Is the closure axiom necessary for algebraic structures defined via a binary operation?

Question

Numerous algebraic structures are often defined as a set $X$ equipped with a binary operation $f:X\times{X}\rightarrow {X}$ that satisfies some set of axioms. Since the image of $f$ is always in $X$ by definition of $f:X\times{X}\rightarrow{X}$ couldn't the closure axiom be dropped from the definition of whatever structure we are talking about. For an example a Magma is a set $M$ along with a binary operation $f:M\times{M}\rightarrow{M}$ which only needs to satisfy the closure axiom which is rather redundant since $f$ can only have values in $M$ by definition. So in instances such as the above couldn't the algebraic structure be defined without requiring closure without losing any of the structure?

Answer 1

It's more important for substructures. A subset $N$ of your magma $M$ (or whatever it is) has a function $N \times N \rightarrow M$ induced on it by the binary operation on $M$, but it isn't certain that the output values return to $N$.