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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 $)?

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  • $\begingroup$ 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). $\endgroup$ – copper.hat Oct 19 '19 at 15:13
  • $\begingroup$ What about a line? How can we show that for a line formally? $\endgroup$ – Pegi Oct 19 '19 at 15:59
  • $\begingroup$ I don't understand your question. Show what for a line? $\endgroup$ – copper.hat Oct 19 '19 at 16:02
  • $\begingroup$ That it is equivalent to the points in a system of linear inequalities in Rn $\endgroup$ – Pegi Oct 19 '19 at 17:21
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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 $x \in L$.

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