Given $2$ sets: $A:=\{1, 2,3\}$ and $B:=\{a, b, c\}.$ Are $A$ and $B$ equivalent? Given $2$ sets: $A:=\{1, 2, 3\}$ and $B:=\{a, b, c\}.$

Are $A$ and $B$ equivalent?

This is one of my question in my homework.
I knew that $2$ sets $A$ and $B$ are equivalent with each other, wrote $A\sim B$, if there exists a bijection $\text f: A\to B$, but I wonder if this is true when one set is not in real number's set.
 A: As sets they are equivalent. The point is: you have a set $\{a,b,c\}$ and a set $\{1,2,3\}$. Imagine you pick each element of the first set and associated a number from the second one. That is, you pair $a$ with $1$, you pair $b$ with $2$ and you pair $c$ with $3$.
What you are essentially doing is "labelling elements of $\{a,b,c\}$ by elements of $\{1,2,3\}$. The sets are then equivalent because if you want to talk about some element of $\{a,b,c\}$ you can refer to it in terms of the label you picked from $\{1,2,3\}$ and everything is fine. In that sense $\{a,b,c\}$ and $\{1,2,3\}$ are the same set (pause and think about this until you get the idea).
Now, bijections need not be between two sets of real numbers. The point with bijections is to do the same thing in a more general case. Whenever the sets are finite you can do the "labeling" as I said. But if the sets are infinite things will get confusing. Bijections capture that same idea in a much more general way which allows us to do essentially the same thing regardless of the set being finite or infinite.
We of course don't need the sets to be composed of numbers to talk about bijections. The general point of view is: If $A$ and $B$ are arbitrary sets, a function $f : A\to B$ is said to be a bijection if it is injective and surjective. Then you need to know what means to be injective and surjective:


*

*Injectivity means that if $x\neq y$ then $f(x)\neq f(y)$, or equivalently, if $f(x)=f(y)$ implies $x=y$. Intuitively, a function is injective when a point of the range is mapped by just on point of the domain.

*Surjectivity means that if $y\in B$ then there is some $x\in A$ such that $f(x) = y$. Intuitively this means that any point of the range is image of some point of the domain.
If you think for a while about the definitions, you'll see that a bijection maps different elements of the domain to different elements of the codomain and covers all of the codomain. If two sets are finite, when there is a bijection means they have the same number of elements.
In other words, every element of the domain can be represented by some element of the range. And every element of the range represents a unique element of the domain. This trivially means the number of elements (here defined just for the finite case) is the same.
With this in mind and assuming $a,b,c$ not necessarily real numbers, but simply three distict objects, define $f : A\to B$ by $f(1)=a$, $f(2)=b$ and $f(3)=c$. It is imediate this obeys the requirements and is thus a bijection.
