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May 24th, 2015, 03:02 PM   #21
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 Originally Posted by Tau I've thought about it and the simple solution is that nobody is going to come for a room because everybody has one already.
I don't understand, could you elaborate more?

 May 24th, 2015, 03:15 PM #22 Member   Joined: Jan 2014 Posts: 86 Thanks: 4 In Hilbert's Hotel, we are told it has infinite rooms all full. Then a new guest comes asking for a room. But that is false, nobody will come since they are already in the hotel. An infinite hotel has everyone in it.
May 24th, 2015, 03:28 PM   #23
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 Originally Posted by Tau In Hilbert's Hotel, we are told it has infinite rooms all full. Then a new guest comes asking for a room. But that is false, nobody will come since they are already in the hotel. An infinite hotel has everyone in it.
LOL, this is a creative answer, I like lateral thinking! But ultimately, it fails to realize the conditions of the puzzle, and that it makes no difference.

Let's then take all the guests out of the odd rooms, let's put them wherever you want, but outside the rooms. Now, let's take down the odd rooms, let's transform them into big kitchens or whatever. Now, the Hotel has an infinite amount of rooms, and the problem is just that the rooms are enumerated with even numbers.

Simple! The new owner of the Hotel, Mr. Tau, says that in fact a room number $\displaystyle 2n$ in his Hotel is an especial distinguishing touch of his own, a nice way to say that all guests are "twice welcome" in each of the 1, 2, 3, 4, 5, ... rooms!! Brilliant!

Last edited by skipjack; May 29th, 2015 at 04:41 PM.

 May 24th, 2015, 03:41 PM #24 Member   Joined: Jan 2014 Posts: 86 Thanks: 4 Clever but still fantasy. You see, at one point we treat the set of rooms like it's finite (take out all the odd rooms), but then when it's suitable for us we go back to "infinite mode" to "show" the "elasticity" of infinity. What I'm saying is that the conditions of the puzzle are contradictory.
May 24th, 2015, 03:54 PM   #25
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 Originally Posted by Tau Clever but still fantasy. You see, at one point we treat the set of rooms like it's finite (take out all the odd rooms), but then when it's suitable for us we go back to "infinite mode" to "show" the "elasticity" of infinity. What I'm saying is that the conditions of the puzzle are contradictory.
No, your internal concepts of how things work when infinities are involved are not yet broadened enough to understand these issues. In the same sense people cannot understand quantum mechanics if they reason in classical basis. Best thing we human beings can do is to verify if our set of beliefs and concepts are contradictory, or at least not good to express all ideas we have in mind and to communicate them to other rational beings. This is the basis of not only language, but also mathematical proofs.

The attitude of attacking is per se not bad, it is good, but it is futile if it is not also done against our own set of internal concepts.

In his book (which I recommend to you) ROADS TO INFINITY, John Stillwell, p. 8, wrote:

Quote:
 The Logic of the Diagonal Argument Many mathematicians aggressively maintain that there can be no doubt of the validity of this proof, whereas others do not admit it. I personally cannot see an iota of appeal in this proof . . .my mind will not do the things that it is obviously expected to do if this is indeed a proof. —P .W. Bridgman (1955), p. 101 P. W. Bridgman was an experimental physicist at Harvard, and winner of the Nobel prize for physics in 1946. He was also, in all probability, one of the smartest people not to understand the diagonal argument. If you had any trouble with the argument above, you can rest assured that a Nobel prize winner was equally troubled.

 May 24th, 2015, 04:00 PM #26 Member   Joined: Jan 2014 Posts: 86 Thanks: 4 Funny, but the best similar case I can remember is that Paul Erdos did not believe the solution to the Monty Hall problem until he saw a computer simulation of it. I understood it fully before trying out a simulation. Everybody has different forte.
May 24th, 2015, 04:05 PM   #27
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 Originally Posted by Tau Funny, but the best similar case I can remember is that Paul Erdos did not believe the solution to the Monty Hall problem until he saw a computer simulation of it. I understood it fully before trying out a simulation. Everybody has different forte.
I could not digest it until I saw a graphic explanation of it!

 May 24th, 2015, 04:09 PM #28 Member   Joined: Jan 2014 Posts: 86 Thanks: 4 It's counter intuitive the first time but the key is to realize that there is a casual link between the events. When the one door is opened the probabilities are reset.
May 24th, 2015, 07:10 PM   #29
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 Originally Posted by Tau The partial list written is incomplete, the imagined "complete" list will include that number. The contradiction is that we postulate completeness but examine incompleteness.
This is false. The number constructed as part of the diagonal proof cannot be in the list, because it is different from every member of the list.

From your answers thus far, I get the impression that your confusion is not so much that the reals should be uncountable than that the naturals should!

The definition of countability is that there exists a 1-1 mapping between a set and some subset of the natural numbers (including the set of natural numbers).

When, in this context, we talk about making a list of elements of a set, we do not mean writing a physical list but setting up a 1-1 mapping between the set and the natural numbers.

Thus, in the diagonal proof, we assume that every real is included in such a mapping and show that this leads to contradiction by pointing out a real number that cannot be part of the mapping, because it is different from every number that is part of the mapping.

Your argument regarding Hilbert's Hotel (that everybody is already in the hotel) is clearly flawed. What about the children of the guests at the hotel that come of age and wish to move into their own room? They certainly do not already have a room, so even if there was nobody available from outside the hotel, there are certainly people requiring rooms from within it.

You claim that we can simply remove the odd numbered rooms, but it's difficult to see why. If we can refer to an infinite number of rooms when assigning them, why can't we do the same to take them away? We can easily describe them all.

Last edited by skipjack; May 29th, 2015 at 04:46 PM.

May 25th, 2015, 04:53 AM   #30
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 Originally Posted by v8archie This is false. The number constructed as part of the diagonal proof cannot be in the list, because it is different from every member of the list.
The red underline in the picture is mine. "Above". This does not prove our "new" number is not in the infinite list (I accept such a thing, I'm not a constructivist). All it proves is that it's not in our finite partial list. With the same method I can "prove" that naturals are not countable.
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