So i was given two questions you either prove or disprove them.
A) For any two sets A and B, if $f: A \rightarrow B$ is injective, then if A is countable, B must be countable.
B) For any two sets A and B, if $f: A \to B$ is surjective, then if A is countable, B must be countable.
I know Every subset of a countable set is countable and I know i have to use injectivity of f to create a bijection for A but i don't know how. Also for B im quite clueless does it follow A?
Best Answer
For (A) consider a function that takes an integer in $\mathbb Z$ to itself , with $\mathbb Z$ living in $\mathbb R$.
For (B), we have hat $B=f(A)$, so that $|B|=|f(A)|$, and since every element in $A$ maps into a single element in $B$, we have $|B|\leq |f(A)|$. Can you use your idea that every subset of a countable set is countable to show how $f(a)$ can be made into a "subset" of $A$? , i.e., find an injection between $A$ and $f(A)$?