# Showing that a set is closed in an inner product space over a field.

I am struggling with these two questions:

Let (X,$$\langle\cdot,\cdot\rangle)$$ be an inner product space over a field K, let $$x_0 \in X$$ and let c $$\in \mathbb{R}$$.

1) Show that the set {$$x \in X$$ : $$\langle x, x_0\rangle$$ = c} is closed.

2) Show that the set {$$x \in X$$ : $$\langle x, x_o \rangle \le c$$} is closed.

I know the properties of the inner product space and so I should probably start here but I'm confused where to begin, thanks.

• What do you know about continuity? Do you know that/if the inner product is continuous? – Mees de Vries Oct 23 '18 at 12:44

For a fixed $$x_0 \in X$$, the function $$f: X \to \mathbb{R}$$ defined by $$f(x) = \langle x,x_0\rangle$$ is continuous by the Cauchy-Schwarz inequality $$|\langle x, x_0 \rangle| \le \|x\|\cdot \|x_0\|$$ so $$f$$ is a Lipschitz map.
The first set is just $$f^{-1}[\{c\}]$$ and the second $$f^{-1}[(-\infty,c]]$$, so inverse images of a closed set (of the reals) under a continuous map.