MHB Upper Bound of Sets and Sequences: Analyzing Logic

AI Thread Summary
The discussion focuses on the definitions of upper bounds for sets and sequences, highlighting the logical transitions between them. It examines how the upper bound of a set, defined as M being greater than or equal to all elements in the set, relates to the upper bound of a sequence generated from that set. The confusion arises in proving the equivalence of the two definitions, particularly in the transition from the set to the sequence. Clarification is provided that while every element of the set corresponds to a specific sequence element, the logical structure must be correctly formalized to establish equivalence. Accurate formalization is emphasized as essential for constructing valid mathematical proofs.
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Upper bound definition for sets: $ M \in \mathbb{R} $ is an upper bound of set $ A $ if $ \forall \alpha\in A. \alpha \leq M$
Upper bound definition for sequences: $ M \in \mathbb{R} $ is an upper bound of sequence $ (a_n)$ if $ \forall n \in \mathbb{N}. a_n \leq M$
Suppose we look at the set $ A = \{ a_n | n \in \mathbb{N} \} $ .

I've been pondering for a while about the following 2 questions related to mathematical-writing , logic and set builder notation:

Questions:
1. How do we get using logic from the defintion of upper bound for sets to the definition of the upper bound for sequences using the set $ A = \{ a_n | n \in \mathbb{N} \} $? My reasoning:
$ \forall \alpha \in A . \alpha \leq M \iff \forall \alpha.( \alpha\in A \rightarrow \alpha \leq M ) \iff $ $\forall \alpha.( \alpha \in A \rightarrow \exists n \in \mathbb{N}. \alpha = a_n \rightarrow \alpha \leq M ) \iff $ $ \forall \alpha.( \alpha \in A \rightarrow \exists n \in \mathbb{N}. \alpha = a_n \rightarrow \alpha \leq M \rightarrow a_n \leq M) $ Hence $ \forall \alpha \in A \exists n \in \mathbb{N}.( \alpha=a_n \land a_n\leq M $ ), now, this is not equivalent to the defintion of upper bound of sequences above ( $ \forall n \in \mathbb{N}. a_n \leq M$ ), why? can you please give correct transitions? I think I made mistakes but It's confusing me to see how to write them correctly.

2. Someone told me that since every element of $ A = \{ a_n | n \in \mathbb{N} \} $ is generated by every element of $ n \in \mathbb{N} $ so therefore $ \forall \alpha\in A. \alpha \leq M$ is equivalent to $ \forall n \in \mathbb{N}. a_n \leq M$ .
How is that possible that the two statements are equivalent? Since for arbitrary $ \alpha \in C $ there exists a specific $ n \in N $ ( not arbitrary ) therefore it appears false to write $ \forall n \in \mathbb{N}. a_n \leq M$ but seems more reasonable to write $ \exists n \in \mathbb{N}. a_n \leq M$ .
 
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The complete and correct formalization of the above statatments are the following :

1)$M\in R$ is an upper bound of $A\subseteq R$ iff (and not if) $\forall a(a\in A\Rightarrow a\leq M)$

2)$M\in R$ is an upper bound of the sequence $(a_n)$ iff $\forall n ( a_n\leq M)$

OR

1)A, $A\subseteq R$ is bounded from above iff ) $\forall a((a\in A\Rightarrow\exists M(M\in R\wedge( a\leq M))$

2) $(a_n)\subseteq R$ is bounded from above iff $\forall n\exists M (M\in R\wedge (a_n\leq M)$

So depending on the words of the definition there different ways to formalise the definition

Also you do not use logic for formalizing a mathematical statement

To prove that two statement are equivalent you have to use logic

Now according to the words expressing that A has an upper bound M you can use the appropriate formalization
 
Thanks, I understand now.
 
I must also point out that only complet and correct formalization will allow you to write a correct formal proof
 
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