# Tagged Questions

37 views

### For a set $\mathbb{X}$ given order relation extended from that of $\mathbb{R}$, if $\mathbb{X} \supseteq \mathbb{R}$ then $\mathbb{X} = \mathbb{R}$?

For a set $\mathbb{X}$ given order relation extended from that of $\mathbb{R}$, if $\mathbb{X} \supseteq \mathbb{R}$ then $\mathbb{X} = \mathbb{R}$ ? Motivation for this question rose from an ...
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### Proof by contradiction: $c=\max (A) \wedge d=\max (B) \to \max (A \cup B ) \in \{c,d\}$

Let be $A,B \subseteq \Bbb{R}$, with $A \neq \emptyset$ and $B \neq \emptyset$ and $\Bbb{R}$ totally ordered by $\leq$, and $c,d \in \Bbb{R}$, with $c=\max (A)$ and $d=\max (B)$. I must proof by ...
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### Are there any known uncountable transfinite increasing sequences of real numbers?

The real numbers are uncountable, so assuming the axiom of choice there is at least transfinite sequence of real numbers $r_0, r_1, r_2, ..., r_\omega, r_{\omega + 1}, ...,$ up to (and possibly ...
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### Let $A$ denote a closed subset of $\mathbb{R}.$ Is it true that for all $B \subseteq A,$ if $B$ has a supremum, then $\mathrm{sup}(B) \in A$?

Let $A$ denote a closed subset of $\mathbb{R}.$ It seems obvious that for all $B \subseteq A$ we have that if $B$ has a supremum, then $\mathrm{sup}(B) \in A$. Perhaps this is trivial, but I cannot ...
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### Dense and complete totally ordered sets

The totally ordered set $\mathbb{R}$ has the following properties. [Density]. Given any two $x,y \in \mathbb{R}$, we can find $\eta \in \mathbb{R}$ such that $x < \eta < y$. [Dedekind ...
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### What is an order reversing bijection on $\Bbb R$?

What is an order reversing bijection on $\Bbb R$ and why must it be continuous? Thanks
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### Prove that $\sup(A+B) = \sup(A) + \sup(B)$ and why does $\sup(A+B)$ exist?

We want to show that $\sup(A)+\sup(B)$ is the least upper bound of the set $A + B$. First, we need to show that $\sup(A) + \sup(B)$ is an upper bound for the set $A + B$. Indeed, if $z\in A + B$, then ...
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### Equality and order in sets

Just started Baby Rudin and got struck in this. While defining order in sets, $<$ was introduced as a relation and for a set to be ordered the condition was: for all $x,y$ belonging to an ordered ...
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$\require{color}$ Question: I need some help in proving that $\color{green}{\text{if$k\geq 0$, then$\sup (kS) = k\sup(S)$and$\inf (kS) = k\inf (S)$}}$, and also that $\color{red}{\text{if ... 2answers 63 views ### “intertwined” subset in linearly ordered set I am trying to solve an exercise in utility representation theory which leads me to the following question. Take any (nonempty) linearly ordered set$(X,\succeq)$. Can we construct a set$Y\subset ...
the definition of Dedekind cut by initial segment is correct: "let $\preceq$ be a linear ordering of a set $A$, and $B \subsetneqq A$, $B \neq \emptyset$, $B$ is Dedekind cut of $A$ under $\preceq$ ...