How $C[a, b]$ satisfies the axioms of vector space

My book states that the set of all continuous functions defined on a closed interval $$[a, b]$$ where $$a ≠ b$$ satisfies the axioms of a vector space.

I am understanding that this means that for the axiom of closure under multiplication, any function within that interval that we would multiply by any scalar would still be a function within that interval. I am a little confused about this. For example how does $$C[2, 4]$$ with scalar $$0$$ satisfy this? If we would multiply any function by scalar $$0$$, how would it still be a function within $$C[2, 4]$$?

Just as a comparison, I have also understood that the set of all quadratic functions is not a vector function because if we would multiply by scalar $$0$$, it would cease to be a quadratic function. Am I wrong about this as well? If not, what is the difference?

• If you multiply a function with the scalar $0$, then the result is the zero function that has function value zero for any $x$ that you plug in. What do you mean by "quadratic function"? If you mean polynomials of order two, the set they consitute is of course a vector space. – amsmath Mar 24 '19 at 18:46
• My book says polynomials of degree two is not a vector space, only polynomials of degree two or less. – agblt Mar 24 '19 at 18:50
• @agblt Is $x^2 + -x^2$ a second order polynomial? – John Douma Mar 24 '19 at 18:52
• agblt Please look again at what you wrote last. You are contradicting yourself. – amsmath Mar 24 '19 at 18:53
• @agblt Ok, the set of polynomials of degree exactly 2 is clearly not a vector space. – amsmath Mar 24 '19 at 20:01

Take, for instance,$$\begin{array}{rccc}f\colon&[2,4]&\longrightarrow&\mathbb R\\&x&\mapsto&x.\end{array}$$Then $$f\in\mathcal C[a,b]$$, right? And $$0\times f$$ is the null function, which is also an element of $$\mathcal C[a,b]$$.