, iff and differ in finitely many digits. (So for example but ). Then this equivalent relation partitions into different equivalent classes. How many equivalent classes there are?
Thanks for the reply. I knew that. But by my definition . What you have is , NOT . I don't really see anything inconsistent with the definition...
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Well, maybe I was wrong and it IS ill-defined (but I'd like to know why?). If that is the case, let me define it this way: , iff and differ in finitely many digits or a (b) ends in ...999... and b (a) ends in ...000...
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I'd like to hear your opinion. Thanks in advance
Edit: Red. sorry for that conspicuous mistake. Corrected in red.
Let me use FernandoRevilla's definition in #7. Please check if there's any problem with the following proofs:
1. Any equivalent class contains countable elements.
Proof: Let be the set of all equivalent classes of in #7. Choose any and .
and such that . Hence , where and for . Since was arbitary, we have . Since is countable for any , we proved that is countable.
Q.E.D.
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2. .
Proof:
(1) If , then E contains finite elements . Now because S is uncountable and , some must contain uncountably many elements, i.e., for some i. Contradiction to proof in 1.
(2) If , then . Define , where for all i. Now we derive a contradiction by constructing but , as follows:
define , then clearly is a bijection from to . Construct y by choosing for and (Or by the Cantor's diagonal method, if you prefer). Clearly , . Which is a contradiction. Q.E.D.
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3.
Proof: Choose from each element of an . Let the set of all such x's be . Clearly . But . Hence . Hence . Q.E.D.
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Edit: all is replaced by
In case this is still not clear: If two objects are equal, then one of them can be replaced by the other in any proposition without changing the truth value of this proposition. (Sometimes this is even taken as the definition of equality.) So here we have
1 ~ 1 (*)
and 1 = 0.999... Replacing the second occurrence 1 in (*) by an equal object 0.999... we get
1 ~ 0.999...
which is false, unlike (*).
Ill-defined definitions typically happen when they involve choosing a representative. Some rational numbers have two decimal expansions, and choosing one or the other gives different results with respect to ~; hence ~ is ill-defined.
I agree with Fernando that considering instead of is probably the most reasonable approach here.