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Ok. My other gripe with his argument is that these aren't equivalence classes. Let the string be 01b, where b is an infinitely long bit string. Then the string

Ok. My other gripe with his argument is that these aren't equivalence classes. Let the string be 01b, where b is an infinitely long bit string. Then the string 01b belongs to equivalence class b, and also equivalence class 1b... and an element cannot belong to multiple equivalence classes at once. The only way out of this is for 1b and b to actually be the same equivalence class, that is, that of b. But, can we do that? I feel like that is wrong.

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thisrod 17y ago
It's like saying 128x and 28x, considered as finite sequences of digits followed by a letter, are in the class of sequences that end in x. Except in the infinite case, there are a lot more than 26 letters.
- jganetsk 17y ago
  They also happen to be in the class of sequences that end in 8x and 28x... also definitely their own classes. So 128x is in at least 4 classes, 28x is in at least 3. Therefore it's not an equivalence relation.
  - ggchappell 17y ago
    Ah, but 8x and 28x are not different classes; they are all the same class. I think it's clearer if you describe things the way the article does. Don't say "these are the classes"; say "here is the relation". As the article says: > Call two such sequences ‘equivalent’ if they are equal after a finite number of entries. To see whether this is an equivalence relation, you only need to check whether it is reflexive, symmetric, and transitive. Figuring out what the classes are, might be interesting, but it is not necessary.
    - jganetsk 17y ago
      If an item belongs to multiple equivalence classes, then transitivity does not hold. In other words, if you show that the classes are not disjoint, then it's not an equivalence relation. 18x belongs in the 8x class, but 28x belongs in both the 28x and 8x classes.
      - jibiki 17y ago
        And 8x belongs in 28x's class, since 8x and 28x share a tail. I think the author meant for you to interpret "equal after a finite number of entries" in the more normal fashion (e.g., 123x = 456x but 1x != 111x, the prefixes have to have equal length.) But your way seems to work fine too.

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thisrod 17y ago

It's like saying 128x and 28x, considered as finite sequences of digits followed by a letter, are in the class of sequences that end in x. Except in the infinite case, there are a lot more than 26 letters.

jganetsk 17y ago
They also happen to be in the class of sequences that end in 8x and 28x... also definitely their own classes. So 128x is in at least 4 classes, 28x is in at least 3. Therefore it's not an equivalence relation.
- ggchappell 17y ago
  Ah, but 8x and 28x are not different classes; they are all the same class. I think it's clearer if you describe things the way the article does. Don't say "these are the classes"; say "here is the relation". As the article says: > Call two such sequences ‘equivalent’ if they are equal after a finite number of entries. To see whether this is an equivalence relation, you only need to check whether it is reflexive, symmetric, and transitive. Figuring out what the classes are, might be interesting, but it is not necessary.
  - jganetsk 17y ago
    If an item belongs to multiple equivalence classes, then transitivity does not hold. In other words, if you show that the classes are not disjoint, then it's not an equivalence relation. 18x belongs in the 8x class, but 28x belongs in both the 28x and 8x classes.
    - jibiki 17y ago
      And 8x belongs in 28x's class, since 8x and 28x share a tail. I think the author meant for you to interpret "equal after a finite number of entries" in the more normal fashion (e.g., 123x = 456x but 1x != 111x, the prefixes have to have equal length.) But your way seems to work fine too.