In the one-dimensional space, suppose the interval $[a, b]$ exists on the real line with boundary points $a$ and $b$.
What is the outward normal vector at the point $a$ and at the point $b$?
The answer is $-1$ for $a$ and $1$ for $b$. But I cannot understand why so.
Usually, the outward normal vector is perpendicular to the line. But in this case, aren't they tangent to the line itself?
I can't digest what's "normal" in the $1$-D case.
I can kind of get that the interval ends at $a$ and at $b$ and so the outward direction is to $-1$ and to $1$, respectively...but is there a mathematical way to express this in the $1$-D case?
A "tangent vector" to the number line is a real number. But the only tangent vector to a point on the number line, viewed as a $0$-dimensional submanifold of a $1$-dimensional ambient space, is the velocity of a constant path, i.e., $0$. Every non-zero real number is therefore a normal vector to any particular point. The numbers $1$ at $b$ and $-1$ at $a$ are tangent vectors to the real line, but are normal vectors to the endpoints viewed as a $0$-manifold, and are the unique choices that are outward pointing from $[a, b]$.
This phenomenon looks weird at first glance, but is correct on closer inspection. Think of a curve in the plane. A normal vector to the curve is still tangent to the plane.