P(N) (N is the natural number set.)
Q
Z
[0, 1] n Q
Its P(N) but it's important you understand why the others are countable. (I take it P means power set).
First of. If a set is countable then a subset of it is also countable.
So... Is countable?
If it is then will be countable.
Surely you can tell that is countable?
Do you know anything about cardinality yet?
Ya no way could I be fully proving that of the top of my head. Could prove the rest are countable though...
To be honest I kinda misinterpreted the question as 'which one is uncountable' so I figured try to show OP how to eliminate the countable ones. Never the less... If OP is learning about cardinality that could be used via Cantors card(x) < card(P(x)) theorem.
I think i can see why P(N) is countable cus usually when you take all the subsets of N right? and thats not countable cus there's alot of sets i really suck at proving stuff I looked at cantors therom
Proof: It suffices to show that [0, 1] is uncountable (see Exercise 7). If
not, then we have a bijection from N to [0, 1]. This is a sequence (x) that
lists all numbers in [0, 1], in some order. By considering the canonical
decimal expansions, we will construct a number not on the list.
Xl = CI,I CI,2 C I.3
X2 = C2, I C2,2 C 2.3
X3 = C3,IC3,2C3.3
Suppose that the expansions appear in order as indicated above. We
build a canonical decimal expansion that disagrees with every expansion
in our list. Let an = 1 if Cn,n = 0, and an = 0 if Cn,n > o. Now (a) disagrees
in position n with the expansion of Xn. Furthermore, since (a) has no 9, (a)
cannot be the alternative expansion of any number in our list. Therefore,
the expansion (a) does not represent a number in our list. By Theorem
13.25, (a) is the canonical expansion of some real number. Thus our list
does not contain expansions for all real numbers in [0, 1]. .
but i dont really get how to use it