# Having trouble finding examples of sets of functions that do NOT form a vector space

In this question someone asks about showing that the set of all functions of the form $$y(t) = c_1\cos\omega t + c_2\sin\omega t$$ is a vector space. But doesn't literally any set of functions of the form $$y(t) = c_1f(t) + c_2g(t) + \ldots$$ form a vector space? After all, there will always be a zero element (coefficients = 0) and an additive inverse (coefficients of opposite sign), and trivially scaling or adding two $$y(t)$$ will yield another function of the same form.

So what is an example of a set of functions that do NOT form a vector space? The most common pedagogic example I've seen is unsatisfyingly contrived: the set of all polynomials of degree N. This is explained to not form a vector space because the zero element is not of degree N. However technically the zero element is still of the form $$c_1 + c_2x + \ldots + c_Nx^N$$, so I'm not sure how comfortable I am with this example.

• The space of all constant functions except $f(x)=3$. The space of all functions such that $F(0)=1$. The space of all polynomials of degree exactly $2$. The space of all polynomials with lead coefficient $1$. And so on. – lulu Oct 11 '19 at 19:09
• Yes, the set of all finite linear combinations of any set of complex- (or real-)valued functions on some domain forms a vector space. That's still a very special set of functions. How about the set of discontinuous real-valued functions on $[0,1]$? That's not a vector space. – saulspatz Oct 11 '19 at 19:18
• Take any set of functions and remove zero (given it's in the set). – amsmath Oct 11 '19 at 19:34
• @saulspatz, isn't not allowing functions to be continuous a more special set of functions than just the set of complex/real valued functions? – user1247 Oct 11 '19 at 20:00
• Yes. Any set of functions is more special than the set of all functions. So what? – saulspatz Oct 11 '19 at 20:04

• @user1247 The statements are equivalent. For example, consider $y^\prime=-y^2$; the solutions are $y=1/(x+c)$. – J.G. Oct 11 '19 at 21:33
• The space of functions being considered would be indexed as $(c_1,c_2,\ldots)$ meaning $1/(x+c_1)$, $1/(x+c_2)$, ...? Just making sure I understand -- we are not talking about linear combinations where the $c_n$ are the coefficients in front of each solution? – user1247 Oct 11 '19 at 21:44