What is the topology on the direct limit of $Bbb R^i rightarrow Bbb R^j$












1












$begingroup$


If we let $X$ be direct limit of
$$Bbb R rightarrow Bbb R^2 rightarrow Bbb R^3 rightarrow cdots $$
where each arrow is inclusion to first coordinates.



What is a choice of $X$ (with a toplogy that coincides with the direct limit topology).



EDIT: I believe it is subspace of $Bbb R^{Bbb N}$, sequence eventually $0$ taking values in $Bbb R$, with the product topology. I wonder if this is correct.






general-topology algebraic-topology







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The direct limit is the set of sequences which are eventually zero.
    $endgroup$
    – Lord Shark the Unknown
    Dec 4 '18 at 7:40










  • $begingroup$
    Yes, sorry, that is right. But how about the topology?
    $endgroup$
    – CL.
    Dec 4 '18 at 7:41


















1












$begingroup$


If we let $X$ be direct limit of
$$Bbb R rightarrow Bbb R^2 rightarrow Bbb R^3 rightarrow cdots $$
where each arrow is inclusion to first coordinates.



What is a choice of $X$ (with a toplogy that coincides with the direct limit topology).



EDIT: I believe it is subspace of $Bbb R^{Bbb N}$, sequence eventually $0$ taking values in $Bbb R$, with the product topology. I wonder if this is correct.






general-topology algebraic-topology







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The direct limit is the set of sequences which are eventually zero.
    $endgroup$
    – Lord Shark the Unknown
    Dec 4 '18 at 7:40










  • $begingroup$
    Yes, sorry, that is right. But how about the topology?
    $endgroup$
    – CL.
    Dec 4 '18 at 7:41
















1












1








1


1



$begingroup$


If we let $X$ be direct limit of
$$Bbb R rightarrow Bbb R^2 rightarrow Bbb R^3 rightarrow cdots $$
where each arrow is inclusion to first coordinates.



What is a choice of $X$ (with a toplogy that coincides with the direct limit topology).



EDIT: I believe it is subspace of $Bbb R^{Bbb N}$, sequence eventually $0$ taking values in $Bbb R$, with the product topology. I wonder if this is correct.






general-topology algebraic-topology







share|cite|improve this question











$endgroup$




If we let $X$ be direct limit of
$$Bbb R rightarrow Bbb R^2 rightarrow Bbb R^3 rightarrow cdots $$
where each arrow is inclusion to first coordinates.



What is a choice of $X$ (with a toplogy that coincides with the direct limit topology).



EDIT: I believe it is subspace of $Bbb R^{Bbb N}$, sequence eventually $0$ taking values in $Bbb R$, with the product topology. I wonder if this is correct.






general-topology algebraic-topology




general-topology algebraic-topology






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 9:18







CL.

















asked Dec 4 '18 at 7:39









CL.CL.

2,2012825




2,2012825








  • 2




    $begingroup$
    The direct limit is the set of sequences which are eventually zero.
    $endgroup$
    – Lord Shark the Unknown
    Dec 4 '18 at 7:40










  • $begingroup$
    Yes, sorry, that is right. But how about the topology?
    $endgroup$
    – CL.
    Dec 4 '18 at 7:41
















  • 2




    $begingroup$
    The direct limit is the set of sequences which are eventually zero.
    $endgroup$
    – Lord Shark the Unknown
    Dec 4 '18 at 7:40










  • $begingroup$
    Yes, sorry, that is right. But how about the topology?
    $endgroup$
    – CL.
    Dec 4 '18 at 7:41










2




2




$begingroup$
The direct limit is the set of sequences which are eventually zero.
$endgroup$
– Lord Shark the Unknown
Dec 4 '18 at 7:40




$begingroup$
The direct limit is the set of sequences which are eventually zero.
$endgroup$
– Lord Shark the Unknown
Dec 4 '18 at 7:40












$begingroup$
Yes, sorry, that is right. But how about the topology?
$endgroup$
– CL.
Dec 4 '18 at 7:41






$begingroup$
Yes, sorry, that is right. But how about the topology?
$endgroup$
– CL.
Dec 4 '18 at 7:41












1 Answer
1






active

oldest

votes


















1












$begingroup$

As mentioned in comments the direct limit is



$$L={(x_1,x_2,x_3,ldots)inmathbb{R}^{mathbb{N}} | x_i=0text{ eventually}}$$



The direct limit topology on $L$ is the topology coherent with ${mathbb{R}^n | ninmathbb{N}}$ treated as subspaces of $L$ via obvious inclusions.



There's another natural choice for topology on $L$: the one induced by the Euclidean norm (note that the Euclidean norm is well defined on sequences that are eventually $0$).



I remember my surprise when I learned that these two are not the same. The example is as follows: let $v_i$ be a vector with $1/i$ on the $i$-th position and $0$ elsewhere. Then $E={e_i}_{i=1}^infty$ is closed in the direct limit (because $Ecapmathbb{R}^n$ is finite for any $n$) but not closed in the Euclidean norm (because $lVert v_irVert=1/i$ and so the sequence converges to $0$).



In fact it can be shown that the direct limit is not metrizable. In particular this also means that $L$ is not a subspace (even up to homeomorphism) of the product topology (at least a product of metrizable spaces) because countable product of metrizable spaces is metrizable.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I wonder what happens if you use the topology where the closed sets are either the full space, or else (images of) closed subsets of some $mathbb{R}^n$. It seems like that would be the smallest topology that makes each $mathbb{R}^n$ have the subspace topology be the Euclidean topology.
    $endgroup$
    – Daniel Schepler
    Dec 6 '18 at 20:48











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3025251%2fwhat-is-the-topology-on-the-direct-limit-of-bbb-ri-rightarrow-bbb-rj%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









1












$begingroup$

As mentioned in comments the direct limit is



$$L={(x_1,x_2,x_3,ldots)inmathbb{R}^{mathbb{N}} | x_i=0text{ eventually}}$$



The direct limit topology on $L$ is the topology coherent with ${mathbb{R}^n | ninmathbb{N}}$ treated as subspaces of $L$ via obvious inclusions.



There's another natural choice for topology on $L$: the one induced by the Euclidean norm (note that the Euclidean norm is well defined on sequences that are eventually $0$).



I remember my surprise when I learned that these two are not the same. The example is as follows: let $v_i$ be a vector with $1/i$ on the $i$-th position and $0$ elsewhere. Then $E={e_i}_{i=1}^infty$ is closed in the direct limit (because $Ecapmathbb{R}^n$ is finite for any $n$) but not closed in the Euclidean norm (because $lVert v_irVert=1/i$ and so the sequence converges to $0$).



In fact it can be shown that the direct limit is not metrizable. In particular this also means that $L$ is not a subspace (even up to homeomorphism) of the product topology (at least a product of metrizable spaces) because countable product of metrizable spaces is metrizable.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I wonder what happens if you use the topology where the closed sets are either the full space, or else (images of) closed subsets of some $mathbb{R}^n$. It seems like that would be the smallest topology that makes each $mathbb{R}^n$ have the subspace topology be the Euclidean topology.
    $endgroup$
    – Daniel Schepler
    Dec 6 '18 at 20:48
















1












$begingroup$

As mentioned in comments the direct limit is



$$L={(x_1,x_2,x_3,ldots)inmathbb{R}^{mathbb{N}} | x_i=0text{ eventually}}$$



The direct limit topology on $L$ is the topology coherent with ${mathbb{R}^n | ninmathbb{N}}$ treated as subspaces of $L$ via obvious inclusions.



There's another natural choice for topology on $L$: the one induced by the Euclidean norm (note that the Euclidean norm is well defined on sequences that are eventually $0$).



I remember my surprise when I learned that these two are not the same. The example is as follows: let $v_i$ be a vector with $1/i$ on the $i$-th position and $0$ elsewhere. Then $E={e_i}_{i=1}^infty$ is closed in the direct limit (because $Ecapmathbb{R}^n$ is finite for any $n$) but not closed in the Euclidean norm (because $lVert v_irVert=1/i$ and so the sequence converges to $0$).



In fact it can be shown that the direct limit is not metrizable. In particular this also means that $L$ is not a subspace (even up to homeomorphism) of the product topology (at least a product of metrizable spaces) because countable product of metrizable spaces is metrizable.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I wonder what happens if you use the topology where the closed sets are either the full space, or else (images of) closed subsets of some $mathbb{R}^n$. It seems like that would be the smallest topology that makes each $mathbb{R}^n$ have the subspace topology be the Euclidean topology.
    $endgroup$
    – Daniel Schepler
    Dec 6 '18 at 20:48














1












1








1





$begingroup$

As mentioned in comments the direct limit is



$$L={(x_1,x_2,x_3,ldots)inmathbb{R}^{mathbb{N}} | x_i=0text{ eventually}}$$



The direct limit topology on $L$ is the topology coherent with ${mathbb{R}^n | ninmathbb{N}}$ treated as subspaces of $L$ via obvious inclusions.



There's another natural choice for topology on $L$: the one induced by the Euclidean norm (note that the Euclidean norm is well defined on sequences that are eventually $0$).



I remember my surprise when I learned that these two are not the same. The example is as follows: let $v_i$ be a vector with $1/i$ on the $i$-th position and $0$ elsewhere. Then $E={e_i}_{i=1}^infty$ is closed in the direct limit (because $Ecapmathbb{R}^n$ is finite for any $n$) but not closed in the Euclidean norm (because $lVert v_irVert=1/i$ and so the sequence converges to $0$).



In fact it can be shown that the direct limit is not metrizable. In particular this also means that $L$ is not a subspace (even up to homeomorphism) of the product topology (at least a product of metrizable spaces) because countable product of metrizable spaces is metrizable.






share|cite|improve this answer











$endgroup$



As mentioned in comments the direct limit is



$$L={(x_1,x_2,x_3,ldots)inmathbb{R}^{mathbb{N}} | x_i=0text{ eventually}}$$



The direct limit topology on $L$ is the topology coherent with ${mathbb{R}^n | ninmathbb{N}}$ treated as subspaces of $L$ via obvious inclusions.



There's another natural choice for topology on $L$: the one induced by the Euclidean norm (note that the Euclidean norm is well defined on sequences that are eventually $0$).



I remember my surprise when I learned that these two are not the same. The example is as follows: let $v_i$ be a vector with $1/i$ on the $i$-th position and $0$ elsewhere. Then $E={e_i}_{i=1}^infty$ is closed in the direct limit (because $Ecapmathbb{R}^n$ is finite for any $n$) but not closed in the Euclidean norm (because $lVert v_irVert=1/i$ and so the sequence converges to $0$).



In fact it can be shown that the direct limit is not metrizable. In particular this also means that $L$ is not a subspace (even up to homeomorphism) of the product topology (at least a product of metrizable spaces) because countable product of metrizable spaces is metrizable.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 6 '18 at 20:37

























answered Dec 6 '18 at 11:34









freakishfreakish

11.9k1629




11.9k1629












  • $begingroup$
    I wonder what happens if you use the topology where the closed sets are either the full space, or else (images of) closed subsets of some $mathbb{R}^n$. It seems like that would be the smallest topology that makes each $mathbb{R}^n$ have the subspace topology be the Euclidean topology.
    $endgroup$
    – Daniel Schepler
    Dec 6 '18 at 20:48


















  • $begingroup$
    I wonder what happens if you use the topology where the closed sets are either the full space, or else (images of) closed subsets of some $mathbb{R}^n$. It seems like that would be the smallest topology that makes each $mathbb{R}^n$ have the subspace topology be the Euclidean topology.
    $endgroup$
    – Daniel Schepler
    Dec 6 '18 at 20:48
















$begingroup$
I wonder what happens if you use the topology where the closed sets are either the full space, or else (images of) closed subsets of some $mathbb{R}^n$. It seems like that would be the smallest topology that makes each $mathbb{R}^n$ have the subspace topology be the Euclidean topology.
$endgroup$
– Daniel Schepler
Dec 6 '18 at 20:48




$begingroup$
I wonder what happens if you use the topology where the closed sets are either the full space, or else (images of) closed subsets of some $mathbb{R}^n$. It seems like that would be the smallest topology that makes each $mathbb{R}^n$ have the subspace topology be the Euclidean topology.
$endgroup$
– Daniel Schepler
Dec 6 '18 at 20:48


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3025251%2fwhat-is-the-topology-on-the-direct-limit-of-bbb-ri-rightarrow-bbb-rj%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Bundesstraße 106

Image
Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
Read more

Verónica Boquete

Image
Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
Read more

Ida-Boy-Ed-Garten

Image
Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
Read more