5 ms·
> Here's how: consider the set of numbers that cannot be described using any finite collection of symbols. Such numbers must exist because there are only a cou
by auggierose 4y ago
> Here's how: consider the set of numbers that cannot be described using any finite collection of symbols. Such numbers must exist because there are only a countably infinite number of numbers that can be described using a finite collection of symbols, but there are an uncountably infinite number of real numbers.
But there are uncountably many finite collections of symbols. Sorry, you will have to take the axiom of choice out of my dead cold hands. Banach-Tarski just tells us if we use un-measurable sets, they don't have much to do with our understanding of physical space. But that's fine, because sets are just collections, and physical space is undoubtedly more than that.
- sebzim4500 4y ago>But there are uncountably many finite collections of symbols How do you mean? Because there are uncountably many symbols? I'm not really sure what a symbol means in this context.
- auggierose 4y agoI just picked on one of the points in the post that make no sense without further explanation. Just take each real number as a symbol.
- davrosthedalek 4y agoIt's normally assumed that the number of symbols is finite.
- ceceron 4y agoThe number of symbols in the language may countable, but the number of sets of symbols will be indeed uncountable, i.e. AFAIR the number of sets of natural numbers is uncountable — you can even construct real numbers as sets on rational (still countable) numbers.
- deleted 4y ago[deleted]
- drdeca 4y agoThe set of all sets of natural numbers is uncountable, but the set of all finite sets of natural numbers, is countable.
- ceceron 4y agoTrue, that's why I haven't used 'finite' in my comment. Apparently I've missed that the parent referred to the finite collections and in fact was wrong making my previous comment misplaced. Thanks for pointing it out :)
- shellothere 4y agoThere are countably many finite strings of a fixed, finite collection of symbols. That’s the point the author is making — it becomes more intuitively obvious that there might be trouble “choosing” when you design the set you’d like to choose from to consist solely of the “nameless” reals.
- karpierz 4y agoThere are countably many finite collections of symbols. There are uncountably many countable collections of symbols. Edit: The above is true if your symbols are countable. If they are uncountable, then obviously there are uncountable finite sets of your symbols (As the singleton sets are a subset, and those are uncountable).
- lisper 4y ago> there are uncountably many finite collections of symbols Only if there are uncountably many symbols. But this is a fair point. I'll need to define what I mean by "symbol" more clearly.
- BeetleB 4y ago> But there are uncountably many finite collections of symbols. As others have pointed out, that depends on how big the set of possible symptoms are. If the set is finite, then there are a countable number of finite collections (the countable union of countable sets is still countable). If the set is countably infinite, then there are still countably many finite collections.[1] If the set is uncountably infinite, then you're right. However, the author is probably talking about all symbols known to humanity, which clearly is not only not uncountable, but it is finite. [1] https://math.stackexchange.com/questions/200389/show-that-the-set-of-all-finite-subsets-of-mathbbn-is-countable https://math.stackexchange.com/questions/200389/show-that-th...
- drdeca 4y agoThere are uncountably many (countably) infinite collections of symbols from within a countably infinite set of possible symbols. The set of finite subsets of a countably infinite set, is countably infinite. There is a computable bijection between the set of natural numbers and the set of finite sequences of natural numbers.