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

Question

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.

Answer 1

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.