Prove that a set is a Linear subspace While reading "Principles and Techniques of Applied Mathematics" from Bernard Friedman I stumbled on an exercise that I don't know how to properly solve... 
The question is as follows:

Prove that the set of vectors $x$ orthogonal to a given vector $y$ forms a linear subspace.

The author gives a tip to use the Cauchy-Schwarz inequality.
Well, to prove that it is a Linear Manifold is trivial but to prove that is closed is not so easy for me...
I thought that I could use the fact that if a sequence converges in the Cauchy sense then it converges. Bearing this in mind, given a sequence $\{x_n\}$ if $|x_n-x_m|<\varepsilon$ then $\langle x_n-x_m,y\rangle\leq |x_n-x_m|\cdot |y|\leq \varepsilon|y|$. Well I know that $\varepsilon$ can be as small as we want but not zero... Is it enough to say that since it is as close to zero as we want it's proved that the considered set forms a linear subspace?
PS: I'm sorry if the question is to trivial but I'm no Mathematician and I never know how to rigoursly prove this kind of maths problems.
 A: I wonder why convergence is being mentioned here at all.  To prove that a set is a linear subspace, it is enough to prove that it's closed under addition and under scalar multiplication.  So suppose $x$ and $y$ are both orthogonal to $z$, then is $x+y$ orthogonal to $z$?  If orthogonality is defined by saying the inner product is $0$, then the question is whether $\langle x+y, z\rangle=0$.  If you know that the inner product satisfies $\langle x+y,z\rangle = \langle x,z\rangle + \langle y, z\rangle$, then you've got it.  Similarly if $c$ is a scalar, you just need the fact that $\langle cx,z\rangle = c\langle x,z\rangle$.
I wonder if you're getting confused about the word "closed" and thinking that you were asked to prove that something is "closed" in the topological sense?
Later note: If you want to show that the space is topologically closed, then the problem is to show that if $\langle x_n-x,x_n-x\rangle\to0$ as $n\to\infty$ and $\langle x_n,z\rangle=0$ for all $n$, then $\langle x,z\rangle=0$.  You have
$$
|\langle x, z\rangle| = |\langle x-x_n,z\rangle + \langle x_n,z\rangle| = |\langle x-x_n,z\rangle| \le \langle x_n-x,x_n-x\rangle \langle z,z\rangle \to 0.
$$
