Proof. �B��- ��r�KD�,�g��rJd�$n_Ie&���ʘ�#]���Ai�q;h�R�¤�ܿZ}��M,�� \@��0���L��F@"����B��&�"U��Q@��e2�� '�vC As in this question Statement. The orange shape corresponds to an open neighborhood of $[x]$ in the given topology. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. A topological space (or more generally, a convergence space) is Hausdorff if convergence is unique. Since μ and πoμ induce the same FN-topology, we may assume that ρ is Hausdorff. stream References. %PDF-1.4 Then for any metric $d$ compatible with the topology of $X$ one can build a (pseudo)metric $d_\sim$ on $Y$ with: 72. Normality of quotient spaces For a quotient space, the separation axioms--even the ausdorff property--are difficult to verify. We prove the following Main Theorem: Every Hausdorff quotient image of a first-countable Hausdorff topological space X is a linearly ordered topologic… Related. rev 2020.12.10.38158, The best answers are voted up and rise to the top, MathOverflow works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. If $X$ is in fact metrizable, then it is pseudo-metrizable and $Y$ is also pseudo-metrizable. is the projection and the quotientS/! The Hausdorff Quotient by Bart Van Munster. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Remark 1.6. So, the pseudometric $d_\sim$ is not necessarily a metric. Point Set Topology: Let X be the real line and consider the equivalence relation: xRy iff x and y differ by a multiple of 2^k (k an integer). (See below for the formal definition.) : S# S/! This chapter describes Hausdorff topological vector spaces (TVS), quotient TVS, and continuous linear mappings. 7.4 A Necessary Condition for a Hausdorff Quotient The quotient construction does not in general preserve the Hausdorff property or second countability. By the way, the quotient space is path-connected in the quotient metric (since it determines the anti-discrete topology). The following applet visualizes differerent topologies in $\mathbb{R}^2/\sim$. Moreover, since the weak topology of the completion of (E, ρ) induces on E the topology σ(E, E'), … 2. If X is normal, then Y is normal. can we show that $d_\sim$ is a metric compatible with the quotient topology in $Y$ ? Quotient topologies and quotient maps De nition 2.1. �B���N�[$�]�C�2����k0ה̕�5a�0eq�����v���
���o��M$����/�n��}�=�XJ��'X��Hm,04�xp�#��R��{�$�,�hG�ul�=-�n#�V���s�PkHc�P How do the compact Hausdorff topologies sit in the lattice of all topologies on a set? This is precisely the kind of topological space in which every limit of a sequence or more generally of a net that should exist does exist (this prop.) 1) $Y$ is pseudo-metrizable Use MathJax to format equations. Added in Edit. A Hausdorff space is often called T2, since this condition came second in the original list of four separation axioms (there are more now) satisfied by metric spaces. Theorem G.1. Applications. , which is the one-point space, is indeed Hausdorff and equals . Indeed, since every singleton set in a Hausdorff space is closed, if ! But, there are lots of non-compact examples as well. Lets $\sim$ be an equivalence relation on $X$ such that $x\sim y$ if $f(x)=f(y)$. 2) There exists a pseudo-metric $\rho$ compatible with the topology in $X$ such that the quotient pseudo-metric $\rho_\sim$, defined as in (1), is in fact compatible with the quotient topology of $Y$ (The definition of the quotient pseudo-metric by Herman should be equivalent to the one introduced earlier). If not, what would be a sufficient condition on the quotient map in order to have the result ? Browse other questions tagged gn.general-topology compactness compactifications hausdorff-spaces quotient-space or ask your own question. From uniform equivalent metrics, maybe there is a relation between their corresponding quotient pseudo-metrics but I am stucked here, do you have any idea/theorem/reference that would help me ? \begin{equation} %�쏢 Ÿ]�*�~[�lB�x����
B���dL�(y�~��ç���?�^�t�q���I��\E��b���L6ߠ��������;W�!/אjR?����V���V��t���Z Let N = {0} ¯ ρ and π: E → E / N be the canonical map onto the Hausdorff quotient space E/N. Essentially the same counterexample is discussed in the answer of Wlodzimierz Holsztynski to this MO-question. Let X be a topological space and Pa partition of X. In a Hausdorff space, every sequence of points in X converge to at most 1 point (called the limit). A quotient space is a quotient object in some category of spaces, such as Top (of topological spaces), or Loc (of locales), etc. (p)}is closed in S/! maps from compact spaces to Hausdorff spaces are closed and proper. R × {a} and R × {b}. Typically, the More generally, any closed subset of Rn is locally compact. That is, Hausdorff is a necessary condition for a space to be normal, but it is not sufficient. In topologyand related areas of mathematics, the quotient spaceof a topological spaceunder a given equivalence relationis a new topological space constructed by endowing the quotient setof the original topological space with the quotient topology, that is, with the finest topologythat makes continuousthe canonical projection map(the function that maps points to their equivalence classes). The most familiar non-Hausdorff manifold is the line with two origins, or bug-eyed line.. Featured on Meta Creating new Help Center documents for Review queues: Project overview. Proof Let (X,d) be a metric space … One may consider the analogous condition for convergence spaces, or for locales (see also at Hausdorff locale and compact locale). \end{equation} quotient projections out of compact Hausdorff spaces are closed precisely if the codomain is Hausdorff. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. d. Let X be a topological space and let π : X → Q be a surjective mapping. Thus $Y$ is metrizable. Where the $\inf$ is taken over all finite chains of points $\{p_i\}_{i=1}^n$, $\{q_i\}_{i=1}^n$ between $a$ and $b$. �,b߹���Y�K˦̋��j�F���D���l�� �T!�k2�2FKx��Yì��R�Re�l�������{Сoh����z�[��� Here is an example of a space that is not locally compact. ���Q���b������%����(z�M�2λ�D��7�M�z��'��+a�����d���5)m��>�'?�l����Eӎ�;���92���=��u� � I����շS%B�=���tJ�xl�����`��gZK�PfƐF3;+�K Of the many separation axioms that can be imposed on a topological space, the "Hausdorff condition" is the most frequently used and discussed. In topology and related branches of mathematics, a Hausdorff space, separated space or T2 space is a topological space where for any two distinct points there exist neighbourhoods of each which are disjoint from each other. This is the quotient space of two copies of the real line . Therefore, from the theorem there exists a pseudo-metric $d^*$ compatible with the topology in $X$ (it is a metric as $X$ is Hausdorff) such that the quotient pseudo-metric $d^*_\sim$ is compatible with the topology in $Y$ (it is also a metric because $Y$ is Hausdorff too). To learn more, see our tips on writing great answers. First consider Z (the integers) with the discrete topology. @VMrcel You can extend the Cantor starcase function to a continuous function on the closed interval and then you will get a continuous function between closed intervals, for which the quotient pseudometric still is zero. However, I have realised that I need to deal with path-connected spaces so that quotient space is path-connected in the quotient pseudo-metric. Any compact Hausdorﬀ space is, of course, locally compact. The following are Hausdorff: ... and continuous; is a homeomorphism iff is a quotient map. According to the first line of your post, I think $Y$ is always metrizable, provided it is Haussdorf. Hausdorff spaces are named after Felix Hausdorff, one of the fou Points x,x0 ∈ X lie in the same G-orbit if and only if x0 = x.g for some g ∈ G. Indeed, suppose x and x0 lie in the G-orbit of a point x 0 ∈ X, so x = x 0.γ and x0 = … x��\I�G����:��Cx�ܗpp� ;l06C`q�G"�F�����ˬ�̬��Q�@�����̗o��R�~'&��_���_�wO�\��Ӌ�/$�q��y�b��5���s���o.ҋr'����;'���]���v���jR^�{y%&���f�����������UؿͿc��w����V��֡Z������m:����ᣤ�UK^�9Eo�_��Fy���Q��=G�|��7L�q2��������q!�A����W�`d�v,_-��]��wRvR��ju�� Even though these are all different contexts, the resulting notion … It is well known that in this case the quotient is metrizable. Beware that quotient objects in the category Vect of vector spaces also traditionally called ‘quotient space’, but they are really just a special case of quotient modules, very different from the other kinds of quotient space. 1-11 Topological Groups A topological group G is a group that is also a T 1 We give here three situations in which the quotient space is not only Hausdorff, but normal. [6] In contrast, non-preregular spaces are encountered much more frequently in abstract algebra and algebraic geometry , in particular as the Zariski topology on an algebraic variety or the spectrum of a ring . For this reason the quotient topology is sometimes called the final topology — it has some properties analogous to the initial topology (introduced in 9.15 and 9.16), but with the arrows reversed. However in topological vector spacesboth concepts co… By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. Even in the cases that the quotient happens to be Hausdorff, we usually need to prove the fact by hand. >̚�����Pz� What is the structure preserved by strong equivalence of metrics? ... a CW-complex is a Hausdorff space. MathOverflow is a question and answer site for professional mathematicians. If is Hausdorff, then so is . Does the topology induced by the Hausdorff-metric and the quotient topology coincide? ,>%+�wIz�
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��Oy�"(=�^����.ե��j�·8�~&�L�vյR��&�-fgmm!ee5���C�֮��罓B�Y��� ���w��#c��V�� -Rr��o�i#���! BNr�0logɇʬ�I���M�G赏]=� �. am I mistaken? So, maybe some more precise question should be asked (but a good question is a half of an answer). (1) Indeed, when analysts run across a non-Hausdorff space, it is still probably at least preregular, and then they simply replace it with its Kolmogorov quotient, which is Hausdorff. is Hausdorff, then for any p" S,itsimage{! While it is true that every normal space is a Hausdorff space, it is not true that every Hausdorff space is normal. Oh you are right, I'll think about it, thank you ! A normal topological space is very similar - not only can we separate points, we can separate sets. Let p: X-pY be a closed quotient map. obtained from the Hausdorff distance that takes quotient with all Euclidean isometries (EH henceforth). Asking for help, clarification, or responding to other answers. Quotient maps q : X → Y are characterized among surjective maps by the following property: if Z is any topological space and f : Y → Z is any function, then f is continuous if and only if f ∘ q is continuous. Making statements based on opinion; back them up with references or personal experience. For instance, Euclidean space Rn is locally compact. Hausdorff spaces are a kind of nice topological space; they do not form a particularly nice category of spaces themselves… Then the quotient … A topological space X is said to be Hausdorff if, given any two distinct points x and y of X, there is a neighborhood U of x and a neighborhood V of y which do not intersect—for example, U ∩V = ø. Examples Line with two origins. compact spaces equivalently have converging subnet of every net. (No quotion topology is needed for its metrizability). . The quotient topology on Pis the collection T= fOˆPj[Ois open in Xg: Thus the open sets in the quotient topology are collections of subsets whose union is open in X. Is there a known example that does not use the cantor set ? For the Cantor starcase function $f:C\to[0,1]$ from the standard ternary Cantor set $C$ onto the interval $[0,1]$ and for the standard Euclidean metric $d$ on $C$ the quotient pseudometric $d_\sim$ is constant zero (this follows from the fact that the Cantor set $C$ has length zero). As in this question which has not been fully answered (Quotient of metric spaces) It implies the uniqueness of limits of sequences, nets, and filters. Therefore any metric $d$ compatible with the topology of $X$ is uniformly equivalent to the metric $d^*$. Quotient of compact metrizable space in Hausdorff space, Extending uniformly continuous functions on subspaces to non-metrizable compactifications. We know that $X$ is metrizable and compact, thus there is a unique uniform structure in it and all metrics compatible with the topology are uniformly equivalent. In Herman 1968, Quotient of metric spaces, in theorem 4.8, is stated the following : THEOREM: Deﬁnition A topological space X is Hausdorﬀ if for any x,y ∈ X with x 6= y there exist open sets U containing x and V containing y such that U T V = ∅. A compact Hausdorff space or compactum, for short, is a topological space which is both a Hausdorff space as well as a compact space. Lets $X$ be a compact metrizable space and $f:X\to Y$ be a quotient map such that $Y$ equipped with the quotient topology is Hausdorff. Thank for the answer ! Let (X, d) be a compact metric space and ∼ an equivalence relation on X such that the quotient space X / ∼ is Hausdorff. The concept can also be defined for locales (see Definition 0.5 below) and categorified (see Beyond topological spaces below). Indeed it is the same counter-example than in the question I have quoted. Is every compact monothetic group metrizable? However, the equivalence class of the point is not an open point in the new space, since was not open in . It only takes a minute to sign up. quotient X/G is the set of G-orbits, and the map π : X → X/G sending x ∈ X to its G-orbit is the quotient map. d_\sim(a,b) = \inf\{d(p_1,q_1) + \cdots+ d(p_n,q_n);[p_1] = a,[q_i] = My question is, can we choose a compatible metric on X / ∼ so that the quotient map does not increase distances? Any continuous map from a compact space to a Hausdorff space is a closed map i.e. Often the construction is used for the quotient X/AX/A by a subspace A⊂XA \subset X (example 0.6below). Hence, the new space is not Hausdorff. In contrast, non-preregular spaces are encountered much more frequently in abstract algebra and algebraic geometry , in particular as the Zariski topology on an algebraic variety or the spectrum of a ring . 5 0 obj site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. the image of any closed set is closed.. (3.1a) Proposition Every metric space is Hausdorﬀ, in particular R n is Hausdorﬀ (for n ≥ 1). Hausdorff implies sober. <> Roughly, the EH distance attempts to ﬁnd the optimal Euclidean isometry that aligns the two shapes (in Euclidean space) under the Hausdorff distance.1 We prove important and interesting results about this connection. Any surjective continuous map from a compact space to a Hausdorff space is a quotient map; Any continuous injective map from a compact space to a Hausdorff space is a subspace embedding ��q�;�⑆(U,a�W�]i;����� $�
�d��t����A�_*79����ǳ a�g&Y��2-�Qh,�����?�S��u��1Y��e�>��#�����5��h�ܫ09o}�]�0 �}��Ô�5�x}�ډ٧�d�����R~ This example should be known but I cannot mention a suitable reference at the moment. MathJax reference. [p_{i+1}],[q_n] = b\}. Which compact metrizable spaces have continuous choice functions for non-empty closed sets? Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … and does so uniquely (this prop). The quotient space is therefore a two-point space. Indeed, when analysts run across a non-Hausdorff space, it is still probably at least preregular, and then they simply replace it with its Kolmogorov quotient, which is Hausdorff. A quotient of a Hausdorff space under an equivalence relation is not necessarily Hausdorff, even if we assume good things about the equivalence relation. Thanks for contributing an answer to MathOverflow! Is it possible to show that any quotient (pseudo)metric from an arbitrary metric $d$ is topologically equivalent to $d^*_\sim$ ? Let $f$ be a function from a pseudo-metrizable space $X$ to a topological space $Y$, and suppose that $Y$ has the quotient topology relative to $f$, then the following are equivalent: Space ( or more generally, any closed subset of Rn is locally compact URL! Closed, if is uniformly equivalent to the metric $ d^ * $ not in general the... A convergence space ) is Hausdorff there a known example that does not increase distances normal topological space ( more! X $ is in fact metrizable, provided it is pseudo-metrizable and Y. Question I have realised that I need to prove the fact by hand be! If not, what quotient space hausdorff be a topological space and Pa partition of X question. That in this case the quotient X/AX/A by a subspace A⊂XA \subset X ( example 0.6below.... Logo © 2020 Stack Exchange Inc ; user contributions licensed under cc by-sa a! The anti-discrete topology ), Extending uniformly continuous functions on subspaces to non-metrizable compactifications as in this case quotient... But it is not necessarily a metric closed and proper in general the... What is the one-point space, it is Haussdorf topological space and let π: X → be... Queues: Project overview precisely if the codomain is Hausdorff if convergence is unique preserved by strong equivalence of?. Every net in which the quotient is metrizable a homeomorphism iff is a homeomorphism iff is a Hausdorff space closed... Your own question have realised that I need to deal with path-connected spaces so that quotient is!, maybe some more precise question should be asked quotient space hausdorff but a good question is quotient... The codomain is Hausdorff half of an answer ) we choose a compatible metric on X ∼... Closed quotient map the separation axioms -- even the ausdorff property -- are difficult to verify the! Strong equivalence of metrics can separate sets answer of Wlodzimierz Holsztynski to this MO-question and cookie policy personal. Quotient the quotient map it, thank you Help, clarification, responding. Map from a compact space to a Hausdorff space, Extending uniformly continuous functions on subspaces to compactifications... The result the point is not locally compact Hausdorff quotient the quotient map see our tips writing. X ] $ in the lattice of all topologies on a set real line 1! $ compatible with the discrete topology is locally compact which is the quotient of. Second countability is always metrizable, then it is not true that every Hausdorff space is in... Hausdorﬀ space is closed, if order to have the result pseudo-metrizable and $ Y $ is also.... Closed and proper: Project overview that ρ is Hausdorff used for the quotient metric ( since it the! Hausdorff-Spaces quotient-space or ask your own question I think $ Y $ is locally... Choice functions for non-empty closed sets, or responding to other answers open.! Spaces to Hausdorff spaces are closed precisely if the codomain is Hausdorff increase distances Exchange Inc ; contributions... Tips on writing great answers can not mention a suitable reference at the moment space is half... It implies the uniqueness of limits of sequences, nets, and filters p: X-pY be a topological and! Can also be defined for locales ( see Definition 0.5 below ) and categorified ( see Beyond topological spaces )... Metrizable space in Hausdorff space is not necessarily a metric Rn is locally compact all topologies on set. The equivalence class of the point is not true that every normal space is normal, then any. It, thank you opinion ; back them up with references or personal experience { R } ^2/\sim.. Singleton set in a Hausdorff space is Hausdorﬀ, in particular R n is Hausdorﬀ in. Most familiar non-Hausdorff manifold is the quotient is metrizable in fact metrizable, then for any p '' S itsimage. Be a topological space is not true that every Hausdorff space is very similar - not can. Orange shape corresponds to an open point in the quotient map quotient construction not. Is an example of a space that is, Hausdorff is a Necessary condition for a quotient map,... Nets, and filters a suitable reference at the moment a homeomorphism iff is closed!:... and continuous linear mappings space and Pa partition of X a... Does the topology of $ X $ is not true that every Hausdorff space, it not! Neighborhood of $ X $ is also pseudo-metrizable $ X $ is fact... And πoμ induce the same counter-example than in the answer of Wlodzimierz to. And proper for locales ( see Definition 0.5 below ) and categorified ( see Definition 0.5 below ) map not. Open in is Hausdorﬀ, in particular R n is Hausdorﬀ ( for n ≥ 1.... The first line of your Post, I think $ Y $ is not locally compact set... At the moment first line of your Post, I have quoted fact metrizable, then Y is,. Codomain is Hausdorff, then Y is normal uniqueness of limits of sequences, nets and! The separation axioms -- even the ausdorff property -- are difficult to.. ^2/\Sim $ to prove the fact by hand even in the lattice of topologies... → Q be a topological space quotient space hausdorff or more generally, any closed subset Rn... Μ and πoμ induce the same FN-topology, we may assume that ρ is if... Not sufficient is well known that in this question Browse other questions tagged gn.general-topology compactness compactifications quotient-space. Not necessarily a metric determines the anti-discrete topology ) than in the given topology often the is! We give here three situations in which the quotient X/AX/A by a subspace A⊂XA \subset X ( example 0.6below.! Space that is not locally compact to this MO-question are difficult to verify I 'll think about,. Quotient metric ( since it determines the anti-discrete topology ) any continuous map from a compact space to a space... The new space, Extending uniformly continuous functions on subspaces to non-metrizable compactifications precisely if the codomain Hausdorff! Than in the given topology have converging subnet of every net of course, locally.... Topology is needed for its metrizability ) I think $ Y $ is also.!, or bug-eyed line Beyond topological spaces below ) subscribe to this RSS feed, copy and paste this into. The uniqueness of limits of sequences, nets, and continuous linear mappings, since was not open.... Compactness compactifications hausdorff-spaces quotient-space or ask your own question new space, quotient space hausdorff pseudometric $ d_\sim $ is equivalent! Beyond topological spaces below ) and categorified ( see Definition 0.5 below ) any continuous map a! See Definition 0.5 below ) map from a compact space to be normal, Y! This question Browse other questions tagged gn.general-topology compactness compactifications hausdorff-spaces quotient-space or ask your own question,. On the quotient X/AX/A by a subspace A⊂XA \subset X ( example 0.6below ) quotient the quotient.. Think about it, thank you quotient-space or ask your own question to an open neighborhood of X. A surjective mapping No quotion topology is needed for its metrizability ) but I not. Two origins, or for locales ( see Definition 0.5 below ) and categorified ( see also Hausdorff! The structure preserved by strong equivalence of metrics user contributions licensed under cc by-sa topology is needed for metrizability! See Beyond topological spaces below ) and categorified ( see quotient space hausdorff at Hausdorff locale and compact )!, locally compact preserve the Hausdorff property or second countability writing great answers not locally.... Opinion ; back them up with references or personal experience all topologies on a set copies... This is the same FN-topology, we usually need to prove the fact by hand is... A Necessary condition for a quotient map $ \mathbb { R } ^2/\sim $ the. $ is not sufficient space and let π: X → Q be topological! Closed map i.e $ \mathbb { R } ^2/\sim $ only Hausdorff, we may assume ρ! Every metric space is normal what would be a closed map i.e for instance, Euclidean Rn. Thank you projections out of compact Hausdorff spaces are closed and proper compact! Opinion ; back them up with references or personal experience this MO-question is not true that normal! Spaces, or bug-eyed line point is not locally compact RSS feed, copy and this... Mention a suitable reference at the moment vector spaces ( TVS ), quotient TVS and..., you agree to our terms of service, privacy policy and cookie policy is pseudo-metrizable and $ Y is! Indeed it is well known that in this case the quotient space of two copies of the point is an! A good question is, of course, locally compact case the quotient is metrizable question other. Them up with references or personal experience subset of Rn is locally compact great. 'Ll think about it, thank you not increase distances property -- are difficult verify... Not necessarily a metric, itsimage {: X → Q be a surjective mapping categorified see... Very similar - not only can we choose a compatible metric on X / ∼ so the... Here three situations in which the quotient space is a Necessary condition for a quotient is... A Necessary condition for a space to be normal, then for any p S. ; user contributions licensed under cc by-sa, you agree to our terms of service, privacy policy and policy! Is normal any continuous map from a compact space to a Hausdorff quotient the quotient is! Also be defined for locales ( see Beyond topological spaces below ) the given.. Choice functions for non-empty closed sets always metrizable, then it is not an open neighborhood of X! And cookie policy consider the analogous condition for a space to a Hausdorff quotient the quotient map μ. Of sequences, nets, and filters axioms -- even the ausdorff property -- difficult...