Results for query "Sup"

Sup preserving maps between distributive lattices

I have been looking at categories of sup semilattices and sup preserving maps.If $A$ and $B$ are two such, the set I denote $[A,B]$ is sup preserving homomorphisms between is also a sup semilattice with pointwise sup.

Is my proof of if $A \subset B$ then $sup A \le sup B$ and $sup (A\cup B) = max(sup A,sup B)$, correct?

Suppose $\exists b_0 \in B$ such that $b_0 \ge sup A$, then $b_0 \notin A$ becausee $sup A \ge a$ for $\forall a \in A$, now we know that $\forall b \in B$, $sup B \ge b$, then $sup B \ge b_0 \ge sup A$, then we get $sup B \ge sup A$, now if theres not that element $b_0 \in B$ such that $b_0 \ge sup A$ and $b_0 \notin A$ , then $sup A \ge b$ for $\forall b \in B$, then $sup A \ge sup B$ $(1)$, and since $a \in B$ then $sup B \ge a$, so $sup B \ge sup A$ $(2)$, from $(1)$ and $(2)$ we get $sup A = sup B$.now for $sup (A\cup B) = max(sup A,sup B)$:

Let $(A\cup B)$ = $\{ x ∣ x\in A \ $or$ \ x \in B \}$

we have $sup A \ge x \ or \ sup B \ge x $ then $max(sup A,supB)\ge x$ then $max(supA,supB) \ge sup (A\cup B)$ $(1)$

now since $x$ can be $a$ or $b$, then $sup (A\cup B) \ge x$ then :

$\sup (A\cup B) \ge a$.

Proving $\sup B=\inf A$

This implies $\exists b\in B, \forall b \in B, b\leq M$.Since $b\leq$ sup$B$, with sup$B$ the largest element in $B$, we have inf$A \leq M =$ sup$B$, so inf$A \leq$ sup$B$.

What does $\sup _n$ mean?

I came across this notation when studying about the Kolmogorov Convergence Criterion, namely

Suppose that $\sup _n |X_n| \leq A$ for some $A>0$,.I understand what does $\sup$ mean, but what does $\sup_n$ mean?

$S$ be a non empty subset of $\mathbb{R}$ having supremum and $T=\{x\in\mathbb{R}: x-a \in S\}$, then $\forall a \in \mathbb{R}$, $\sup T=a+\sup S$.

Let $S$ be a non empty subset of $\mathbb{R}$ having supremum and $T=\{x\in\mathbb{R}: x-a \in S\}$, then $\forall a \in \mathbb{R}$ , $\sup T=a+\sup S$.My proof:- Let $x\in T$

$\implies x-a≤ \sup S$

$\implies x≤ \sup S+a$

$\because \sup T $ act as an least upper bound of $T $

$\therefore \sup T≤a+\sup S$

How to prove the converse?

When does Max=Sup ?

I know the following statement is true.$\forall S,$ (S is compact and S$\neq\emptyset \to \max(S)=\sup(S))$

In what else situations can be Max=Sup?

How to compare sums of infimum (or supremum) of two sets?

$ \sup\{x_n+y_n\}⩽\sup\{x_n\}+\sup\{y_n\} $
$ \sup\{x_n+y_n\}⩾\sup\{x_n\}+\sup\{y_n\} $
$ \sup\{x_n−y_n\}⩽\sup\{x_n\}−\sup\{y_n\} $
$ \sup\{x_n−y_n\}⩾\sup\{x_n\}−\sup\{y_n\} $
$ \sup\{x_n+y_n\}⩽\sup\{x_n\}+\inf\{y_n\} $
$ \sup\{x_n+y_n\}⩾\sup\{x_n\}+\inf\{y_n\}$

Which of this expression are correct?Why or why not?

Proving that $sup(S+T)=sup(S)+sup(T)$

Let $S$ and $T$ be nonempty sets of real numbers and define
$$S+T = \{s+t| s \in S, t \in T\}$$
show that $sup(S+T)=sup(S)+sup(T)$ if S and T are bounded above.I am thinking about a proof by contradiction such as:

Let's suppose that $sup(S+T) \neq sup(S) +sup(T)$, then either $sup(S+T) > sup(S) +sup(T)$ OR $sup(S+T) < sup(S) +sup(T)$
If $$sup(S+T) > sup(S) +sup(T)$$
$$ sup(S+T) -2\epsilon > sup(S) -\epsilon +sup(T) -\epsilon $$
with $\epsilon >0$
$ \exists x_3 \in(S+T)$ , $ \exists x_2 \in S$ , $ \exists x_1 \in T$ such that

$$ x_3 >sup(S+T) -2\epsilon > x_1 + x_2> sup(S) -\epsilon +sup(T) -\epsilon $$
(using the property of supremum) statement A

Also, if $x_2 \in S$ and $x_1 \in T$, then $ x_2,x_1 \in (S+T)$

and, as $x_3 > x_2+x_1$ , using the theorem of supremum, $x_2+x_1 > sup(S+T)- 2\epsilon$ statement B

However, statement B is in contradiction with statement A.