Yes, and your inability to get it is the reason of why you don't get my answers about this case.
Ok, if you really consider those three statements equivalent, then it is time to move on. I bet you can't find anyone in this forum that would honestly agree with your position, though.
...I am talking about the immediate predecessor of y...
Fine, we can focus on just that. Why do you assume any real number, Y, must have an immediate predecessor (or immediate successor)?
My claim is this, if y is an immediate successor of some R member, then this member is the immediate predecessor of y.
Since no real number has an immediate successor, your statement is trivially true.
IF <proposition that is false> THEN <any proposition whatsoever>.
This claim must be true both in [x,y) or [x,y] cases, since the universal quantifier "for all" is used in both cases.
Umm, did you miss the fact that intervals are not elements of the set of real numbers? And just how do you connect the universal qualifier to this and make it responsible for your claim.
Maybe this will help as a starting point. Here's a half-open interval expressed as an equivalent set:
[latex]$$$ [X, Y) \equiv \{Z : X \le Z < Y\} $$$[/latex]
Standard Math uses the twisted legend, that any representation method is limited to aleph0
Where'd you get that idea?
...and as a result there is no way to represent the immediate successor or the immediate predecessor of any arbitrary given R member along the real-line.
No, that's wrong. It is not for lack of representation; it is for lack of existence. No real number has an immediate predecessor or immediate successor.
But this twisted legend is collapsed because Standard Math (as you wrote) can't show the immediate successor or the immediate predecessor of any arbitrary given Q member along the real-line, even if there are aleph0 Q members along the real-line.
Yes, rational numbers don't have immediate predecessors or immediate successors, either. Why would you expect otherwise?
We do not need more than that in order to show that the use of the universal quantifier "for all" on a collection of non-finite elements, does not hold.
So, what you are saying is that because the reals (and the rationals) don't have a property they shouldn't have, this proves the universal qualifier doesn't work? Curious.
By the way, the word is qualifier, not quantifier. Also, appending the "for all" to "universal qualifier" is redundant. It weakens your case when you can't even get these simple things correct.
you have failed to get the fact that Cantor explicitly used a way to define the exact member that is not mapped with any member of N, and by using this method, the conclusion and the premise are under a circular reasoning.
Again, you demonstrate you have no understanding of Cantor's second uncountability proof. You have expanded it, though, to show you have no understanding of proof by contradiction nor the term, circular reasoning.