# Re: Cantor's Diagonal

Bruno Marchal skrev:
```Le 20-nov.-07, à 23:39, Barry Brent wrote :

```
```You're saying that, just because you can *write down* the missing
sequence (at the beginning, middle or anywhere else in the list), it
follows that there *is* no missing sequence.  Looks pretty wrong to me.```
```
Cantor's proof disqualifies any candidate enumeration.  You respond
by saying, "well, here's another candidate!"  But Cantor's procedure
disqualified *any*, repeat *any* candidate enumeration.

Barry Brent
```
```

Torgny, I do agree with Barry. Any bijection leads to a contradiction,
even in some effective way, and that is enough (for a classical
logician).
```

What do you think of this "proof"?:

Let us have the bijection:

0 -------- {0,0,0,0,0,0,0,...}
1 -------- {1,0,0,0,0,0,0,...}
2 -------- {0,1,0,0,0,0,0,...}
3 -------- {1,1,0,0,0,0,0,...}
4 -------- {0,0,1,0,0,0,0,...}
5 -------- {1,0,1,0,0,0,0,...}
6 -------- {0,1,1,0,0,0,0,...}
7 -------- {1,1,1,0,0,0,0,...}
8 -------- {0,0,0,1,0,0,0,...}
...
omega --- {1,1,1,1,1,1,1,...}

What do we get if we apply Cantor's Diagonal to this?

--
Torgny

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