I'm new to mathematical logic and to real analysis. I have studied Daniel Velleman's How to Prove It and that's how I got my first experience with logical structures. Now, I'm studying real analysis (Abbot's Understanding Analysis) and I figure out that I did not grasp 100% of the concepts behind a logical structure when I was studying the convergence of sequences.
Therefore, I was wondering if someone can help me understand the fundamentals behind the logical structure that I present below. Let me show how I got there:
Definition of Convergence of Sequence: A sequence $({ a }_{ n })$ converges to a real number $a$ if, for every positive number $\varepsilon$, there exists a $N\in\Bbb{N}$ such that whenever $n\ge\Bbb{N}$ it follows that $\left| { a }_{ n }-a \right| <\varepsilon$.
I translated this definition into a logical structure and I got the following: $\left( \forall \varepsilon >0 \right) \left( \exists N\in { N } \right) \left( \forall n\ge { N } \right) \left| { a }_{ n }-a \right| <\varepsilon$
However, when I tried to write a formal proof for a simple convergence example, I noticed that I have to rewrite the logical structure like the one below so that I could follow what some authors call "The template for some proof".
So my question is: Why does the logical structure above can be translated to the one below? What concept and I missing? WHat should I now in order to understand this? $\left( \forall \varepsilon >0 \right) \left( \exists N\in { N } \right) \forall n\left( n\ge N\Rightarrow \left| { a }_{ n }-a \right| <\varepsilon \right)$
With my question in mind, I just want to show why am I focusing so much on transforming into this last logical structure. The reason why is because I can easily follow steps that usually help me prove things. (I have made a lot of progress since I learned about logical structures. When I was trying to study real analysis without it, I could not prove anything at all, just very simple things.)
So, let me give an example in case I had figured out the last logical structure at first. I would just do the following steps:
Example: Prove that lim$\left( \frac { 1 }{ \sqrt { n } } \right)=0$
1) Write the "right" logical structure: $\left( \forall \varepsilon >0 \right) \left( \exists N\in { N } \right) \forall n\left( n\ge N\Rightarrow \left| { a }_{ n }-a \right| <\varepsilon \right)$
2) Change the first parenthesis for: "Let $\varepsilon >0$ be an arbitrary positive real number". Then I would be left with: $\left( \exists N\in { N } \right) \left( \forall n\ge { N } \right) \left| { a }_{ n }-a \right| <\varepsilon$.
3) I learned that I need to pick just one good example for N in order to complete the proof. So, after some scratch work, I would say: "Choose a natural number $N$ satisfying $N>\frac { 1 }{ { \varepsilon }^{ 2 } }$"
3) Now, I would change the "$\forall n$" for: "Let $n$ be an arbitrary natural number such that $n\ge { N }$. Thus, assume $n\ge { N }$".
4) Now, my goal would be to prove $\left| { a }_{ n }-a \right| <\varepsilon$.
5) I would complete the proof and show that: $n\ge N>\frac { 1 }{ { \varepsilon }^{ 2 } }$ implies that $n>\frac { 1 }{ { \varepsilon }^{ 2 } }$ which implies that $\frac { 1 }{ \sqrt { n } } <\varepsilon$, which completes the proof.
Therefore, just because I was not able to (at first) note that $\left( \forall \varepsilon >0 \right) \left( \exists N\in { N } \right) \left( \forall n\ge { N } \right) \left| { a }_{ n }-a \right| <\varepsilon$ can be transformed into this $\left( \forall \varepsilon >0 \right) \left( \exists N\in { N } \right) \forall n\left( n\ge N\Rightarrow \left| { a }_{ n }-a \right| <\varepsilon \right)$ it has prevented me from safely writing a proof for this example.
I know that I am a beginner and that mathematicians do not focus on the logical structures to prove things, but at least it has been helping me a lot.
