Prove this completely obvious question

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To prove that α = sup A if and only if α - 1/n is not an upper bound of A while α + 1/n is an upper bound for all n in N, a contradiction approach is suggested. The proof requires demonstrating both implications of the "if and only if" statement, which the original poster only partially completed. Understanding the definitions and theorems related to supremums is crucial for grasping such concepts, as they often build on previous results. For further study, resources like Rudin's "Principles of Mathematical Analysis" and mathematical methods for physics books are recommended. Rigorous practice and a solid grasp of foundational definitions are essential for mastering these "obvious" mathematical questions.
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The Question -

Let ##\mathbb{A}## be a non-empty subset of ##\mathbb{R}## and ##\alpha \in \mathbb{R}##. Show that ##\alpha = sup\mathbb{A}## if and only if ##\alpha - \frac{1}{n}## is not an upper bound of ##\mathbb{A}## but ##\alpha + \frac{1}{n}## is an upper bound of ##\mathbb{A}\;\forall \;n \in \mathbb{N}##.

My solution -
I tried to approach this completely obvious question (as per to me) with two contradictions - one for each case.

Let ##\alpha## be an upper bound but ##\alpha + \frac{1}{n}## isn't upper bound. Then, ##\exists \; a\in \mathbb{A}## such that
$$\alpha + \frac{1}{n} \leq a \Rightarrow a - \alpha \geq \frac{1}{n}\; \forall \;n \in \mathbb{N}$$
but from the first condition,
$$\alpha \geq a \;\forall\;a\in\mathbb{A} \Rightarrow \Leftarrow $$
Hence, our assumption is flawed. Hence ##\alpha - \frac{1}{n}## must be upper bound if ##\alpha## is to be the supremum of ##\mathbb{A}##.

Similarly for the second condition (by contradiction)

Help required -
The thing is, I simply can't get to solve such "simple" and "obvious" question in such methods without toiling for hours. If possible, can someone refer me to course materials where these "obvious" questions are solved by such "not-so-obvious" methods? I am really troubled and unable to understand things like these in class, furthermore, my exams are drawing near.
 
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"Obvious" questions in mathematics often depend on definitions - and definitions often come from a previous result, such as a theorem which proves the existence of some "thing" ... if that thing is useful it has a ready-made definition.

So you need to know the definitions and understand how they originated. This (hopefully) provides a more intuitive understanding ... and makes many things obvious.

Lots of practice makes things more obvious ...
 
Practice is fine, but is there any reference course material where I can collectively see such theorems (like Archimedes Principle, Bolzano-Wierstrass Theorem, Cauchy criterion, etc.) with ease. I have a hard time to understand them from pages like on wikipedia.
 
For some good advice try Terrance Tao:
http://terrytao.wordpress.com/career-advice/there%E2%80%99s-more-to-mathematics-than-rigour-and-proofs/

If you are looking for a general reference work with lots of stuff in it - try any of the books especially written for "mathematical methods for physics" ... they include lots of topics, and usually have good references.
 
Keep in mind that in addition to learning mathematics, you're also (likely) learning how to prove things as well as write those proofs. It's not uncommon for students to be asked to prove "obvious" statements so that they can practice various methods of proof.

In this case, you've really only done about 25% of the work that needs to be done. The statement that you're being asked to prove is an "if and only if" statement, and you've only demonstrated part of one of those implications. Furthermore, there is an incorrect (though easily fixable) statement as well as a typo in the part of the proof that you have done.

It's not always the case that statements which are intuitively "obvious" are easy to prove or even true. The point of doing rigorous mathematics is to be, well, rigorous.

I use Rudin's Principles of Mathematical Analysis as a reference for topics like the ones that you have mentioned.
 
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