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Questions tagged [order-topology]

Ordered sets have a natural topology generated by the open intervals. This tag is meant for questions about this topology.

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A second countable GO-space can be homeomorphically embedded in a second countable LOTS

A GO-space (generalized ordered space) is a topological space $X$ with topology $\tau$ together with a linear order $<$ such that $X$ is $T_1$ and every point has a local base of $\tau$-open ...
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Equivalent definitions for GO-spaces (generalized ordered spaces)

GO-spaces (= generalized ordered spaces) are subspaces of LOTS (linearly ordered topological spaces). There are several definitions in use and I am wondering how to show the equivalence between them. ...
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Why isn't every subspace of $\mathbb R$ a linearly ordered topological space (LOTS)?

(Asking the rhetorical question to provide a reference to begin building up GO-space theory on π-Base.) The real numbers are a linearlly ordered topological space, generated by the subbasis of rays $\{...
Steven Clontz's user avatar
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Open Set based on Function Image in Order Topology

I was reading through this stackexchange post, and the user Pete L. Clark stated that, given a continuous function $f:X\to X$, for $X$ being completely linearly ordered with the order topology, the ...
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About compactness of simply ordered set with least upper bound property (Theorem 27.1 of Munkres)

I'm studying topology by Munkres. Theorem 27.1: Let $X$ be a simply ordered set having the least upper bound property. In the order topology, each closed interval in $X$ is compact. What makes me ...
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In an order topology, are connected sets convex, and are they intervals?

Problem: $X$ is an ordered set with order topology. Is it true that (1) $A\subseteq X$ is connected $\implies$ $A$ is convex (2) $A\subseteq X$ is connected $\implies$ $A$ is an interval ? (Here ...
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Order topology in $\omega^\alpha + 1$ and dynamical systems

Let $f \colon \omega^\alpha + 1 \to \omega^\alpha + 1$, where $\alpha \geq 1$ a continous funciton such that, exists $w \in \omega^\alpha + 1$ with $\mathcal{O}_f(w)$ dense in $\omega^\alpha + 1$. The ...
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A set with a topology and a partial ordering question?

Suppose I have a topological space whose underlying set $X$ has a partial ordering on it. A prior this partial ordering has no relation to the topology. Now, take any meet semi-lattice $I \subseteq X,$...
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Basis of order topology on $\mathbb{R} × \mathbb{R} $

On James r Monkers Topology book in the second chapter the section of order topology, in the example 2 of order topology on $\mathbb{R} × \mathbb{R}$ he defined the order topology the one whose the ...
Ifielmodes's user avatar
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Does the locally compactness imply the least upper bound property?

When we say that a totally ordered set $X$ has the least upper bound property, we mean that every upper bounded subset of $X$ has a least upper bound. I know that a totally ordered set with the least ...
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Ordinal embedding in $\mathbb{R}$, and the use of $\mathsf{AC}$

Prove that if an ordinal $\alpha$ can be embedded (in an order preserving manner) into $\Bbb{R}$, then $\alpha$ is countable. I managed to prove this using $\mathsf{AC}$ (countable choice is ...
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Exercise 3, Section 31 of Munkres’ Topology

Show that every order topology is regular. My attempt: Approach(1): Let $X$ be an ordered set equipped with $\mathcal{T}_o$ order topology. Let $\{ x\}$ be a singleton set in $X$. Then $(x,+\infty)\...
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closed set and the supremum in order topology

Suppose $\alpha$ is an ordinal endowed with the order topology, i.e. its basic open sets are generalized open intervals. Given $C \subseteq \alpha$, I want to show (1) implies (2) : (1) For all ...
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Find an element which is not in the topology.

The order topology is defined on $\{0,1\} \times \mathbb{N}$. Need to find a non-singleton set which is not an element of order topology. I Choose $A=\{(0,10), (1,1)\}$. How to prove this? Is my ...
Madhan Kumar's user avatar
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1 answer
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Prove: If $A$ and $B$ are closed subsets of $[0,\Omega]$ then at least $A$ or $B$ is bounded

As usual, I am self studying topology and my knowledge of ordinals is meagre. Have done some research on it. Theorem 5.1 Any countable subset of $[0,\Omega)$ is bounded above. (This exercise requires ...
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Doubt related to example 4 of section 14 chapter 2 of Munkres Topology book

The example goes as follows The set X = {1,2} × $\mathbb{Z^+}$ in the dictionary order is another example of an ordered set with a smallest element. Denoting $1×n$ by $a_n$ and $2×n$ by $b_n$ We can ...
Sam Hoffman's user avatar
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Is every ordered field a topological field?

Let $F$ be an ordered field and give $F$ the order-topology. Then is $F$ a topological field (that is, are the operations \begin{equation} \begin{split} +&:F\times F\to F\\ -&:F\to F\\ \times&...
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Limit points and order topology

Let $X = [0,1) \cup \{2\} \subset \mathbb{R}$. Consider two topologies on X. The subspace topology on X induced by the standard topology on $\mathbb{R}$, and the order topology on X induced by the ...
appy's user avatar
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What is the topology defined on $[0, \omega_1)$ it and how do we define its basis?

The set $[0, \omega_1)$ is a countably compact topological space. How do we prove it to be a topological space. How does its basis elements look like?
Mir Aaliya's user avatar
3 votes
3 answers
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If f: $\mathbb{R}_{l} \rightarrow S_{\Omega}$ is continuous, then f is not injective.

If f: $\mathbb{R}_{l} \rightarrow S_{\Omega}$ is continuous, then f is not injective. I've been trying to solve this problem for a few days, but I haven't been able to see how can I do it. First, $\...
george45's user avatar
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2 answers
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Supremum in a dictionary ordered set.

Take double arrow space $[0,1]\times \{1,2\}$ Where topology is dictionary topology defined by linear order as $(a,b)<(c,d)$ iff $a<c$; or ;$a=c \;\&\; b<d$ open sets are in the form $$((...
Jale'de jaled's user avatar
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3 answers
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Why is the subspace topology same as the Order topology?

Hi i am reading Topology by Munkres and there in the proof of theorem 27.1 it is written that Given $a<b$, let $\mathcal A$ be a covering of $[a,b]$ by sets open in $[a,b]$ in the subspace ...
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Separation axioms of topology

Let $X=Z_+×Z_2, Y=Z_2×Z_+$ with the dictionary order topologies, are $X$ and $Y$,$T_1$,$T_2$ and $T_3$? $Z_+=\{1,2,3,...\}$ $Z_2 =\{0,1\}$ The open set on the dictionary ordering will be $I=\{x×y∈\...
Ahmad qasem's user avatar
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0 answers
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Conclusion about corollary V.6.1 in Schaeffer & Wolff (Topological Vector Spaces 2nd edition)

Currently, I am stuck with following the final conclusion in the proof of corollary V.6.1 in Schaeffer & Wolff's Topological Vector Spaces, 2nd edition, pp. 230-231. Probably it is trivial and I ...
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1 answer
503 views

Questions about completely normal spaces.

I'm trying to solve the next problem: A topological space $(X,\tau)$ is called completely normal if, and only if, every subspace is normal. Prove that the following conditions are equivalent: a) $X$ ...
Dendrilo's user avatar
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4 answers
675 views

Munkres order topology difference between definition of simple order using $<$ instead of $\leq$

Munkres in his Topology 2ed section 14 p. 84 defines a simply ordered set as in this math.stack question here definition of simply ordered set by Munkres I recalled that I have come across the ...
simpl0s's user avatar
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What is definition of order topology?

I am doing self study of topology, the book I am following has used term order topology in problems and few theorems. What is order topology? Kindly give explanation with some examples that will be ...
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3 answers
374 views

How is the closed ordinal space compact hausdorff?

So the original question in the paper was whether every compact hausdorff space automatically metrizable? I found out that closed ordinal space with order topology is a counter example. Since the ...
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1 answer
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Open sets in the order topology

I'm having trouble understanding the order topology. If we consider $R_{\ge 0} \times R_{\ge 0}$ given the order topology coming from the lexicographic order, my understanding is that the sets $[2,3] \...
selfstudy's user avatar
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1 answer
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Linear continuum of $I\times I $ under subspace topology of $\mathbb{R}^2$ with dictionary topology on it, where $I=[a,b]$.

Let $I\times I$ be subspace of space $\mathbb{R}^2$ with dictionary order, where $I=[a,b]$. What can you say about the linear continuum of $I\times I$ with the subspace topology. { I've proved that ...
TBourne's user avatar
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1 answer
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If Y is a subset of X, is always true that then Y inherits (always) a total order from X?

If Y is a subset of X, then Y inherits a total order from X. The set Y therefore has an order topology, the induced order topology What does it mean "inherits" ? Is this a a Kripke model M ...
Oscar's user avatar
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0 answers
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What are the limit points of $X \times \mathbb{N}$?

Let $X=\{0,1,2 \}$ with natural order topology and give $\mathbb{N}$ its natural order topology. Consider $X \times \mathbb{N}$ with the dictionary order topology. What are the limit points of $X \...
MAS's user avatar
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Help me to visualise limit points of $S$. [duplicate]

Consider the ordered square $I^2,$ then the set $[0,1]\times [0,1]$ with the dictionary order. Let the general element of $I^2$ be denoted by $x\times y,$ where $x,y\in[0,1].$ The closure of the ...
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2 answers
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Order Topology on a Preorder

While looking at the definition of the order topology defined on a total order (https://en.wikipedia.org/wiki/Order_topology), I realized I needed a generalization to preorders. So ultimately the ...
user512716's user avatar
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1 answer
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How is this definition of closedness compatible with the order topology?

Let $\kappa$ be a limit ordinal. Taken from the definition of a closed unbounded set, we say a subset $C\subseteq\kappa$ is closed in $\kappa$ if and only if $\sup(C\cap\alpha)=\alpha<\kappa\...
Lilalas's user avatar
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If $f\colon X\to Y$ is continuous and $X$ is compact Hausdorff and connected, $Y$ is given an order topology. Is $f$ necessarily onto?

I want to ask you for following statement $f$ is continuous from $X$ to $Y$, where $X$ is compact Hausdorff, connected and $Y$ is ordered set in order topology, then $f$ is onto? My attempt For ...
fivestar's user avatar
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8 votes
2 answers
372 views

Can the long line be embedded in the ordered plane?

It is a well known result that the long line (namely, the topological space $S_\omega \times [0, 1)$ in the order topology, where $S_\omega$ is the minimal uncountable well-ordered set) cannot be ...
Niki Di Giano's user avatar
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1 answer
158 views

$\omega_1$ is a limit point of the subset $[0,\omega_1)$

In the Wiki of order topology, I encounter the following statement. $\omega_1$ is a limit point of the subset $[0,\omega_1)$ even though no sequence of elements in $[0,\omega_1)$ has the element $\...
Idonknow's user avatar
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2 votes
2 answers
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Right continuous functions when considered with lower limit topology

Exercise 17.7 (a) from James Munker's Topology says: Suppose that $f : \mathbb R \rightarrow \mathbb R$ is "continuous from the right", i.e. $$ \forall_{a \in \mathbb R} \lim_{x \rightarrow a^+} f(x)...
enedil's user avatar
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1 answer
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Closed subsets of regular cardinals

Let $\kappa$ be a regular cardinal, equipped with order topology, and $C\subseteq\kappa$ a closed unbounded subset. If $\alpha<\kappa$ is another regular cardinal and a limit point of $C$, is the ...
Ajeje's user avatar
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1 answer
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Munkres Example 16.3

The example says: Let $I=[0,1]$. The dictionary order on $ I \times I$ is just the restriction to $I\times I$ of the dictionary order on the plane $\mathbb{R} \times \mathbb{R}$. However, the ...
Amphiaraos's user avatar
1 vote
1 answer
200 views

The lub of the usual and co-countable topology on $\mathbb R$

This exercise comes from Steven A. Gaal's Point Set Topology(1964,Academic Press), Page 37, exercise 4. It is Let $\mathscr T$ be the least upper bound of the usual topology and of the topology ...
W.Leywon's user avatar
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6 votes
2 answers
229 views

Open sets on ordered topology in $\mathbb{R\times R}$

I am having some difficulty grasping the concept of an ordered topology in $\mathbb R \times\mathbb R $. The definition I was given is that this is the topology where $(a,b) < (c,d)$ if $a<c$ or ...
Sergei's user avatar
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2 answers
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Separating Closed Sets in a Well-Ordered Set in the Order Topology

Theorem 32.4 from Munkres' Topology book: Every well-ordered set $X$ is normal in the order topology This is something a continuation of this post, where, evidently, I forgot what the definition ...
user193319's user avatar
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3 votes
2 answers
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Prob. 7 (a), Sec. 24, in Munkres' TOPOLOGY, 2nd ed: Any order preserving surjective map between ordered sets is a homeomorphism

Here is Prob. 7 (a), Sec. 24, in the book Topology by James R. Munkres, 2nd edition: Let $X$ and $Y$ be ordered sets in the order topology. Show that if $f \colon X \to Y$ is order preserving and ...
Saaqib Mahmood's user avatar
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1 answer
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Is the interval $((a,b), (a,d))$ is open in $\mathbb{R}\times \mathbb{R}$ which is equipped with the dictionary order topology

Is the interval $((a,b), (a,d))$ is open in $\mathbb{R}\times \mathbb{R}$ which is equipped with the dictionary order topology. For me, since $(a,b)$ and $(a,c)$ are points in $\mathbb{R^2}$ with $(...
Our's user avatar
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1 vote
1 answer
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questions about subspace topology, order topology and convex subset

I am learning Munkres' s topology, and have some questions about subspace topology, order topology and convex subset. I have read 3 similar questions here [ref]:One question on subspace topology and ...
ZJX's user avatar
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1 answer
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Show that R^2 with the dictionary order topology is homeomorphic to Rd * R.

I am currently working topology and I can not prove this exercise. Let $Rd$ denote the set $R$ with the discrete topology. Show that $R^2$ with the dictionary order topology is homeomorphic to $Rd \...
M. Umut Kazancıoğlu's user avatar
2 votes
1 answer
618 views

The closure of $(a;b)$ in an order topology [duplicate]

Consider a linearly ordered set $(X; \prec)$ with its order topology. Show that closure of $(a;b)$ is a subset of $[a;b]$. Under what conditions does equality hold? Give an example of a strict ...
M. Umut Kazancıoğlu's user avatar
2 votes
1 answer
1k views

$X$ and $Y$ are ordered set, $f:X\rightarrow Y$ is order preserving and surjective, then $f$ is a homeomorphism.

Let $X$ and $Y$ are ordered set in the order topology. Show that if $f:X\rightarrow Y$ is order preserving and surjective, then $f$ is a homeomorphism. My attempt: Simply $f$ is bijective (take $x&...
topology_001's user avatar