atinymonkey Posted December 30, 2004 Posted December 30, 2004 Blarg: - http://dictionary.reference.com/search?q=bijection
lama Posted December 30, 2004 Author Posted December 30, 2004 No Blarg, I am waiting to Matt's explanation of the bijection concept, in a very simple English.
bloodhound Posted December 30, 2004 Posted December 30, 2004 a bijection is a mapping which is both injective and surjective. a map is injective iff f(a)=f(b) implies a = b a map is surjective iff for all x in the Target there is a y in the Source which is mapped into x by the map
matt grime Posted December 30, 2004 Posted December 30, 2004 equivalently a map is a beijction between sets if and only if it possesses an inverse. As you've said the bijections are fundamentally wrong (as are cardinals) why am I explaining this to you? surely you know the definition (otherwise, how can you have formed an opinion). nb, the inverse of a set function f from X to Y is a map g from Y to X statisfying gf(x)=x, and fg(y)=y whenever x is an element of X and y is an element of Y.
lama Posted December 30, 2004 Author Posted December 30, 2004 Bloodhond are you also Matt Grime? I am still waiting to Matt's very simple English exaplanation of the Bijection concept.
lama Posted December 30, 2004 Author Posted December 30, 2004 equivalently a map is a beijction between sets if and only if it possesses an inverse. As you've said the bijections are fundamentally wrong (as are cardinals) why am I explaining this to you? surely you know the definition (otherwise' date=' how can you have formed an opinion). nb, the inverse of a set function f from X to Y is a map g from Y to X statisfying gf(x)=x, and fg(y)=y whenever x is an element of X and y is an element of Y. [/quote'] Matt, Let us say that I am a 5 years old and you want to explain to me what is a bijection. Can you do that?
lama Posted December 30, 2004 Author Posted December 30, 2004 A very simple question that needs a very simple answer.
bloodhound Posted December 30, 2004 Posted December 30, 2004 well, don't know about matt, but i definitely couldn't. cos a 5 year old wouldn't have enough foundation on basic mathematics. so why dont YOU explain to us (assuming that we are 5 years old ) Why f(x)=x is not a bijection.
matt grime Posted December 30, 2004 Posted December 30, 2004 what kind of question is that???? A stupid one? Who on earth says maths must be understandable by 5 year olds? Mind you I suspect a 5 year old might be able to spot the holes in Doron's arguments. We have defined a bijection in two simpel and equivalent ways. I can add a third if you like that generalizes to the notion of isomoprhism which is an invertible arrow in a category. As I say, you claim to have spotted an egregious error in the definition, so one presumes you already know the definition. So, is the map fomr the ring of integers to itself given by f(z)=z-1 a bijection? If not why not, since every one else can see it is a bijection.
lama Posted December 30, 2004 Author Posted December 30, 2004 Ok, If Matt cannot do it, then let us continue. Now Matt as I clearly explained, the existence of a non-finite collection depends on the 'existence of the next element' as we can clearly show, by using Cantor's second-diagonal method. In a non-finite collection, the diagonal number must be permanently added to the collection. In a finite collection, the diagonal number must not be added to the collection. We clearly know that we cannot define a bijection in the second diagonal method, because each time we define a bijection, we discover that there is a new element which is out of the domain of our 1-1 correspondence mapping (a permanent next element). Conclusion: No bijection can be found in a non-finite collection, and this property can be found in any given non-finite collection and only in a non-finite collection. Therefore the cardinality of a non-finite collection, like set N (for example) is |N|-NEXT.
matt grime Posted December 30, 2004 Posted December 30, 2004 Doron, this has nothing to do with anything. Cantor's argument is in essence about the fact that there is no bijection between N and its powerset P(N). It the does not define a beijction, it shows that one cannot exist, exactly as we want. It is nothing to do with whether or not f(z)=z-1 is a bijection from Z to Z. If it isn't a bijection, as you claim it cannot be, then demonstrate that it isn't a biejction, that it isn't invertible.
matt grime Posted December 30, 2004 Posted December 30, 2004 Actually, you're contradicting yourself here too. You claim that you cannot quantify for all for infinite sets. Yet you're saying, from one example that has nothing to do with the matter in hand that all maps between all infinite sets are not bijections. How can you say that? YOu've jsut qunatified an infinite set with a for all.
matt grime Posted December 30, 2004 Posted December 30, 2004 I did, it makes no sense at all. Demonstrate why the map f(z)=z-1 is not a bijection from Z to Z. INcidentally, here is the proof that it is a bijection: let g(z)=z+1, then fg(z)=gf(z)=z, hence f is invertible, and thus a bijection. Where is this wrong? For any moderators: If Doron cannot do this, please can you lock the thread again. We have defined bijection for him, and he has not acknowledged the definition, nor has he shown why simple maps that we can see are bijections, cannot be bijections. The introduction of cardinals is completely spurious and is yet another attempt to distract from the fact that he doesn't know what he is talking about by dressing it up in more pointless complicated waffle.
lama Posted December 30, 2004 Author Posted December 30, 2004 You claim that you cannot quantify for all for infinite sets. Yet you're saying' date=' from one example that has nothing to do with the matter in hand that all maps between all infinite sets are not bijections. How can you say that? YOu've jsut qunatified an infinite set with a for all. [/quote'] If I said it, then it is my mistake. Again: The difference between a finite set and a non-finite set is: A bijection cannot be found in a non-finite set. Proof: Cantor's second-Diagonal method.
JaKiri Posted December 30, 2004 Posted December 30, 2004 Ok, If Matt cannot do it, then let us continue. Nicely dragging the argument down to an even lower level. Frankly, I've had enough of this. No progress can ever be made if both sides in the argument feel that they are immutably correct. Thread closed, and do NOT start another one on this topic. Take it to PM if you like, but this really is pointless.
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