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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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 ...
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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 ...
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1 vote
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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 ...
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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 ...
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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 ...
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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 ...
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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 ...
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Existence of continuous order-preserving function

Let $X$ be a countable simple ordered set. I am trying to prove that there exists on X a real, order-preserving function, continuous in the order topology. I construct by induction an order-...
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A totally ordered ring with its order topology is not a topological ring.

There is a theorem that, if $R$ is a totally ordered ring which is also a division ring, then $R$ is a topological division ring with respect to the order topology on $R$. I am certain that, in ...
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Linearly ordered X is regular

Prove that every linearly ordered space X is regular. Can anyone please help me with this proof? I started with letting $x$ belong to $x$ and take a nbhd $U=(a,b)$ of $x$ and then taking $A=(a,x)$, ...
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Is the Function a Homeomorphism?

This an example from a book I am reading. The author claims that there are two ways to see that the bijection $f : (-1,1) \rightarrow \mathbb{R}$ defined by $f(x) = \frac{x}{1-x^2}$ is a homeomorphism....
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Complete ordered space

In their paper A Method for Constructing Ordered Continua , Hart and van Mill give the following definition of ordered continuum: An ordered continuum is a compact, connected linearly ordered ...
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Is $[0,1]^2$ with the dictionary order topology separable? First countable?

I have question about order square $I^2_0$ which is $[0,1]\times[0,1]$ with dictionary order this is what I know about it: compact Hausdorff, hence normal and regular as well. and also it is Lindelof....
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Homeomorphism between evenly spaced integer topology and the rationals

The evenly spaced integer topology is countable, metrizable, and has no isolated points, and hence is homeomorphic to the rationals with the order topology. But what is an explicit construction for ...
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Continuous surjective function from $[0,1] \rightarrow$ lexicographical order

This is a follow up question to the question I asked here: The range of a continuous function on the order topology is convex Let $(X,\mathcal{T})$ denote the unit square with the lexicographical ...
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The range of a continuous function on the order topology is convex

Let $(W, \leq)$ be a linear order, and let $f : [0, 1] \rightarrow W$ be a continuous function (where [0, 1] has the usual topology and W has its order topology). Show that the range of f is convex. ...
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I'm conflicted: If we consider the set $\{x\} \times (0,1)$, for $x \in [0,1]$, these are open in the unit square, uncountable and disjoint, but what about open intervals of the form ((a,b), (c,d)) ...