@Stupidquestioninc That phrase refers to the final inequality. In general, given any real r > 0, we have r/2 ≥ 1/n as n → ∞, and hence r/2 ≥ 1/n for all except finitely many n, which implies r − 1/n ≥ r/2 for all except finitely many n.
@Stupidquestioninc Err? They showed that if the perfect set is countable then they get a contradiction. Why say "countable implies contradiction" when that is not what they did?
The point is that if you have a countable set then you can enumerate it by a bijection with ℕ. If you do not know that the set is countable, you cannot do so.
If you want to learn mathematics properly, you are really going to have to understand first-order logic because proofs are based on that. I think it is clear that you do not understand the logical structure of this proof, and you need to deal with that issue first before you try to go on to more difficult mathematics.
Truth tables are not enough to capture first-order logic (with quantifiers), so we use inference rules instead. Each inference rule is chosen to be sound, meaning that if you start with true statements and use the rule you will deduce only true statements. We say that these rules are truth-preser...
My suggestion is to stick to my system, because it contains just the core of what is needed (many other introductory-level books are hundreds of pages most of which are pointless). You probably have seen the rules up to "Boolean Operations", which are for PL (propositional logic). You may not have seen the rules for "Quantifiers", or have never really used them before.
I don't know if I have told you before, but it is easy to check whether you know how to use a deductive system or not; simply prove all the exercises here.
@Stupidquestioninc: To demonstrate that your issue is really with the logical structure of proofs, let me show you what the proof really looks like in the Fitch-style reasoning that I told you to learn first:
Given any perfect subset S of ℝ^k:
If S is countable:
...
Contradiction.
Therefore S is uncountable.
Every perfect subset S of ℝ^k is uncountable.
"⊥" in the system stands for "false" / "contradiction".
Let me know if you have questions about the system. Also, post your attempts of those exercises I gave you in this room and I will check them for you.
There is no other way to learn to use the system without doing exercises.
@user21820 Sorry My school bell rang and I needed to switch off my phone 😅
@user21820 Yes I understand it after forcefully thinking for while
So what the proof does it prove it is uncountable uncountable to do this it needs to be infinite and not countable
we suppose it is countable
then if it is countable then we can construct the $V_n$ s but then we do contradiction so we can't do that
@user21820 Yes I swear to check these when day after tomorrow since I have to be in school for whole day and I only arrive home or have freetime at night which is resting time 😭
Though note that many mathematicians consider "countable" to mean "injects into ℕ", and we would say "countably infinite" if we wanted to say "bijects with ℕ".
Just take note in case you see a different use of "countable" somewhere else.
@Stupidquestioninc Sure take your time, and see you!
