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Is this possible? Given that a set $A$ of infinite cardinality must have property $p$...

To prove that a infinite set $A$ cannot exist if it is to have property $p$, we can start from one arbitrary element in $A$ and whilst constructing $A$ while keeping property $p$ true, we enter a contradiction that shows property $p$ isn't possible.

Stone
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    generally sounds reasonable, would want to see a particular example to judge if it is sensible or not – gt6989b Oct 24 '17 at 09:42
  • The part i'm confused about is the construction of $A$. How do I know that this general method or algorithm/steps will construct $A$ as specified? – Stone Oct 24 '17 at 09:43
  • No, we can't, at least not in general. For instance, if this were the case, we would be able to show that there is a natural number that is larger than all natural numbers: Start with ${0}$: It has the property that there is a natural number larger than any of its elements. Keep adding the natural numbers to the set, one by one, and the property remains true. It fails when we pass to infinity. – Arthur Oct 24 '17 at 09:43
  • @Arthur. Well, wouldn't that be good? We show that when constructing the set of natural numbers (simply just adding a new element that is +1 larger than the original sets largest natural number), we don't enter any contradictions to the idea there is no largest natural number because in the process, there is always a next step. – Stone Oct 24 '17 at 09:46
  • It like a question that asks if $A$ exists with certain properties, in this case, property $p$. I'm trying to see that for an infinite set, can I show that during the process of constructing $A$ (realistically not possible but if I tried to...), if I enter some contradiction while sticking to the properties, then this proves that $A$ can't exist. – Stone Oct 24 '17 at 10:19

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