Topology- Hyperplane proof don't understand

[zeion]

Homework Statement

Let $$\mathbf {a} \in R^n$$ be a non zero vector, and define {$$S = \mathbf {x} \in R^n : \mathbf {a} \cdot \mathbf {x} = 0$$ }. Prove that S interior $$= {\o}$$

Homework Equations

The Attempt at a Solution

Intuitively I understand that if a is a vector in R^3, S would be a plane. And if I place an open ball on any point on the plane it would include points that are not on the plane. So S interior would be empty.

But I'm not sure how to prove this rigorously?
I'm supposed to somehow use the hint that a hyperplane in R^n will have 1 dimension less than R^n (like the plane in R^3 has 2 dimensions)
But I am confused about the dimension of the plane? All vectors on a plane in R^3 will have 1 (the same) component = 0? So does this mean that the dimension of the plane is 2?

 

[zeion]

Can I just show that there is a point in the ball that has n components ?
Instead of n-1

 

[Dick]

Can I just show that there is a point in the ball that has n components ?
Instead of n-1

If a.x=0 you can pretty easily find a point y in any ball around x such that a.y is not zero.

 

[zeion]

I'm a little confused about that.
Do I need to differentiate between a point and a vector?
Like if I show that there are points in the ball that are not on the plane, then do I need to specify that the vector corresponding to that point is not perp a ?

 

[Dick]

I'm a little confused about that.
Do I need to differentiate between a point and a vector?
Like if I show that there are points in the ball that are not on the plane, then do I need to specify that the vector corresponding to that point is not perp a ?

Now I'm confused about that. Any point y in the ball around x of radius r is given by y=x+a where 'a' is any vector whose length is less than r. An awful lot of them don't satisfy a.y=0. Can you find just one?