# Prove that a line is a polyhedral set (can be made by a finite number of inequalities)

I know that to prove this I have to shown that a set of finite inequalities make a line in $$\mathbb R^n$$ that is

$$L = \{ x_0 + \lambda d : \lambda \in \mathbb R^n \}$$

But can we say a line is the intersection of 3 hyperplanes in $$\mathbb R^n$$ (since each hyperplane can be shown with 2 inequalities). Can we say the same things about a half-line (that is when $$\lambda \geq 0$$)?

• No. The intersection of affine sets is affine, in particular, the intersection must either be a point or (at least) contain a line. An affine set is a translate of a linear subspace (there are other characterisations). – copper.hat Oct 19 '19 at 15:13
• What about a line? How can we show that for a line formally? – Pegi Oct 19 '19 at 15:59
• I don't understand your question. Show what for a line? – copper.hat Oct 19 '19 at 16:02
• That it is equivalent to the points in a system of linear inequalities in Rn – Pegi Oct 19 '19 at 17:21

I am assuming that $$d \neq 0$$ below:

Let $$v_2,...,v_n$$ form a basis for $$(\operatorname{sp}\{d\})^\bot$$.

Define $$H_k = \{x| \langle v_k, x-x_0 \rangle = 0 \}$$.

Then $$L= \cap_k H_k$$.

Suppose $$x \in L$$, then $$x=x_0+\lambda d$$, and it is easy to check that $$x \in H_k$$ for all $$k$$.

In general, given any $$x$$, there are unique $$\lambda, \alpha_k$$ such that $$x=x_0+\lambda d + \sum_k \alpha_k v_k$$.

Suppose $$x \in H_k$$, then we must have $$\alpha_k = 0$$.

Hence if $$x \in \cap_k H_k$$, then $$x=x_0 + \lambda d$$ and so $$x \in L$$.