Cantor Confusion
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From: Poker Joker on 28 Sep 2006 23:08 "Randy Poe" <poespam-trap(a)yahoo.com> wrote in message news:1159494111.724651.95600(a)i3g2000cwc.googlegroups.com... > That's incorrect. You don't have to assume none map onto R in order to > prove none map onto R. > > The direct argument starts this way: Let f be any such function, from > naturals to reals. Certainly we should assume that f *MIGHT* have R as its image, right? > Now, are you saying that somehow that misses some possible functions > from naturals to reals? How so? No, but we haven't proven that the image of f can't be R in step #1, right? So step #2 isn't valid, right? It's only *VALID* if R isn't the image of f. But without step #2 being valid under *ALL* conditions, the proof as a whole breaks down. > Given only that assumption about f (which includes ALL POSSIBLE > SUCH FUNCTIONS) it follows that f misses some reals. It follows > just from assuming f maps naturals to reals. Under the most general assumption, we can't count out that R is f's image, so defining a real in terms of the image of f *MIGHT* be self-referential, and it certainly is if the image of f is R. So, in general, the best we can say is that the real from step #2 might be self-referential and hasn't been shown not to be so without assuming the conclusion is true. I don't usually accept proofs that *REQUIRE* you to assume the conclusion is true for the proof to be valid. At best, the proof might show the conclusion to be independent of the premises. > What function from naturals to reals is missed by that argument? I'm not missing any, because I'm considering functions with R as their image. It seems that you are the one not considering those functions and their implication for step #2.
From: Poker Joker on 28 Sep 2006 23:15 "William Hughes" <wpihughes(a)hotmail.com> wrote in message news:1159499239.811924.305030(a)c28g2000cwb.googlegroups.com... > Define a procedure D, that takes a list M and produces > a real number r=D(M). This procedure D will work on any > list M If M contains all reals, it contains r, so: r = D( M : r in M) See the self reference?
From: William Hughes on 28 Sep 2006 23:31 Poker Joker wrote: > "William Hughes" <wpihughes(a)hotmail.com> wrote in message > news:1159499239.811924.305030(a)c28g2000cwb.googlegroups.com... > > > Define a procedure D, that takes a list M and produces > > a real number r=D(M). This procedure D will work on any > > list M > > If M contains all reals, it contains r, so: > > r = D( M : r in M) > > See the self reference? No I suppose what you mean is r = D(M, with M being a list that contains r) but we never have this. What we have is r = D(M, with M being a list that may contain r) later we learn that M did not contain r (no contradiction we only said may). - William Hughes
From: Ross A. Finlayson on 28 Sep 2006 23:43 Poker Joker wrote: > <cbrown(a)cbrownsystems.com> wrote in message > news:1159495306.456537.79590(a)e3g2000cwe.googlegroups.com... > > > Apply your logic to "oddven". > > > > A natural number is even if it is of the form 2*n, and odd if it is not > > of this form. A number is oddven if it is both even and odd. > > > > 1) Assume that n is oddven. > > 2) If it is oddven, then it is even; so it is of the form 2*n. > > 3) But if it is oddven, then it is odd; and therefore it is not of the > > form 2*n. > > 4) Contradiction; therefore no such n exists. > > First, the OP isn't saying anything is wrong with "proofs > by contradiction". He's saying that there is a meaningless > definition like "This statement is false." > > There is no meaningless definition in your step 2. > > Defining a real number in terms of all real numbers in a > set is self referential if the set contains all the real > numbers. The same is true for lists. So under the > assumption that the list *MIGHT* contain all the real > numbers, we can't say that the definition is not > self-referential. Therefore it is meaningless. Poker Joker: Bravo! You're damn right. Dang, I wanted to be 39. I must have seen ten thousand usenet messages. Ross
From: cbrown on 28 Sep 2006 23:53
Poker Joker wrote: > "William Hughes" <wpihughes(a)hotmail.com> wrote in message > news:1159499239.811924.305030(a)c28g2000cwb.googlegroups.com... > > > Define a procedure D, that takes a list M and produces > > a real number r=D(M). This procedure D will work on any > > list M > > If M contains all reals, it contains r, so:... Likewise, the definition given implies that M is a complete list of reals if and only if M is complete, and M is a list of reals. If M is a list of reals, we can construct a real which is not in M; i.e, M is not complete. Therefore, the assertion "M is a complete list of reals" is only true if the assertion "M is complete, and M is not complete" is true. (A and ~A) = false. Therefore, the assertion "M is a complete list of reals" is logically false for any M. There is no need to assume that an object such as a complete list of all reals exists in order to prove that such an object does not exist. Cheers - Chas |