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2 votes
1 answer
169 views

When does a set of real numbers have a minimal element?

I Believe the answer is when the set is closed in the sense of standard topology (we exclude $\mathbb{R}$) itself. Examples: Point sets have a minimum, and they are closed. Closed interval also have ...
Cathartic Encephalopathy's user avatar
1 vote
1 answer
78 views

Complete iff Compact in Well-Ordered space

Let $T=(S, \leq, \tau)$ a well-ordered set equipped with order topology (defined here). Definition 1: $T$ is called complete iff every non-empty subset of $T$ has a greatest lower bound (inferior) and ...
Manuel Bonanno's user avatar
0 votes
0 answers
37 views

The arithmetic of first uncountable ordinal number

I think, I know the proof of 1+ω0 = ω0. (ω0 is countable ordinal s.t ω0=[N]). To prove this, I can define a function f: {-1,0,1,2,...}->{0,1,2,...} by f(x)=x+1. But if ω1 is first uncountable ...
Mathforest's user avatar
2 votes
1 answer
48 views

Intersection of a chain of closed sets

I was recently attempting proving a conjecture of mine about the existence of certain minimal nonempty closed sets in a topological space. I opted to use Zorn's Lemma, and the proof would go through ...
Nick F's user avatar
  • 1,249
3 votes
2 answers
324 views

Product of Countable Well-Ordered Set with $[0,1)$ is Homeomorphic to $[0,1)$

As part of a proof that the long line is locally Euclidean, I'd like to prove the following: Proposition. If $A$ is a countable well-ordered set, then $A \times [0,1)$ with the dictionary order is ...
Nick A.'s user avatar
  • 2,221
0 votes
0 answers
170 views

Cantor-Bendixson derivative sets

I try to show that a compact subset $A\subset \mathbb{C}$ is at most countable if and only if there exists a countable ordinal number $\alpha$ (i.e $\alpha <\omega_{1},$ where $\omega_{1}$ is ...
pospos's user avatar
  • 51
1 vote
0 answers
60 views

Convex subspace of linear continuum is connected [Theorem 24.1, Munkres]

My questions from the same excerpt quoted here, Proof. $\ \ $ Recall that a subspace $Y$ of $L$ is said to be convex if for every pair of points $a,b$ of $Y$ with $a<b$, the entire interval $[a,b]$...
Cathartic Encephalopathy's user avatar
1 vote
0 answers
46 views

Discussing Ex 13. Section 3. p-29, Munkres' Topology 2E. [duplicate]

In Exercise 13, section 3 on page 29 of Munkres’ Topology 2E, the problem is stated as follows. Prove the following: Theorem. If an order set $A$ has the least upper bound property, then it has the ...
Myo Nyunt's user avatar
  • 307
1 vote
1 answer
63 views

Open sets in $S_\Omega$ (minimal uncountable well ordered set) with the order topology.

$X$ is a well ordered set. $S_\Omega= \{ x:x\in X\text{ and } x<\Omega\}$ such that it is the minimal uncountable well ordered set. The section $S_\Omega$ of $X$ is uncountable and any other ...
Fabrizio G's user avatar
  • 2,117
2 votes
1 answer
78 views

If $Y$ is a closed and bounded subset of an ordered space $X$ then is it compact?

Let be $(X,\preccurlyeq)$ a totally ordered set. So if $Y$ is a bounded and closed subset of $X$ with respect the order topology then is it compact? if this is not generally true then is it true when $...
Antonio Maria Di Mauro's user avatar
0 votes
1 answer
189 views

Zorn's lemma and maximal elements

This continues a discussion begun at checking Zorn's lemma on an example where an example was offered to help understand maximal elements and Zorn's lemma. That example used the set {1,...,100} ...
crabtree's user avatar
0 votes
1 answer
256 views

First countability requirement of the Sequence Lemma

Consider the Sequence Lemma: Let $X$ be a topological space, $A\subseteq X$ any subset and $x\in X$. If there is a sequence of points in $A$ converging to $x$, then $x\in\bar A$; the converse holds ...
Daniel Turizo's user avatar
0 votes
1 answer
42 views

Is the proper segment of a TOSET is initial segment? [closed]

I can't fimd its answer. I have learnt that for a well ordered set proper segment is the initial segment. But i am unable to find segment of a toset which is proper and not initial.
Abhya Gupta's user avatar
0 votes
2 answers
43 views

Order topology related question

Say I have two topological spaces $X,Y$ s.t $Y$ has the order topology, and $f,g$ are two continuous functions $g,f:X\rightarrow Y$. I want to show that the set $O=\{x:f(x)> g(x)\}$ is open in $X$. ...
Math101's user avatar
  • 4,653
0 votes
0 answers
120 views

On an exercise concerning well-ordering

Working on set theory I stumbled across this exercise: Let $\{A_\alpha: \alpha \in \mathcal{A}\}$ be a family of well-ordered sets. If for every pair of sets, one is an ideal of the other, prove that ...
Sultan's user avatar
  • 175

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