Thread: Question involving showing there's a bijection

My question is:

Let A be a finite set with m elements, for some . And suppose x is an object that is not a member of A.

Prove, using the definitions, that has m+1 elements.

(All you are told about A is that it has m elements. You need to show that there is a bijection from to .)

Any help would be appreciated

Do you need to show there's a bijection?

You could use induction as well. Start with the empty set (base case m=0). empty set union with {x} is {x} by definition and has 0+1 = 1 elements. so the base case is true.
then assume P(k): "A union with {x} has m+1 elements for "

then prove P(k+1)

Yeah, I need to show that there's a bijection rather than doing a proof by induction. Would you be able to help me? Thanks

Maybe it's easier than I originally thought, maybe. Since A is a finite set, we can label the m elements a₀ a₁, a₂ ... a_m

then by definition
has elements a₀ a₁, a₂ ... a_m, x

Let f be a bijection from to N_m+1 such that a_n -> n for and x -> m+1