Use the bijection , .
I need to prove this question: 2 the set of all even natural numbers. Prove that 2 is countably infinite.
I know that I can combine each even number with an number so that 2=1, 4=2, 6=3,....
={1,2,3,4,5,6,.....,n} (n∈ )
2={2,4,6,8,10,12,...2n} (n∈ )
I know how to write it down but not how to prove it proper.
And ik know I need to prove its bijective but I dont know how.
f: N-->2N f(n)=2n
So if I want to prove its bijective I say:
n1,n2 ∈N f(n1)=f(n2) so 2n1=2n2 so n1=n2 so f is injective
n∈N f(n)=2n so n lays in the image of f so f is surjective
I say f is bijective. Is my prove correct?
Can I now say that 2N is countably finite ?
I need to prove this question: 2N is the set of all even natural numbers. Prove that 2N is countably infinite. ( N= natural numbers I cant copy the right notation)
In the forum girdav said that I need to use f;N->2N and f=2n
I thought I needed to prove that f is bijective to say that 2N is countably infinite
I understand all this. I agreed that you have proved injectivity of f. We were talking about surjectivity. In post #6 I asked what you are proving without using the word "surjective." It is very important to get the statement right, including the quantifiers ("there exists," "for all"). A proof must match the structure of the claim. For example, there are specific ways to prove claims that start with "For all." Also, a proof must not contain undefined objects like n. All identifiers should be properly introduced.
So, what exactly are you proving at this point?
If you are using n as the argument of f, you cannot require that n be in 2N. And by asserting that n is in N, you are going the wrong way.
Rather, to prove f is "surjective" (one-to-one) you need to prove "If y is in 2N then there exist n in N such that y= f(x)". If y is even then, by definition of "even", there exist n in N such that y= 2n= f(n).