Example of a locally compact + Hausdorff + ¬normal + connected spaceIs dependent choice necessary to prove every perfect compact Hausdorff space is uncountable?Proof help. Core-compactness, Hausdorff, Locally CompactConnected and Compact preserving function is not continuous example?In a locally compact Hausdorff space, why are open subsets locally compact?Global structure of the Gromov-Hausdorff spaceClosed and open subsets of locally compact Hausdorff space are locally compactA locally compact space having a non-locally-compact open subsetCompact iff connectedNon-Hausdorff locally compact topological fieldAn open connected subset of a Peano space is arcwise connected.
Does multi-classing into Fighter give you heavy armor?
"Words were different when they (lived / were living) inside of you"
Are Roman Catholic priests ever addressed as pastor
Are ETF trackers fundamentally better than individual stocks?
How to explain that I do not want to visit a country due to personal safety concern?
Python if-else code style for reduced code for rounding floats
Knife as defense against stray dogs
Is it insecure to send a password in a `curl` command?
What is "focus distance lower/upper" and how is it different from depth of field?
et qui - how do you really understand that kind of phraseology?
How to make healing in an exploration game interesting
Why does overlay work only on the first tcolorbox?
How should I state my peer review experience in the CV?
What is the significance behind "40 days" that often appears in the Bible?
Is it true that good novels will automatically sell themselves on Amazon (and so on) and there is no need for one to waste time promoting?
When to use a slotted vs. solid turner?
Meme-controlled people
Are all passive ability checks floors for active ability checks?
How could an airship be repaired midflight?
About the actual radiative impact of greenhouse gas emission over time
What exactly is this small puffer fish doing and how did it manage to accomplish such a feat?
Recruiter wants very extensive technical details about all of my previous work
Why does a Star of David appear at a rally with Francisco Franco?
I got the following comment from a reputed math journal. What does it mean?
Example of a locally compact + Hausdorff + ¬normal + connected space
Is dependent choice necessary to prove every perfect compact Hausdorff space is uncountable?Proof help. Core-compactness, Hausdorff, Locally CompactConnected and Compact preserving function is not continuous example?In a locally compact Hausdorff space, why are open subsets locally compact?Global structure of the Gromov-Hausdorff spaceClosed and open subsets of locally compact Hausdorff space are locally compactA locally compact space having a non-locally-compact open subsetCompact iff connectedNon-Hausdorff locally compact topological fieldAn open connected subset of a Peano space is arcwise connected.
$begingroup$
Playing with $pi$-base for a while, I found that
In fact, only four spaces on $pi$-base satisfy the first three properties.
My question is:
Can we prove or disprove the existence of a locally compact + Hausdorff + ¬normal + connected space?
Many thanks!
Related:
- https://mathoverflow.net/questions/53300/locally-compact-hausdorff-space-that-is-not-normal
general-topology examples-counterexamples
$endgroup$
add a comment |
$begingroup$
Playing with $pi$-base for a while, I found that
In fact, only four spaces on $pi$-base satisfy the first three properties.
My question is:
Can we prove or disprove the existence of a locally compact + Hausdorff + ¬normal + connected space?
Many thanks!
Related:
- https://mathoverflow.net/questions/53300/locally-compact-hausdorff-space-that-is-not-normal
general-topology examples-counterexamples
$endgroup$
1
$begingroup$
What happens if you take the Tychonoff-plank-minus-corner example and make it connected by filling in intervals between consecutive points in both $omega_1+1$ and $omega+1$. In other words, take the one-point compactification $bar L=Lcup*$ of the long line $L$, form the product $bar Ltimes[0,1]$, and delete the corner point $(*,1)$? This is locally compact and $T_2$ because it's an open subspace of the compact $T_2$ space $bar Ltimes[0,1]$. It seems to not be normal as $Ltimes1$ and $*times[0,1)$ can't be separated, and it looks connected.
$endgroup$
– Andreas Blass
Mar 12 at 18:53
$begingroup$
@AndreasBlass a separation for those sets would imply one for the deleted Tychonoff plank and so cannot exist. The example works I think.
$endgroup$
– Henno Brandsma
Mar 12 at 21:32
add a comment |
$begingroup$
Playing with $pi$-base for a while, I found that
In fact, only four spaces on $pi$-base satisfy the first three properties.
My question is:
Can we prove or disprove the existence of a locally compact + Hausdorff + ¬normal + connected space?
Many thanks!
Related:
- https://mathoverflow.net/questions/53300/locally-compact-hausdorff-space-that-is-not-normal
general-topology examples-counterexamples
$endgroup$
Playing with $pi$-base for a while, I found that
In fact, only four spaces on $pi$-base satisfy the first three properties.
My question is:
Can we prove or disprove the existence of a locally compact + Hausdorff + ¬normal + connected space?
Many thanks!
Related:
- https://mathoverflow.net/questions/53300/locally-compact-hausdorff-space-that-is-not-normal
general-topology examples-counterexamples
general-topology examples-counterexamples
asked Mar 12 at 10:07
YuiTo ChengYuiTo Cheng
2,0532637
2,0532637
1
$begingroup$
What happens if you take the Tychonoff-plank-minus-corner example and make it connected by filling in intervals between consecutive points in both $omega_1+1$ and $omega+1$. In other words, take the one-point compactification $bar L=Lcup*$ of the long line $L$, form the product $bar Ltimes[0,1]$, and delete the corner point $(*,1)$? This is locally compact and $T_2$ because it's an open subspace of the compact $T_2$ space $bar Ltimes[0,1]$. It seems to not be normal as $Ltimes1$ and $*times[0,1)$ can't be separated, and it looks connected.
$endgroup$
– Andreas Blass
Mar 12 at 18:53
$begingroup$
@AndreasBlass a separation for those sets would imply one for the deleted Tychonoff plank and so cannot exist. The example works I think.
$endgroup$
– Henno Brandsma
Mar 12 at 21:32
add a comment |
1
$begingroup$
What happens if you take the Tychonoff-plank-minus-corner example and make it connected by filling in intervals between consecutive points in both $omega_1+1$ and $omega+1$. In other words, take the one-point compactification $bar L=Lcup*$ of the long line $L$, form the product $bar Ltimes[0,1]$, and delete the corner point $(*,1)$? This is locally compact and $T_2$ because it's an open subspace of the compact $T_2$ space $bar Ltimes[0,1]$. It seems to not be normal as $Ltimes1$ and $*times[0,1)$ can't be separated, and it looks connected.
$endgroup$
– Andreas Blass
Mar 12 at 18:53
$begingroup$
@AndreasBlass a separation for those sets would imply one for the deleted Tychonoff plank and so cannot exist. The example works I think.
$endgroup$
– Henno Brandsma
Mar 12 at 21:32
1
1
$begingroup$
What happens if you take the Tychonoff-plank-minus-corner example and make it connected by filling in intervals between consecutive points in both $omega_1+1$ and $omega+1$. In other words, take the one-point compactification $bar L=Lcup*$ of the long line $L$, form the product $bar Ltimes[0,1]$, and delete the corner point $(*,1)$? This is locally compact and $T_2$ because it's an open subspace of the compact $T_2$ space $bar Ltimes[0,1]$. It seems to not be normal as $Ltimes1$ and $*times[0,1)$ can't be separated, and it looks connected.
$endgroup$
– Andreas Blass
Mar 12 at 18:53
$begingroup$
What happens if you take the Tychonoff-plank-minus-corner example and make it connected by filling in intervals between consecutive points in both $omega_1+1$ and $omega+1$. In other words, take the one-point compactification $bar L=Lcup*$ of the long line $L$, form the product $bar Ltimes[0,1]$, and delete the corner point $(*,1)$? This is locally compact and $T_2$ because it's an open subspace of the compact $T_2$ space $bar Ltimes[0,1]$. It seems to not be normal as $Ltimes1$ and $*times[0,1)$ can't be separated, and it looks connected.
$endgroup$
– Andreas Blass
Mar 12 at 18:53
$begingroup$
@AndreasBlass a separation for those sets would imply one for the deleted Tychonoff plank and so cannot exist. The example works I think.
$endgroup$
– Henno Brandsma
Mar 12 at 21:32
$begingroup$
@AndreasBlass a separation for those sets would imply one for the deleted Tychonoff plank and so cannot exist. The example works I think.
$endgroup$
– Henno Brandsma
Mar 12 at 21:32
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The fourth example is easily "fixable":
Let $X= L(omega_1) times [0,1]^[0,1]$ (we could also take $[0,1]^omega_1$ as the second factor if we care about minimal weight, with the same result), and where $L(omega_1)$ is the so-called long line, which is a "connectified" version of $omega_1$ (or $[0,Omega)$, as $pi$-base calls it; the first uncountable ordinal (in the order topology)).
$X$ is Hausdorff as the product of two Hausdorff spaces.
$X$ is locally compact as the product of the locally compact $L(omega_1)$ and the (locally) compact $[0,1]^[0,1]$. (local compactness is finitely productive.)
$X$ is connected as the product of two connected spaces.
$X$ is not normal, as it contains $L(omega_1) times beta L(omega_1)$ as a closed subspace, where $beta L(omega_1)$ is the Cech-Stone compactification of $L(omega_1)$ and equals its one-point compactfication (which is homeomorphic to the extended long line ($L^ast$ on the linked page)), because $[0,1]^[0,1]$ is a universal space for all spaces of weight $le aleph_1$ (even for all spaces of weight $le mathfrakc$) by Tychonoff's embedding theorem) and so contains (a homeomorphic copy of) $beta L(omega_1)$ as a (necessarily closed) subspace. Tamano's theorem that $X$ is paracompact Hausdorff iff $X times beta X$ is normal, plus the fact that $L(omega_1)$ is not paracompact implies that $X$ is not normal. This is in essence the same argument (with $omega_1$ replaced by its connected variant $L(omega_1)$) that we also can hold to show non-normality of the fourth space from the $pi$-base list (which has no special name that I know of).
So we could also use a space of the form $Y times beta Y$, where $Y$ is a locally compact, connected, Hausdorff space. This makes $beta Y$ connected (as $Y$ is dense in it) and we have the same essential argument (minus the universal space argument, but we need Tamano's theorem) that such a product will be an example.
So $mathbbR times beta mathbbR$ works too, e.g. (but has a much larger weight). So maybe you can make more special examples (with extra properties) this way.
I would have liked to have been able to make a connected version of Mrówka's $Psi$-space, which is also pseudocompact and not countably compact. But I did not see how to do that...
$endgroup$
add a comment |
$begingroup$
The hunt for interesting counterexample is always a fun one.
Note: This post is long and ultimately doesn't answer OP's question. I just wanted to suggest a line of attack.
One thing that you might try first is the following, every locally compact Hausdorff space is $T_3$. We then might first broaden the objective to finding a connected $T_3$ space that isn't normal. The $pi$-base does has some examples of these.
Only two though
With one of these spaces in hand, you can attempt to "locally compactify" the space. That is you can embed such a space into a locally compact Hausdorff space in such a way that it is a dense subspace. As connectedness is closed under closure this may be do the job so long as the local compactification doesn't introduce normality. There are then two questions, one of which there is an answer for:
- How you locally compactify a ($T_3$) space?
- How do you make sure that a local compactification is not normal?
I will admit that I do not know how to answer the second question. The first question however has a reasonably well studied answer. Namely, every separated local proximity space admits a local compactification. All of the following information was pulled from Naimpally and Warrack's book "Proximity Spaces".
Definition: A nonempty collection $mathcalB$ of subsets of a topological space $X$ is called a boundedness (this might be old terminology) iff
(i) $AinmathcalB$ and $Bsubseteq A$ implies $BinmathcalB$
(ii) $mathcalB$ is closed under finite unions.
Basically a boundedness is a bornology without the requirement that the collection covers the space.
Definition: A local proximity psace is a triple $(X,alpha,mathcalB)$ where $X$ is a set, $mathcalB$ is a boundedness on $X$ and $alpha$ is a binary relation on the power set of t$X$ satisfing:
(i) $Aalpha Bimplies Balpha A$
(ii) $(Acup B)alpha Ciff$ $Aalpha C$ or $Balpha C$
(iii) $Aalpha Bimplies A,Bneqemptyset$
(iv) $Acap Bneqemptysetimplies Aalpha B$
(v) Let $Asubseteq X$ and $BinmathcalB$. If for every $CinmathcalB$ either $Aalpha C$ or $(Xsetminus C)alpha B$, then $Aalpha B$.
(vi) If $Aalpha B$, then there is a $DinmathcalB$ such that $Dsubseteq B$ and $Aalpha D$.
If in addition to these axioms the relation $alpha$ satisfies
$$xalphayiff x=y$$
then the local proximity space $(X,alpha,mathcalB)$ is called separated.
The topology on a local proximity space $(X,alpha,mathcalB)$ is defined by means of the closure operator
$$overlineA:=xin Xmid xalpha A$$
Some good facts to have about local proximity spaces are the following:
(i) The closure of bounded sets are bounded.
(ii) Every compact subset of a local proximity space is bounded.
Finally, how do we locally compactify? Effectively we define a means of getting a one-point compactification.
Definition: Let $(X,alpha,mathcalB)$ be a local proximity space an define a binary relation $delta$ on the power set of $X$ by
$$Adelta Biff Aalpha B,text or A,BnotinmathcalB$$
Then $delta$ is called the Alexandroff extension of $alpha$, and is a proximity on $X$. Moreover, if $alpha$ was a separated localy proximity, then $delta$ is separated.
Every separated proximity space has a compactification called its Smirnov compactification. There is a certain point in the Smirnov compactification that you can delete, leaving you with a locally compact Hausdorff space. If your original local proximity space is connected, then this locally compact hausdorff space will be connected as well. The parts of the theorem that would matter for you are the following:
Theorem: Given a separated local proximity space $(X,alpha,mathcalB)$, there is a locally compact Hausdorff space $L$ and a one-to-one map $f:Xrightarrow L$ such that:
(i) $Aalpha Biffoverlinef(A)capoverlinef(B)=emptyset$ in $L$
(ii) $BinmathcalBiffoverlinef(B)$ is compact in $L$
(iii) $overlinef(X)=L$
This is probably where I would start the search for the space you want if I didn't have an obvious example. I am not at all sure how to make sure that normality isn't gained in this process.
EDIT: I forgot to mention, every completely regular space admits a proximity relation whose induced topology is the original topology.
$endgroup$
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%2f3144897%2fexample-of-a-locally-compact-hausdorff-%25c2%25acnormal-connected-space%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The fourth example is easily "fixable":
Let $X= L(omega_1) times [0,1]^[0,1]$ (we could also take $[0,1]^omega_1$ as the second factor if we care about minimal weight, with the same result), and where $L(omega_1)$ is the so-called long line, which is a "connectified" version of $omega_1$ (or $[0,Omega)$, as $pi$-base calls it; the first uncountable ordinal (in the order topology)).
$X$ is Hausdorff as the product of two Hausdorff spaces.
$X$ is locally compact as the product of the locally compact $L(omega_1)$ and the (locally) compact $[0,1]^[0,1]$. (local compactness is finitely productive.)
$X$ is connected as the product of two connected spaces.
$X$ is not normal, as it contains $L(omega_1) times beta L(omega_1)$ as a closed subspace, where $beta L(omega_1)$ is the Cech-Stone compactification of $L(omega_1)$ and equals its one-point compactfication (which is homeomorphic to the extended long line ($L^ast$ on the linked page)), because $[0,1]^[0,1]$ is a universal space for all spaces of weight $le aleph_1$ (even for all spaces of weight $le mathfrakc$) by Tychonoff's embedding theorem) and so contains (a homeomorphic copy of) $beta L(omega_1)$ as a (necessarily closed) subspace. Tamano's theorem that $X$ is paracompact Hausdorff iff $X times beta X$ is normal, plus the fact that $L(omega_1)$ is not paracompact implies that $X$ is not normal. This is in essence the same argument (with $omega_1$ replaced by its connected variant $L(omega_1)$) that we also can hold to show non-normality of the fourth space from the $pi$-base list (which has no special name that I know of).
So we could also use a space of the form $Y times beta Y$, where $Y$ is a locally compact, connected, Hausdorff space. This makes $beta Y$ connected (as $Y$ is dense in it) and we have the same essential argument (minus the universal space argument, but we need Tamano's theorem) that such a product will be an example.
So $mathbbR times beta mathbbR$ works too, e.g. (but has a much larger weight). So maybe you can make more special examples (with extra properties) this way.
I would have liked to have been able to make a connected version of Mrówka's $Psi$-space, which is also pseudocompact and not countably compact. But I did not see how to do that...
$endgroup$
add a comment |
$begingroup$
The fourth example is easily "fixable":
Let $X= L(omega_1) times [0,1]^[0,1]$ (we could also take $[0,1]^omega_1$ as the second factor if we care about minimal weight, with the same result), and where $L(omega_1)$ is the so-called long line, which is a "connectified" version of $omega_1$ (or $[0,Omega)$, as $pi$-base calls it; the first uncountable ordinal (in the order topology)).
$X$ is Hausdorff as the product of two Hausdorff spaces.
$X$ is locally compact as the product of the locally compact $L(omega_1)$ and the (locally) compact $[0,1]^[0,1]$. (local compactness is finitely productive.)
$X$ is connected as the product of two connected spaces.
$X$ is not normal, as it contains $L(omega_1) times beta L(omega_1)$ as a closed subspace, where $beta L(omega_1)$ is the Cech-Stone compactification of $L(omega_1)$ and equals its one-point compactfication (which is homeomorphic to the extended long line ($L^ast$ on the linked page)), because $[0,1]^[0,1]$ is a universal space for all spaces of weight $le aleph_1$ (even for all spaces of weight $le mathfrakc$) by Tychonoff's embedding theorem) and so contains (a homeomorphic copy of) $beta L(omega_1)$ as a (necessarily closed) subspace. Tamano's theorem that $X$ is paracompact Hausdorff iff $X times beta X$ is normal, plus the fact that $L(omega_1)$ is not paracompact implies that $X$ is not normal. This is in essence the same argument (with $omega_1$ replaced by its connected variant $L(omega_1)$) that we also can hold to show non-normality of the fourth space from the $pi$-base list (which has no special name that I know of).
So we could also use a space of the form $Y times beta Y$, where $Y$ is a locally compact, connected, Hausdorff space. This makes $beta Y$ connected (as $Y$ is dense in it) and we have the same essential argument (minus the universal space argument, but we need Tamano's theorem) that such a product will be an example.
So $mathbbR times beta mathbbR$ works too, e.g. (but has a much larger weight). So maybe you can make more special examples (with extra properties) this way.
I would have liked to have been able to make a connected version of Mrówka's $Psi$-space, which is also pseudocompact and not countably compact. But I did not see how to do that...
$endgroup$
add a comment |
$begingroup$
The fourth example is easily "fixable":
Let $X= L(omega_1) times [0,1]^[0,1]$ (we could also take $[0,1]^omega_1$ as the second factor if we care about minimal weight, with the same result), and where $L(omega_1)$ is the so-called long line, which is a "connectified" version of $omega_1$ (or $[0,Omega)$, as $pi$-base calls it; the first uncountable ordinal (in the order topology)).
$X$ is Hausdorff as the product of two Hausdorff spaces.
$X$ is locally compact as the product of the locally compact $L(omega_1)$ and the (locally) compact $[0,1]^[0,1]$. (local compactness is finitely productive.)
$X$ is connected as the product of two connected spaces.
$X$ is not normal, as it contains $L(omega_1) times beta L(omega_1)$ as a closed subspace, where $beta L(omega_1)$ is the Cech-Stone compactification of $L(omega_1)$ and equals its one-point compactfication (which is homeomorphic to the extended long line ($L^ast$ on the linked page)), because $[0,1]^[0,1]$ is a universal space for all spaces of weight $le aleph_1$ (even for all spaces of weight $le mathfrakc$) by Tychonoff's embedding theorem) and so contains (a homeomorphic copy of) $beta L(omega_1)$ as a (necessarily closed) subspace. Tamano's theorem that $X$ is paracompact Hausdorff iff $X times beta X$ is normal, plus the fact that $L(omega_1)$ is not paracompact implies that $X$ is not normal. This is in essence the same argument (with $omega_1$ replaced by its connected variant $L(omega_1)$) that we also can hold to show non-normality of the fourth space from the $pi$-base list (which has no special name that I know of).
So we could also use a space of the form $Y times beta Y$, where $Y$ is a locally compact, connected, Hausdorff space. This makes $beta Y$ connected (as $Y$ is dense in it) and we have the same essential argument (minus the universal space argument, but we need Tamano's theorem) that such a product will be an example.
So $mathbbR times beta mathbbR$ works too, e.g. (but has a much larger weight). So maybe you can make more special examples (with extra properties) this way.
I would have liked to have been able to make a connected version of Mrówka's $Psi$-space, which is also pseudocompact and not countably compact. But I did not see how to do that...
$endgroup$
The fourth example is easily "fixable":
Let $X= L(omega_1) times [0,1]^[0,1]$ (we could also take $[0,1]^omega_1$ as the second factor if we care about minimal weight, with the same result), and where $L(omega_1)$ is the so-called long line, which is a "connectified" version of $omega_1$ (or $[0,Omega)$, as $pi$-base calls it; the first uncountable ordinal (in the order topology)).
$X$ is Hausdorff as the product of two Hausdorff spaces.
$X$ is locally compact as the product of the locally compact $L(omega_1)$ and the (locally) compact $[0,1]^[0,1]$. (local compactness is finitely productive.)
$X$ is connected as the product of two connected spaces.
$X$ is not normal, as it contains $L(omega_1) times beta L(omega_1)$ as a closed subspace, where $beta L(omega_1)$ is the Cech-Stone compactification of $L(omega_1)$ and equals its one-point compactfication (which is homeomorphic to the extended long line ($L^ast$ on the linked page)), because $[0,1]^[0,1]$ is a universal space for all spaces of weight $le aleph_1$ (even for all spaces of weight $le mathfrakc$) by Tychonoff's embedding theorem) and so contains (a homeomorphic copy of) $beta L(omega_1)$ as a (necessarily closed) subspace. Tamano's theorem that $X$ is paracompact Hausdorff iff $X times beta X$ is normal, plus the fact that $L(omega_1)$ is not paracompact implies that $X$ is not normal. This is in essence the same argument (with $omega_1$ replaced by its connected variant $L(omega_1)$) that we also can hold to show non-normality of the fourth space from the $pi$-base list (which has no special name that I know of).
So we could also use a space of the form $Y times beta Y$, where $Y$ is a locally compact, connected, Hausdorff space. This makes $beta Y$ connected (as $Y$ is dense in it) and we have the same essential argument (minus the universal space argument, but we need Tamano's theorem) that such a product will be an example.
So $mathbbR times beta mathbbR$ works too, e.g. (but has a much larger weight). So maybe you can make more special examples (with extra properties) this way.
I would have liked to have been able to make a connected version of Mrówka's $Psi$-space, which is also pseudocompact and not countably compact. But I did not see how to do that...
edited Mar 12 at 23:13
answered Mar 12 at 21:30
Henno BrandsmaHenno Brandsma
112k348121
112k348121
add a comment |
add a comment |
$begingroup$
The hunt for interesting counterexample is always a fun one.
Note: This post is long and ultimately doesn't answer OP's question. I just wanted to suggest a line of attack.
One thing that you might try first is the following, every locally compact Hausdorff space is $T_3$. We then might first broaden the objective to finding a connected $T_3$ space that isn't normal. The $pi$-base does has some examples of these.
Only two though
With one of these spaces in hand, you can attempt to "locally compactify" the space. That is you can embed such a space into a locally compact Hausdorff space in such a way that it is a dense subspace. As connectedness is closed under closure this may be do the job so long as the local compactification doesn't introduce normality. There are then two questions, one of which there is an answer for:
- How you locally compactify a ($T_3$) space?
- How do you make sure that a local compactification is not normal?
I will admit that I do not know how to answer the second question. The first question however has a reasonably well studied answer. Namely, every separated local proximity space admits a local compactification. All of the following information was pulled from Naimpally and Warrack's book "Proximity Spaces".
Definition: A nonempty collection $mathcalB$ of subsets of a topological space $X$ is called a boundedness (this might be old terminology) iff
(i) $AinmathcalB$ and $Bsubseteq A$ implies $BinmathcalB$
(ii) $mathcalB$ is closed under finite unions.
Basically a boundedness is a bornology without the requirement that the collection covers the space.
Definition: A local proximity psace is a triple $(X,alpha,mathcalB)$ where $X$ is a set, $mathcalB$ is a boundedness on $X$ and $alpha$ is a binary relation on the power set of t$X$ satisfing:
(i) $Aalpha Bimplies Balpha A$
(ii) $(Acup B)alpha Ciff$ $Aalpha C$ or $Balpha C$
(iii) $Aalpha Bimplies A,Bneqemptyset$
(iv) $Acap Bneqemptysetimplies Aalpha B$
(v) Let $Asubseteq X$ and $BinmathcalB$. If for every $CinmathcalB$ either $Aalpha C$ or $(Xsetminus C)alpha B$, then $Aalpha B$.
(vi) If $Aalpha B$, then there is a $DinmathcalB$ such that $Dsubseteq B$ and $Aalpha D$.
If in addition to these axioms the relation $alpha$ satisfies
$$xalphayiff x=y$$
then the local proximity space $(X,alpha,mathcalB)$ is called separated.
The topology on a local proximity space $(X,alpha,mathcalB)$ is defined by means of the closure operator
$$overlineA:=xin Xmid xalpha A$$
Some good facts to have about local proximity spaces are the following:
(i) The closure of bounded sets are bounded.
(ii) Every compact subset of a local proximity space is bounded.
Finally, how do we locally compactify? Effectively we define a means of getting a one-point compactification.
Definition: Let $(X,alpha,mathcalB)$ be a local proximity space an define a binary relation $delta$ on the power set of $X$ by
$$Adelta Biff Aalpha B,text or A,BnotinmathcalB$$
Then $delta$ is called the Alexandroff extension of $alpha$, and is a proximity on $X$. Moreover, if $alpha$ was a separated localy proximity, then $delta$ is separated.
Every separated proximity space has a compactification called its Smirnov compactification. There is a certain point in the Smirnov compactification that you can delete, leaving you with a locally compact Hausdorff space. If your original local proximity space is connected, then this locally compact hausdorff space will be connected as well. The parts of the theorem that would matter for you are the following:
Theorem: Given a separated local proximity space $(X,alpha,mathcalB)$, there is a locally compact Hausdorff space $L$ and a one-to-one map $f:Xrightarrow L$ such that:
(i) $Aalpha Biffoverlinef(A)capoverlinef(B)=emptyset$ in $L$
(ii) $BinmathcalBiffoverlinef(B)$ is compact in $L$
(iii) $overlinef(X)=L$
This is probably where I would start the search for the space you want if I didn't have an obvious example. I am not at all sure how to make sure that normality isn't gained in this process.
EDIT: I forgot to mention, every completely regular space admits a proximity relation whose induced topology is the original topology.
$endgroup$
add a comment |
$begingroup$
The hunt for interesting counterexample is always a fun one.
Note: This post is long and ultimately doesn't answer OP's question. I just wanted to suggest a line of attack.
One thing that you might try first is the following, every locally compact Hausdorff space is $T_3$. We then might first broaden the objective to finding a connected $T_3$ space that isn't normal. The $pi$-base does has some examples of these.
Only two though
With one of these spaces in hand, you can attempt to "locally compactify" the space. That is you can embed such a space into a locally compact Hausdorff space in such a way that it is a dense subspace. As connectedness is closed under closure this may be do the job so long as the local compactification doesn't introduce normality. There are then two questions, one of which there is an answer for:
- How you locally compactify a ($T_3$) space?
- How do you make sure that a local compactification is not normal?
I will admit that I do not know how to answer the second question. The first question however has a reasonably well studied answer. Namely, every separated local proximity space admits a local compactification. All of the following information was pulled from Naimpally and Warrack's book "Proximity Spaces".
Definition: A nonempty collection $mathcalB$ of subsets of a topological space $X$ is called a boundedness (this might be old terminology) iff
(i) $AinmathcalB$ and $Bsubseteq A$ implies $BinmathcalB$
(ii) $mathcalB$ is closed under finite unions.
Basically a boundedness is a bornology without the requirement that the collection covers the space.
Definition: A local proximity psace is a triple $(X,alpha,mathcalB)$ where $X$ is a set, $mathcalB$ is a boundedness on $X$ and $alpha$ is a binary relation on the power set of t$X$ satisfing:
(i) $Aalpha Bimplies Balpha A$
(ii) $(Acup B)alpha Ciff$ $Aalpha C$ or $Balpha C$
(iii) $Aalpha Bimplies A,Bneqemptyset$
(iv) $Acap Bneqemptysetimplies Aalpha B$
(v) Let $Asubseteq X$ and $BinmathcalB$. If for every $CinmathcalB$ either $Aalpha C$ or $(Xsetminus C)alpha B$, then $Aalpha B$.
(vi) If $Aalpha B$, then there is a $DinmathcalB$ such that $Dsubseteq B$ and $Aalpha D$.
If in addition to these axioms the relation $alpha$ satisfies
$$xalphayiff x=y$$
then the local proximity space $(X,alpha,mathcalB)$ is called separated.
The topology on a local proximity space $(X,alpha,mathcalB)$ is defined by means of the closure operator
$$overlineA:=xin Xmid xalpha A$$
Some good facts to have about local proximity spaces are the following:
(i) The closure of bounded sets are bounded.
(ii) Every compact subset of a local proximity space is bounded.
Finally, how do we locally compactify? Effectively we define a means of getting a one-point compactification.
Definition: Let $(X,alpha,mathcalB)$ be a local proximity space an define a binary relation $delta$ on the power set of $X$ by
$$Adelta Biff Aalpha B,text or A,BnotinmathcalB$$
Then $delta$ is called the Alexandroff extension of $alpha$, and is a proximity on $X$. Moreover, if $alpha$ was a separated localy proximity, then $delta$ is separated.
Every separated proximity space has a compactification called its Smirnov compactification. There is a certain point in the Smirnov compactification that you can delete, leaving you with a locally compact Hausdorff space. If your original local proximity space is connected, then this locally compact hausdorff space will be connected as well. The parts of the theorem that would matter for you are the following:
Theorem: Given a separated local proximity space $(X,alpha,mathcalB)$, there is a locally compact Hausdorff space $L$ and a one-to-one map $f:Xrightarrow L$ such that:
(i) $Aalpha Biffoverlinef(A)capoverlinef(B)=emptyset$ in $L$
(ii) $BinmathcalBiffoverlinef(B)$ is compact in $L$
(iii) $overlinef(X)=L$
This is probably where I would start the search for the space you want if I didn't have an obvious example. I am not at all sure how to make sure that normality isn't gained in this process.
EDIT: I forgot to mention, every completely regular space admits a proximity relation whose induced topology is the original topology.
$endgroup$
add a comment |
$begingroup$
The hunt for interesting counterexample is always a fun one.
Note: This post is long and ultimately doesn't answer OP's question. I just wanted to suggest a line of attack.
One thing that you might try first is the following, every locally compact Hausdorff space is $T_3$. We then might first broaden the objective to finding a connected $T_3$ space that isn't normal. The $pi$-base does has some examples of these.
Only two though
With one of these spaces in hand, you can attempt to "locally compactify" the space. That is you can embed such a space into a locally compact Hausdorff space in such a way that it is a dense subspace. As connectedness is closed under closure this may be do the job so long as the local compactification doesn't introduce normality. There are then two questions, one of which there is an answer for:
- How you locally compactify a ($T_3$) space?
- How do you make sure that a local compactification is not normal?
I will admit that I do not know how to answer the second question. The first question however has a reasonably well studied answer. Namely, every separated local proximity space admits a local compactification. All of the following information was pulled from Naimpally and Warrack's book "Proximity Spaces".
Definition: A nonempty collection $mathcalB$ of subsets of a topological space $X$ is called a boundedness (this might be old terminology) iff
(i) $AinmathcalB$ and $Bsubseteq A$ implies $BinmathcalB$
(ii) $mathcalB$ is closed under finite unions.
Basically a boundedness is a bornology without the requirement that the collection covers the space.
Definition: A local proximity psace is a triple $(X,alpha,mathcalB)$ where $X$ is a set, $mathcalB$ is a boundedness on $X$ and $alpha$ is a binary relation on the power set of t$X$ satisfing:
(i) $Aalpha Bimplies Balpha A$
(ii) $(Acup B)alpha Ciff$ $Aalpha C$ or $Balpha C$
(iii) $Aalpha Bimplies A,Bneqemptyset$
(iv) $Acap Bneqemptysetimplies Aalpha B$
(v) Let $Asubseteq X$ and $BinmathcalB$. If for every $CinmathcalB$ either $Aalpha C$ or $(Xsetminus C)alpha B$, then $Aalpha B$.
(vi) If $Aalpha B$, then there is a $DinmathcalB$ such that $Dsubseteq B$ and $Aalpha D$.
If in addition to these axioms the relation $alpha$ satisfies
$$xalphayiff x=y$$
then the local proximity space $(X,alpha,mathcalB)$ is called separated.
The topology on a local proximity space $(X,alpha,mathcalB)$ is defined by means of the closure operator
$$overlineA:=xin Xmid xalpha A$$
Some good facts to have about local proximity spaces are the following:
(i) The closure of bounded sets are bounded.
(ii) Every compact subset of a local proximity space is bounded.
Finally, how do we locally compactify? Effectively we define a means of getting a one-point compactification.
Definition: Let $(X,alpha,mathcalB)$ be a local proximity space an define a binary relation $delta$ on the power set of $X$ by
$$Adelta Biff Aalpha B,text or A,BnotinmathcalB$$
Then $delta$ is called the Alexandroff extension of $alpha$, and is a proximity on $X$. Moreover, if $alpha$ was a separated localy proximity, then $delta$ is separated.
Every separated proximity space has a compactification called its Smirnov compactification. There is a certain point in the Smirnov compactification that you can delete, leaving you with a locally compact Hausdorff space. If your original local proximity space is connected, then this locally compact hausdorff space will be connected as well. The parts of the theorem that would matter for you are the following:
Theorem: Given a separated local proximity space $(X,alpha,mathcalB)$, there is a locally compact Hausdorff space $L$ and a one-to-one map $f:Xrightarrow L$ such that:
(i) $Aalpha Biffoverlinef(A)capoverlinef(B)=emptyset$ in $L$
(ii) $BinmathcalBiffoverlinef(B)$ is compact in $L$
(iii) $overlinef(X)=L$
This is probably where I would start the search for the space you want if I didn't have an obvious example. I am not at all sure how to make sure that normality isn't gained in this process.
EDIT: I forgot to mention, every completely regular space admits a proximity relation whose induced topology is the original topology.
$endgroup$
The hunt for interesting counterexample is always a fun one.
Note: This post is long and ultimately doesn't answer OP's question. I just wanted to suggest a line of attack.
One thing that you might try first is the following, every locally compact Hausdorff space is $T_3$. We then might first broaden the objective to finding a connected $T_3$ space that isn't normal. The $pi$-base does has some examples of these.
Only two though
With one of these spaces in hand, you can attempt to "locally compactify" the space. That is you can embed such a space into a locally compact Hausdorff space in such a way that it is a dense subspace. As connectedness is closed under closure this may be do the job so long as the local compactification doesn't introduce normality. There are then two questions, one of which there is an answer for:
- How you locally compactify a ($T_3$) space?
- How do you make sure that a local compactification is not normal?
I will admit that I do not know how to answer the second question. The first question however has a reasonably well studied answer. Namely, every separated local proximity space admits a local compactification. All of the following information was pulled from Naimpally and Warrack's book "Proximity Spaces".
Definition: A nonempty collection $mathcalB$ of subsets of a topological space $X$ is called a boundedness (this might be old terminology) iff
(i) $AinmathcalB$ and $Bsubseteq A$ implies $BinmathcalB$
(ii) $mathcalB$ is closed under finite unions.
Basically a boundedness is a bornology without the requirement that the collection covers the space.
Definition: A local proximity psace is a triple $(X,alpha,mathcalB)$ where $X$ is a set, $mathcalB$ is a boundedness on $X$ and $alpha$ is a binary relation on the power set of t$X$ satisfing:
(i) $Aalpha Bimplies Balpha A$
(ii) $(Acup B)alpha Ciff$ $Aalpha C$ or $Balpha C$
(iii) $Aalpha Bimplies A,Bneqemptyset$
(iv) $Acap Bneqemptysetimplies Aalpha B$
(v) Let $Asubseteq X$ and $BinmathcalB$. If for every $CinmathcalB$ either $Aalpha C$ or $(Xsetminus C)alpha B$, then $Aalpha B$.
(vi) If $Aalpha B$, then there is a $DinmathcalB$ such that $Dsubseteq B$ and $Aalpha D$.
If in addition to these axioms the relation $alpha$ satisfies
$$xalphayiff x=y$$
then the local proximity space $(X,alpha,mathcalB)$ is called separated.
The topology on a local proximity space $(X,alpha,mathcalB)$ is defined by means of the closure operator
$$overlineA:=xin Xmid xalpha A$$
Some good facts to have about local proximity spaces are the following:
(i) The closure of bounded sets are bounded.
(ii) Every compact subset of a local proximity space is bounded.
Finally, how do we locally compactify? Effectively we define a means of getting a one-point compactification.
Definition: Let $(X,alpha,mathcalB)$ be a local proximity space an define a binary relation $delta$ on the power set of $X$ by
$$Adelta Biff Aalpha B,text or A,BnotinmathcalB$$
Then $delta$ is called the Alexandroff extension of $alpha$, and is a proximity on $X$. Moreover, if $alpha$ was a separated localy proximity, then $delta$ is separated.
Every separated proximity space has a compactification called its Smirnov compactification. There is a certain point in the Smirnov compactification that you can delete, leaving you with a locally compact Hausdorff space. If your original local proximity space is connected, then this locally compact hausdorff space will be connected as well. The parts of the theorem that would matter for you are the following:
Theorem: Given a separated local proximity space $(X,alpha,mathcalB)$, there is a locally compact Hausdorff space $L$ and a one-to-one map $f:Xrightarrow L$ such that:
(i) $Aalpha Biffoverlinef(A)capoverlinef(B)=emptyset$ in $L$
(ii) $BinmathcalBiffoverlinef(B)$ is compact in $L$
(iii) $overlinef(X)=L$
This is probably where I would start the search for the space you want if I didn't have an obvious example. I am not at all sure how to make sure that normality isn't gained in this process.
EDIT: I forgot to mention, every completely regular space admits a proximity relation whose induced topology is the original topology.
answered Mar 12 at 18:12
Robert ThingumRobert Thingum
9011317
9011317
add a comment |
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%2f3144897%2fexample-of-a-locally-compact-hausdorff-%25c2%25acnormal-connected-space%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
1
$begingroup$
What happens if you take the Tychonoff-plank-minus-corner example and make it connected by filling in intervals between consecutive points in both $omega_1+1$ and $omega+1$. In other words, take the one-point compactification $bar L=Lcup*$ of the long line $L$, form the product $bar Ltimes[0,1]$, and delete the corner point $(*,1)$? This is locally compact and $T_2$ because it's an open subspace of the compact $T_2$ space $bar Ltimes[0,1]$. It seems to not be normal as $Ltimes1$ and $*times[0,1)$ can't be separated, and it looks connected.
$endgroup$
– Andreas Blass
Mar 12 at 18:53
$begingroup$
@AndreasBlass a separation for those sets would imply one for the deleted Tychonoff plank and so cannot exist. The example works I think.
$endgroup$
– Henno Brandsma
Mar 12 at 21:32