $X$ either $A$ or its complement is in $mathcal{F}$ (this is a classic ultrafilter property). This both implies that $X$ is $T_1$ ( as the ultrafilter is free it contains all cofinite sets) and that it is a door space.
This construction gives us a lot of non-homeomorphic spaces, as there are many so-called types of ultrafilters; See set theory for more on this.
answered Jan 27 at 8:16
Henno BrandsmaHenno Brandsma
113k348123
answered Jan 27 at 8:16
Henno BrandsmaHenno Brandsma
113k348123
answered Jan 27 at 8:16
Henno BrandsmaHenno Brandsma
113k348123
answered Jan 27 at 8:16
Henno BrandsmaHenno Brandsma
113k348123
113k348123
1
$begingroup$
And this is the most general example: if $X$ is a nonempty door space in which the intersection of any two nonempty open sets is nonempty, then the set of all nonempty open sets is an ultrafilter (a free ultrafilter if $X$ is a $text{T}_1$-space).
$endgroup$
– bof
Jan 27 at 8:55
$begingroup$
This connection between ultrafilters and topologies is most certainly interesting. If such ultrafilter $mathcal{F}$ is principal instead of free, one gets one of the better known door spaces: a particular point space. But not all door spaces come from ultrafilters, as any excluded point topology shows. Henno, do you know if someone has ever investigated this topic any further?
$endgroup$
– Rlos
Jan 27 at 19:20
$begingroup$
Also, in your previous comment you mentioned that textit{there can be no explicit examples} of free ultrafilters. Does this mean that it has been proven that their existence dependes solely on AC?
$endgroup$
– Rlos
Jan 27 at 19:24
$begingroup$
@Rlos I believe there are models of ZF that have no free ultrafilters. E.g. there are models of ZF and dependent choice in which all real subsets are measurable, if we assume some large cardinal (Solovay model). This implies that even countable choice doesn’t give us an ultrafilter on the natural numbers in that case.
$endgroup$
– Henno Brandsma
Jan 27 at 19:29
$begingroup$
@Rlos I believe there aren’t any other notable papers on door spaces that I know of (but I don’t follow that area of topology very closely). I had seen your referenced paper before. Look in MR to find more, e.g.
$endgroup$
– Henno Brandsma
Jan 27 at 19:32
add a comment |
1
$begingroup$
And this is the most general example: if $X$ is a nonempty door space in which the intersection of any two nonempty open sets is nonempty, then the set of all nonempty open sets is an ultrafilter (a free ultrafilter if $X$ is a $text{T}_1$-space).
$endgroup$
– bof
Jan 27 at 8:55
$begingroup$
This connection between ultrafilters and topologies is most certainly interesting. If such ultrafilter $mathcal{F}$ is principal instead of free, one gets one of the better known door spaces: a particular point space. But not all door spaces come from ultrafilters, as any excluded point topology shows. Henno, do you know if someone has ever investigated this topic any further?
$endgroup$
– Rlos
Jan 27 at 19:20
$begingroup$
Also, in your previous comment you mentioned that textit{there can be no explicit examples} of free ultrafilters. Does this mean that it has been proven that their existence dependes solely on AC?
$endgroup$
– Rlos
Jan 27 at 19:24
$begingroup$
@Rlos I believe there are models of ZF that have no free ultrafilters. E.g. there are models of ZF and dependent choice in which all real subsets are measurable, if we assume some large cardinal (Solovay model). This implies that even countable choice doesn’t give us an ultrafilter on the natural numbers in that case.
$endgroup$
– Henno Brandsma
Jan 27 at 19:29
$begingroup$
@Rlos I believe there aren’t any other notable papers on door spaces that I know of (but I don’t follow that area of topology very closely). I had seen your referenced paper before. Look in MR to find more, e.g.
$endgroup$
– Henno Brandsma
Jan 27 at 19:32
1
1
$begingroup$
And this is the most general example: if $X$ is a nonempty door space in which the intersection of any two nonempty open sets is nonempty, then the set of all nonempty open sets is an ultrafilter (a free ultrafilter if $X$ is a $text{T}_1$-space).
$endgroup$
– bof
Jan 27 at 8:55
$begingroup$
And this is the most general example: if $X$ is a nonempty door space in which the intersection of any two nonempty open sets is nonempty, then the set of all nonempty open sets is an ultrafilter (a free ultrafilter if $X$ is a $text{T}_1$-space).
$endgroup$
– bof
Jan 27 at 8:55
$begingroup$
This connection between ultrafilters and topologies is most certainly interesting. If such ultrafilter $mathcal{F}$ is principal instead of free, one gets one of the better known door spaces: a particular point space. But not all door spaces come from ultrafilters, as any excluded point topology shows. Henno, do you know if someone has ever investigated this topic any further?
$endgroup$
– Rlos
Jan 27 at 19:20
$begingroup$
This connection between ultrafilters and topologies is most certainly interesting. If such ultrafilter $mathcal{F}$ is principal instead of free, one gets one of the better known door spaces: a particular point space. But not all door spaces come from ultrafilters, as any excluded point topology shows. Henno, do you know if someone has ever investigated this topic any further?
$endgroup$
– Rlos
Jan 27 at 19:20
$begingroup$
Also, in your previous comment you mentioned that textit{there can be no explicit examples} of free ultrafilters. Does this mean that it has been proven that their existence dependes solely on AC?
$endgroup$
– Rlos
Jan 27 at 19:24
$begingroup$
Also, in your previous comment you mentioned that textit{there can be no explicit examples} of free ultrafilters. Does this mean that it has been proven that their existence dependes solely on AC?
$endgroup$
– Rlos
Jan 27 at 19:24
$begingroup$
@Rlos I believe there are models of ZF that have no free ultrafilters. E.g. there are models of ZF and dependent choice in which all real subsets are measurable, if we assume some large cardinal (Solovay model). This implies that even countable choice doesn’t give us an ultrafilter on the natural numbers in that case.
$endgroup$
– Henno Brandsma
Jan 27 at 19:29
$begingroup$
@Rlos I believe there are models of ZF that have no free ultrafilters. E.g. there are models of ZF and dependent choice in which all real subsets are measurable, if we assume some large cardinal (Solovay model). This implies that even countable choice doesn’t give us an ultrafilter on the natural numbers in that case.
$endgroup$
– Henno Brandsma
Jan 27 at 19:29
$begingroup$
@Rlos I believe there aren’t any other notable papers on door spaces that I know of (but I don’t follow that area of topology very closely). I had seen your referenced paper before. Look in MR to find more, e.g.
$endgroup$
– Henno Brandsma
Jan 27 at 19:32
$begingroup$
@Rlos I believe there aren’t any other notable papers on door spaces that I know of (but I don’t follow that area of topology very closely). I had seen your referenced paper before. Look in MR to find more, e.g.
$endgroup$
– Henno Brandsma
Jan 27 at 19:32
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3088994%2fexamples-of-connected-door-spaces%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
The name 'door space' is so brilliant, I love it!
$endgroup$
– SmileyCraft
Jan 27 at 3:24
$begingroup$
Just saying, the Euclidean topology also satisfies the third type, but is not a door space.
$endgroup$
– SmileyCraft
Jan 27 at 3:30
$begingroup$
Before anyone else wastes their time with this, the straight forward Zorn's lemma approach does not work. There is a chain of connected topologies for which any upper bound is disconnected. In particular, let $mathcal{T}_r$ be the connected topology on $mathbb{R}$ generated by ${(-infty,l):lleq-r}cup{(u,infty):rleq u}cup{{0}}$. Then any upper bound of ${mathcal{T}_r:r>0}$ must have $(-infty,0]$ and $(0,infty)$ both as open sets.
$endgroup$
– SmileyCraft
Jan 27 at 4:12
1
$begingroup$
Yes, I love the concept as well and that's essentially why I became interested in these spaces haha. Unfortunately (or fortunately?) I just found out that a door space is of the third type described above if and only if excluding the empty set from the topology gives a free ultrafilter. To my knowledge, no one has ever constructed an explicit free ultrafilter, so I suppose that that answers my own question...
$endgroup$
– Rlos
Jan 27 at 4:40
$begingroup$
Indeed an ultrafilter is necessary and there can be no "explicit" example of an ultrafilter, as is well-known. But just use a little bit of AC and we can prove one exists...
$endgroup$
– Henno Brandsma
Jan 27 at 10:31