Upper Bound of Sets and Sequences: Analyzing Logic

In summary, an upper bound for a set A is a real number M such that every element in A is less than or equal to M. Similarly, an upper bound for a sequence (a_n) is a real number M such that every term in the sequence is less than or equal to M. To show that these definitions are equivalent, we use logic to write the formalizations of the statements and prove their equivalence. The exact formalization depends on the words used in the definitions.f
  • #1
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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$ .
 
  • #2
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
 
  • #3
Thanks, I understand now.
 
  • #4
I must also point out that only complet and correct formalization will allow you to write a correct formal proof
 

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