Let $V$ be a vector space over $\mathbb{R}$ and $\langle\,,\rangle$ an inner product over $V$. Let $d$ be the metric induced by $\langle\,,\rangle$.
Prove that the set $A := \{\bar{x} \in V | \langle\bar{x}, \bar{v}\rangle \geq c \}$ is connected in $(V, d)$, where $c \in \mathbb{R}$ and $\bar{0} \neq \bar{v} \in V$.
I thought of using the union theorem somehow (splitting $A$ into a set of connected subsets that have a non-empty intersection), but, drawing the $\mathbb{R}^2$ case for inspiration, there doesn't seem to be a natural way to accomplish this.
I also thought of proving it is path-connected, but, again looking at $\mathbb{R}^2$, I'm not certain how this could be accomplished.
Finding a continuous function from $[0,1]$ to $A$ that utilises $d$ somehow also doesn't seem relevant.
I must be missing something... would be helpful to hear others' thoughts/suggestions. Thanks!
Path-connectedness is quite easy: Just use $$ t \mapsto tv_1 + (1-t)v_2, $$ and use $$ \langle tv_1 + (1-t)v_2, x \rangle = t \langle v_1, x \rangle + (1 - t) \langle v_2, x \rangle \ge tc + (1 - t)c = c. $$